TWI272618B - Reduced size multi-port register cell - Google Patents

Reduced size multi-port register cell Download PDF

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Publication number
TWI272618B
TWI272618B TW092100941A TW92100941A TWI272618B TW I272618 B TWI272618 B TW I272618B TW 092100941 A TW092100941 A TW 092100941A TW 92100941 A TW92100941 A TW 92100941A TW I272618 B TWI272618 B TW I272618B
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Taiwan
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transistors
register unit
coupled
transistor
patent application
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TW092100941A
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Chinese (zh)
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TW200406782A (en
Inventor
Gene K Frydel
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Ip First Llc
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Priority claimed from US10/279,209 external-priority patent/US6834024B2/en
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Publication of TWI272618B publication Critical patent/TWI272618B/en

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Abstract

A multi-ported register cell that reduces the number of metal wires and/or transistors per write port. The cell includes a storage element that stores a bit. Each write port includes three transistors and two wires. The first transistor is coupled to a true input of the storage element. The second transistor is coupled to a complement input of the storage element. The first wire selectively turns on the first and second transistors of one of the ports. The second wire provides the update value. The third transistor selectively couples the second transistor to ground depending upon whether the second wire turns on the third transistor, thereby providing a complement of the update value to the second transistor. The cell also includes one or more read ports for reading the storage element bit. A multi-ported register file may be created from the cells.

Description

1272618 五、發明說明(1) ----- [ 0 0 0 1 ]本申請案主張以下美國申請案之 號1 0/279,209,申請日為2〇〇2年1〇月22日。无榷·累 【發明所屬之技術領域】 ’特別是有關 semiconductor [0 0 0 2 ]本發明係有關於半導體的領域 於多埠半導體暫存器單元(fflulti_ported register cell ) ° 【先前技術】 [0 0 0 3 ]數位電路一般會使用暫存器來儲存資料。尤 其,微處理器通常會包括一組暫存器(一般稱為暫存器槽 案(register file)),用以儲存指令運算元及結果。 微處理器暫存器檔案的一例為浮點(fl〇ating — p〇int)暫 存器檔案,其為一暫存器的陣列,用以保存浮點單元的運 算元及結果。浮點暫存器檔案可能是相當大的。例如,用 於x86架構的浮點單元之使用者可見(user — visible)的 浮點暫存器播案,即包括八個8 〇位元的暫存器。 [0004]通常’在微處理器内,有多個功能方塊 (functional block)需要存取暫存器檔案。原因之一在 於’現代的微處理器通常是管線化的。亦即,處理器包括 多重階段’當一指令行經其中時,每一階段皆會執行該指 令的一部份。因此,處理器會同時執行多個指令。結果 是’在管線各個階段内的多重功能方塊可能需要從暫存器 檔案中讀取資料,以及將資料寫入暫存器檔案。例如,處1272618 V. INSTRUCTIONS (1) ----- [ 0 0 0 1 ] This application claims the following U.S. application No. 1 0/279,209, the filing date is 2〇〇2 years, 1 month, 22 days. Innocent and tired [Technical field to which the invention pertains] 'Especially related to semiconductors [0 0 0 2 ] The present invention relates to the field of semiconductors in a multi-turn semiconductor register unit (fflulti_ported register cell) ° [Prior Art] [0 0 0 3 ] Digital circuits typically use a scratchpad to store data. In particular, microprocessors typically include a set of registers (generally referred to as register files) for storing instruction operands and results. An example of a microprocessor scratchpad file is a floating point (fl〇ating) p暂int register file, which is an array of scratchpads that holds the operands and results of the floating point unit. The floating point register file can be quite large. For example, a user-visible floating-point register for a floating-point unit of the x86 architecture, including eight 8-bit scratchpads. [0004] Typically, within a microprocessor, there are multiple functional blocks that require access to the scratchpad file. One reason for this is that modern microprocessors are usually pipelined. That is, the processor includes multiple stages. When an instruction passes through it, each stage executes a portion of the instruction. Therefore, the processor executes multiple instructions at the same time. The result is that multiple function blocks in various stages of the pipeline may need to read data from the scratchpad file and write the data to the scratchpad file. For example, at

1272618 五、發明說明(2) ---- 理器内的快取記憶體可能需要將資料寫入暫存器檔案,或 2 Ϊ存态檔案中讀取資料。同樣地,處理器内的算術與邏 =早=也需要從暫存器檔案中讀取資料,以及將資料寫入 暫存器檔案。再者,在處理器與記憶體間傳送資料的階 段’亦需存取暫存器檔案。 [00〇5]通常,處理器内的各種功能方塊會需要同時 时亦即,在相同時脈週期内)存取暫存器檔案。如果暫存 态檔案設計成一次只能允許一個功能方塊來存取暫存器檔 案,則需要存取的其他功能方塊就必須等待。如此可能景3 響效能,因為管線中的各個階段會暫停運作,以二 ^塊對暫存器槽案進行存取,而無法發揮處理器化^ 性的優點。 [ 0006 ]為了克服這個問題,處理器通常 暫存器檔案。多埠暫存器檔案具有多個讀取及,旱 =使暫存器檔案由多個功能方塊同時進行讀取及寫入動’、 [ 0007 ]例如,假設暫存器檔案具有至少四個 ίΓΓϋί。’資料快取記憶體可將資料寫入暫存 裔檔案中之弟一暫存器,第一算術邏輯單元 存 果寫入第二暫存器,第二算術邏輯 一二I結 寫:第::存11,第三算術邏輯單元可=指㈡ΐ 果,以寫入記憶體’以及位址產生器Hss結 generator )可從第六暫存器讀取一位址運算元,以上全1272618 V. Description of the invention (2) ---- The cache memory in the processor may need to write data to the scratchpad file, or 2 to store the data in the file. Similarly, the arithmetic and logic in the processor = early = also need to read data from the scratchpad file and write the data to the scratchpad file. Furthermore, the stage of transferring data between the processor and the memory also needs to access the scratchpad file. [00〇5] Typically, various functional blocks within the processor will need to access the scratchpad file at the same time, i.e., during the same clock cycle. If the scratchpad file is designed to allow only one function block to access the scratchpad file at a time, then other function blocks that need to be accessed must wait. This is likely to be effective because the various stages in the pipeline will be suspended and the access to the scratchpad slot will be accessed in two blocks, which will not take advantage of the processor. [0006] To overcome this problem, the processor usually has a scratchpad file. Multiple buffer files have multiple reads, and drought = causes the scratchpad file to be read and written simultaneously by multiple function blocks. [0007] For example, suppose the scratchpad file has at least four ίΓΓϋί . 'Data cache memory can write data into the temporary register of the temporary file, the first arithmetic logic unit writes the result to the second register, and the second arithmetic logic writes the first one: :Save 11, the third arithmetic logic unit can = (2) result, to write to the memory 'and the address generator Hss junction generator) can read the address operator from the sixth register, the above

I272618__ 五、發明說明(3) 都在相同的時脈週期中進行。 成。埠多暫蜂存暫4器元播储案广係由广阜暫存器單元所組 存器檔案。每個存器會配置在一起而成為暫 取埠。這些埠利用八i存态早兀具有多個寫入埠及多個讀 到暫存器^元,2屬ϊ線的動作將資料及控制訊號傳送 係耦接至暫存器單元入早兀。資料及控制訊號 值或用做控制邏輯,以體’這些電晶體可儲存位元 [〇〇〇9]用來產//,疋哪些埠將讀取及寫入單元。 單元的一個問題是夕土暫木存為播案之習用多蟑暫存器 埠,所以會造成暫存器播^ = 2接線會伴隨相當多的 存器檔案中,擁塞會產孟接線極度地擁塞。在暫 [0010]對典型的暫存哭m間J1用的問題。 半導體層次的尺寸即決·早兀而岁,其内含電晶體之 存器單元的另一問題θ…,i 70的大小。某些習用多璋暫 元的大小’而非由半ί體i::f f接線可能即決定了單 元藉由減少金屬接線,試“U些習用暫存器單 題,但此舉會增加電曰:jI此問喊及接線擁塞的問 [0 0U]目此,:們所數需!,而严暫存器單元變大。 少金屬接線的暫存器單元。 、疋’一種尺寸較小且減 發明内容】 種暫存器單元,可減少大多數 [0 0 1 2 ]本發明提供I272618__ V. Inventions (3) are all performed in the same clock cycle. to make. The 暂 暂 暂 暂 暂 暂 器 器 器 器 器 器 器 器 器 器 器 元 元 元 元 元 元 元 元 元 元 元Each register will be configured together to become a temporary buffer. These 埠 埠 八 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀 兀The data and control signal values are used as control logic to the body 'these transistors can be used to store bits [〇〇〇9] for production, /, which will be read and written to the unit. One problem with the unit is that the temporary soil is stored as a multi-scratch register for the broadcast case, so it will cause the register to be broadcast. The ^ 2 wiring will be accompanied by a considerable number of memory files. The congestion will be extremely high. congestion. In the temporary [0010] for the typical temporary storage crying m between J1. The size of the semiconductor level is determined by the age of the transistor, and the size of the memory cell of the transistor is θ..., i 70 . Some of the more common use of the size of the temporary ' instead of the half-length i::ff wiring may determine the unit by reducing the metal wiring, try "U some use the scratchpad single question, but this will increase the power :jI This question is called and the wiring is congested [0 0U] This is the case, the number of the registers is larger, and the strict register unit is larger. The register unit with less metal wiring. According to the invention, the scratchpad unit can reduce most of the [0 0 1 2] provided by the present invention.

1272618 五、發明說明(4) I用::存:單元中的金屬接線’而不會增加電晶體的數 -種44: 本發明的一項特徵是提供 補數於入°r ,、匕括/、有真值輸入(true input )及 里_ j! complement lnput)之一儲存元件。該暫存3g 包括N個寫入電路,其中每一個皆耦接至 子: 件。母一個寫入電路包括一第一輸入 f兀 2以寫入該儲存元件中。每個寫入電路還士;二: =括:該第:輸入,接至真值輸入。丄= 蛀1一及第二電晶體,其係串接至補數輸入。該第 ΓΓΓ值的補數至第二電晶體。每 路遏匕括一弟二輸入接線,係耦接至該第一電晶體 電 電晶體’用㈣擇性地使第—電晶體及第二電晶 ^一 以選擇性地使該二進位值能寫入儲存元件。 、, [0013]另一方面’本發明的一項特徵是,提供— 暫存器單元’丨包括具有真值輸入及補數輸入之一儲種 件。該暫存器單元還包括N個寫入電路,其中每一個1存元 接至該儲存元件。每一寫入電路基本上由第一與第二私輪 接線及第一、第二和第三電晶體所組成。第一輸入:=入 达一一進位值以寫入該儲存元件。第一電晶體將第—弘傳 接線耦接至真值輸入。第二與第三電晶體係串接至補 入。第一輸入接線亦耦接至第三電晶體,用於選擇=輪 第三電晶體導通,以提供該二進位值的補數至第二電曰使 體。第二輸入接線係耦接至第一與第二電晶體,用於= 、k擇1272618 V. INSTRUCTIONS (4) I use:: store: metal wiring in the unit without increasing the number of transistors - 44: A feature of the present invention is to provide a complement in the °r, /, has a true input (true input) and _ j! complement lnput) one of the storage components. The temporary storage 3g includes N write circuits, each of which is coupled to a sub-piece. The parent write circuit includes a first input f 兀 2 for writing into the storage element. Each write circuit is also returned; two: = bracket: the first: input, connected to the true value input.丄 = 蛀1 and the second transistor, which are connected in series to the complement input. The complement of the first value is to the second transistor. Each of the channels includes a second input wiring coupled to the first transistor, and (4) selectively causes the first transistor and the second transistor to selectively enable the binary value Write to the storage component. [0013] In another aspect, a feature of the present invention is that the provisioning register unit 丨 includes a storage unit having a truth input and a complement input. The register unit also includes N write circuits, each of which is connected to the storage element. Each write circuit consists essentially of first and second private wheel connections and first, second and third transistors. The first input: = enters a carry value to write to the storage element. The first transistor couples the first-to-high pass wiring to the true value input. The second and third electro-crystalline systems are connected in series to the complement. The first input wiring is also coupled to the third transistor for selecting the third transistor to be turned on to provide the complement of the binary value to the second electrical actuator. The second input wiring is coupled to the first and second transistors for =, k

1272618 五 發明說明(5) 性地使第一盥裳_ 能寫入儲存^件了電晶體導通,以選擇性地使該二進位值 [0 0 1 4 ]另— 暫存器單元,复I ,本發明的一項特徵是,提供一種 及補數輪入之儲二本士由^列部分組成··一具有真值輸入 -個_至該儲G件個每寫上電路與M個讀取電路,每 接線,用於傳送___ 固寫入電路包括一第一輸入 入電路還包括一第— 以:入該儲存元件中。每個寫 值輸入。二括Γ第一輸入接線耗接至真 接至補數輸入。第一輸入接線亦耗接f;電二其:、串 選擇性地使第三電晶體導通,以提供該:::::、’用於 第二電晶體。每個寫入電路還包括二^ 、補數至 至第-及第二⑽,用於ί;;地輸:接線,接 導通’以選擇性地使該二進位值能寫入儲存元二:電晶體 [〇 〇 1 5 ]另一方面,本發明的一項特 暫存器單元。1¾暫存器單元包括_用'以'提供一種 元件。該暫存器單it還包括N個寫人埠。每位70之儲存 =二條金屬接線。該暫存器單元還包_個==好 母個寫入電路將其所對應之其中一個寫入埠寫入電路。 兀件。每個寫入電路恰好包含三個電晶體。接至忒儲存 [0016]另一方面,本發明的一項牿矜9 多埠暫存器單元。$多埠暫存器單元包括二疋’提供-種 元之儲存元件。該多埠暫存器單元還包括N組以曰儲存一—位 組僅有=個電晶體,其用以將$杨- 、、’電晶體,每 ,—個電曰曰遐〃用μ將3亥位疋寫入該儲存元件。每 五、發明說明(6) 至儲存元件。該多蟑暫存器單元還包括n 、,且孟屬接線,母組僅有二條金屬接線,其用以將該位元 :,件。每組金屬接線係耦接至該N組電晶體_對應’ 哭^Λ00/1本發明的優點是’可顯著減少某些習用暫存 :早二上:個寫人蟑所需之金屬接線。對於 ,组=暫存器檔案而言,具有降低金屬接線擁=;早 盈:此外’本發明可使其他習用之暫存器單元i,每個寫 η;:::;;,:的半導體層空間顯著減少。 案的尺寸變小小可使整個暫存器檀 系wtd、’因而潛在地使包含本 理器之整個晶粒(die)尺寸縮小。 f仔為早兀的處 [0 0 1 8 ]本發明之其它特徵及優, 之其餘部分及所附圖示後,將更能突顯出來 說明書 【實施方式】 [ = 022 ]在本說明書中’ t電路元件被描 ί接:接謂元2藉由如金屬接線或半導體材料* 直接連接在一起。「耦接」的另一種涵 =一 另一電路元件而間接地連接。 牛係經由1272618 5 invention description (5) sexually make the first _ _ can write to the storage device to turn on the transistor to selectively make the binary value [0 0 1 4 ] another - register unit, complex I A feature of the present invention is that a storage and a second complement of the second round are composed of a column portion, a true value input - a _ to the storage G piece, each write circuit and M readings. Taking the circuit, each wiring, for transmitting ___ the solid write circuit includes a first input-in circuit and a first--to: into the storage element. Each write value is entered. The second input line is connected to the true input to the complement input. The first input wiring also consumes f; the second: the string selectively turns on the third transistor to provide the :::::, ' for the second transistor. Each write circuit further includes two ^, a complement to the first and the second (10), for the ground: wire, connect to 'to selectively enable the binary value can be written to the storage element two: A transistor [〇〇1 5 ], on the other hand, is a special register unit of the present invention. The 13⁄4 register unit includes a component provided by 'to'. The register unit it also includes N writers. Storage of 70 per seat = two metal connections. The register unit also includes _ == good mother write circuit writes one of its corresponding write 埠 write circuits. Mail. Each write circuit happens to contain three transistors. Connected to 忒 Storage [0016] In another aspect, a 牿矜9 multi-register register unit of the present invention. The multi-tap register unit includes two storage elements of the 'provided-type. The multi-slot register unit further includes N groups to store one-bit group only = one transistor, which is used to convert $yang-, 'transistor, each, one-electrode with μ The 3H position is written to the storage element. Every five, invention description (6) to storage components. The multi-tap register unit further includes n, and the Meng is wired, and the parent group has only two metal wires for using the bit: Each set of metal wiring is coupled to the N sets of transistors _ corresponding 'Cry' 00/1 The advantage of the present invention is that it can significantly reduce some of the conventional temporary storage: the second two: a metal wiring required for writing. For the group=scratch file, there is a reduced metal wiring. =; early profit: In addition, the present invention can make other conventional register units i, each writing η;:::;;,: semiconductor The layer space is significantly reduced. The smaller size of the case allows the entire register to be wtd, thus potentially reducing the size of the entire die containing the processor. The other features and advantages of the present invention will be more prominent, and the rest of the invention and the accompanying drawings will be more prominent. [Embodiment] [= 022] In this specification The t circuit components are depicted: the junction elements 2 are directly connected together by, for example, metal wiring or semiconductor material*. Another type of "coupling" is indirectly connected to another circuit component. Cattle

[0023]圖一係為本發明之多埠暫哭 圖。圖-所示之暫存器單元10。的實施例:括:的方: 及二個讀取埠。然而’暫存器單元1〇〇的電晶體組U 1272618[0023] Figure 1 is a sneak peek of the present invention. Figure - shows the register unit 10. Example: The side of: and the two read 埠. However, the transistor group U 1272618 of the register unit 1〇〇

而仍具有本發 改為包含其他不同數量的讀取埠及寫入璋 明的優點。 [ 0024 ] 電晶體1 0 2 < 檔案中的暫 一真值的輸 及DB來表示 可儲存多個 成之暫存器 [ 0025 ] 能方塊會同 別是,由圖 檔案,可允 存器檔案中 理器内的兩 個不同暫存 的特定暫存 :J器單元100包括一儲存元件1〇2 儲存元件102係組態為儲存暫存器(如暫存器 存裔)之單一位元的資料。儲存元件102包括 入/輸出線及一補數的輸入/輪出線,分別以0 。多個暫存器單元100可耦接在一起,组 位元之多埠暫存器。多個由暫存器單元100组 可耦接在一起,以組成一多埠暫存器檔案。 暫存器槽案之所以為多埠的,係因有多個功 時寫入及/或讀取暫存器檔案中的暫存器。特 一實施例中之暫存器單元所組成的多埠暫存器 斗Μ處理器内之四個不同功能方塊,同時對暫 之四個不同暫存器作寫入動作。同樣地,微處 個不同功能方塊可同時讀取暫存器檔案中之兩 器。再者,亦可同時讀取及寫入暫存器檔案中 器。 [0026]圖二係為圖一之儲存元件1〇2的方塊圖。儲存 元件102包括二個Ν通道MOS元件(標示為^7及^8 ),以 及二個Ρ通道MOS元件(標示為Ρ1及P2 ) °Ν17及Ν18的源極 摩馬接至Vss或接地。Ρ1及Ρ2的源極耦接至vdd。Ν18的汲極 係麵接至P2的汲極,這兩者則都耦接到真值輸入/輸出d以 及P1與N1 7的閘極。N1 7的汲極係耦接至pi的汲極,這兩者 則都耦接到補數輸入/輸出DB以及ρ 2與N18的閘極。依此組There is still the advantage that the present invention is changed to include other different numbers of read and write instructions. [0024] The transistor 1 0 2 < the temporary value of the file in the file and the DB to indicate that a plurality of registers can be stored [ 0025 ] can be the same as the block file, the file can be saved Two different temporary storage specific temporary storages in the middle controller: J device unit 100 includes a storage component 1〇2 storage component 102 is configured to store a single bit of a temporary storage device (such as a temporary storage device) data. The storage element 102 includes an input/output line and a complement input/round line, respectively, at zero. The plurality of register units 100 can be coupled together, and the plurality of registers of the group bits. A plurality of sets of register units 100 can be coupled together to form a multi-byte register file. The reason why the scratchpad slot is cumbersome is that there are multiple functions to write and/or read the scratchpad in the scratchpad file. In the embodiment, the buffer unit composed of the register unit has four different function blocks in the processor, and simultaneously writes to the four different registers. Similarly, a different function block can read two of the registers in the scratchpad file at the same time. Furthermore, the scratchpad file can be read and written simultaneously. 2 is a block diagram of the storage element 1〇2 of FIG. The memory component 102 includes two germanium channel MOS components (labeled as ^7 and ^8) and two germanium channel MOS components (labeled Ρ1 and P2). The source of the Ν18 and Ν18 is connected to Vss or ground. The sources of Ρ1 and Ρ2 are coupled to vdd. The drain of the Ν18 is connected to the drain of P2, both of which are coupled to the true input/output d and the gates of P1 and N1 7. The N1 7's drain is coupled to the pi's drain, both of which are coupled to the complement input/output DB and the gates of ρ 2 and N18. According to this group

12726181272618

’以儲存單一位元 入DB來寫入,及藉 由 態,N17、N18、P1以及P2會共同運作 值’其會藉由真值輸入D及/或補數輸 真值輸出D來讀取,如上所述。 „ [ 0027 ] “7與1>1之閘極與汲極的耦接會產生一反相 器’其輸入為D,輸出為DB。同樣地,N1^p2之閘極 極的搞接亦會產生-反相器,其輸入為⑽,輸出為〇。每 個反相器⑲出會搞接至另一反相器的輸入,冑以在節點 D上產生用以儲存一位元值的電路。 — [ 0028 ]請再參照圖一,暫存器單元1〇〇亦包括四條寫 ^字組線(write wordiine,WWL)的水平金屬接線,標 示為WWL1、WWL2、WWL3以及WWL4。WWL1-4為控制訊號/用 以寫入由類似單元1〇〇的單元所組成之暫存器。在典型的 暫存中,WWL1-4會耦接至每一個單元1〇〇。處理器於 WWL卜4其中一條接線上產生高值(high value),以將 值寫入暫存器。 [0029] 暫存器單元1〇〇也包括二條讀取字組線 wordl ine ’ RWL )的水平金屬接線,標示為”!^及 RWL2。RWL1-2為控制訊號,用以讀取由類似單元1〇〇的單 =所組成之暫存器。在典型的暫存器中,RWL1 — 2會耦接至 每一個單元100。處理器KRWL1—2其中一條接線上產生高 值’以讀取儲存於暫存器中之值。 [0030] 暫存器單元1〇〇也包括四條寫入位元線 (write bitline,WBL)的垂直金屬接線,標示為WBL1、 WBL2、WBL3及WBL4。WBL1 -4為資料訊號,用以將位元值寫'Storing a single bit into the DB to write, and by means of the state, N17, N18, P1 and P2 will work together with the value' which will be read by the true value input D and / or the complement value output value D, As mentioned above. „ [ 0027 ] The coupling of the gate and the drain of “7 and 1” will produce an inverter whose input is D and the output is DB. Similarly, the gate of N1^p2 will also generate an inverter with an input of (10) and an output of 〇. Each inverter 19 outputs an input to another inverter to generate a circuit for storing a one-bit value at node D. — [ 0028 ] Referring again to FIG. 1 , the register unit 1〇〇 also includes four horizontal metal lines of write word meaning (WWL), denoted as WWL1, WWL2, WWL3, and WWL4. WWL 1-4 is a control signal / a register used to write a unit of similar unit 1 。. In a typical staging, WWL1-4 is coupled to each unit. The processor generates a high value on one of the WWL blocks to write the value to the scratchpad. [0029] The register unit 1〇〇 also includes horizontal metal lines of two read word lines wordl ine ' RWL ), labeled as “!^ and RWL2. RWL1-2 is a control signal for reading by a similar unit. A single register consists of a temporary register. In a typical register, RWL1 - 2 is coupled to each unit 100. The processor KRWL1 - 2 generates a high value on one of the wires to read the storage. The value in the register. [0030] The register unit 1〇〇 also includes four vertical metal lines for writing bit lines (WBL), denoted as WBL1, WBL2, WBL3, and WBL4. WBL1 -4 Data signal to write the bit value

第13頁 1272618Page 13 1272618

^由類似單元1 〇〇的單元所組成之暫存器的儲存元件〗〇2 中。多組WBL1-4會配置在一起而成為:^料 值)寫入暫存器。對於資料匯流排之“ ,、且WBL1-4,處理器會在每一組之其中一條接線上,產生! 或0的值,以將64位元值寫入暫存器。 [ 003 1 ]暫存器單元100也包括二條讀取位元線 tMtlme,RBL)的垂直金屬接線,標示為⑽以及 RBL2 °RBLl-2為資料訊號,用以從由類似單元1〇〇的單元 所組成之暫存器的儲存元件102中,讀取其位元值。多組 RBL1 -2會配置在一起而成為資料匯流排,以從暫存器讀取 資料值(如64位元值)。對於資料匯流排之64組^11_2, 處理器會在每一組之其中一或兩條接線上讀取丨或〇的值, 以從暫存器讀取6 4位元值。 [ 0 032 ]整體來看,WWL1&WBU構成寫入埠1。同樣 地,WWL2及WBL2構成寫入埠2,WWL3及WBL3構成寫入埠3, 而WWL4及WBL4構成寫入埠4。整體來看,RWU &RBU構成 讀取埠1,而RWL2及RBL2構成讀取埠2。 [0033]暫存器單元1〇〇也包括四個n通道M〇s元件,標 示為N5到N8。N5到N8的汲極係耦接至儲存元件丨〇2的D輸 入/輸出。N5到N8的閘極分別耦接至WWL1至〇[4。N5到㈣ 的源極則分別耦接至WBL1至WBL4。 [ 0034 ]暫存器單元1〇〇也包括四個n通道M〇s元件,標 示為N9到N12。N9到N12的汲極係耦接至儲存元件1〇2的db" 輸入/輸出。N9到N1 2的閘極係分別耦接至WWL1至驛[4。 1272618 五、發明說明(10) [0035] 暫存器單元1〇〇也包括四個n通道M〇s元件,標 示為N1 3到N1 6。N1 3到N1 6的汲極係分別耦接至n 9到N1 2的 源極。亦即,N 9到N1 2係分別串接至N1 3到N1 6。N1 3到N1 6 的閘極分別耦接至WBL1至WBL4。N1 3到N1 6的源極則耦接至 Vss 〇 [0036] N5到N16總共構成四個寫入電路1〇8。其中一 個寫入電路108係由N5、N9及N13所組成。另一個寫入電路 108係由N6、N10及N14所組成。再一個寫入電路係由 N7、N11及N15所組成。最後一個寫入電路1〇8則由N8、N12 及N16所組成。 [ 0037 ] N5 到 N16、儲存元件 1〇2、WBLl-4 及 WWL1-4 之 前述叙接方式,使儲存元件1 〇 2能藉由四個功能方塊中的 任一個來寫入。例如,假設耦接至寫入埠3的功能方塊想 要寫入一高值(即二進位的1)至儲存元件1〇2。功能方塊 會將WWL3與WBL3設定為高值,亦即功能方塊會將WWL3及 WBL3充電到實質上接近Md的值。WBL3上的高值會使N15導 通’而WWL3上的高值會使Nl 1導通。因此,接地的放電路 徑會提供給儲存元件1〇2的補數DB輸入/輸出,致使一高值 存入儲存元件102中。此因圖二儲存元件1〇2的N17及?1會 將補數DB輸入/輸出端所收到的低值予以反相,且儲存元 件102會儲存反相後的值,亦即會將高值存於真值〇輸入/ 輸f端上。換言之,由於WBL3為高,所以N15在其汲極上 會提供低值。WWL3上的高值會使Ni 1導通,而Ni 1則將其源 極上從N1 5之汲極所收到的低值,送至儲存元件丨〇 2的補數^ The storage element of the register consisting of units like unit 1 〇 is in 〇2. Multiple sets of WBL1-4 will be configured together to become: *values) written to the scratchpad. For the data bus, ", and WBL1-4, the processor will generate a value of ! or 0 on one of each group's wiring to write the 64-bit value to the scratchpad. [003 1] The memory unit 100 also includes vertical metal lines of two read bit lines tMtlme, RBL), labeled as (10) and RBL2 °RBLl-2 as data signals for temporary storage from units of similar units The storage element 102 of the device reads its bit value. The multiple sets of RBL1 - 2 are configured together to become a data bus to read data values (such as 64 bit values) from the scratchpad. The 64 groups ^11_2, the processor will read the value of 丨 or 在 on one or two of each group to read the 6.4-bit value from the scratchpad. [ 0 032 ] Overall, WWL1 & WBU constitutes write 埠 1. Similarly, WWL2 and WBL2 constitute write 埠 2, WWL3 and WBL3 constitute write 埠 3, and WWL4 and WBL4 constitute write 埠 4. Overall, RWU & RBU constitutes read Take 埠1, and RWL2 and RBL2 form read 埠 2. [0033] The register unit 1 〇〇 also includes four n-channel M 〇 s components, labeled N5 to N8 The N5 to N8 gates are coupled to the D input/output of the storage device 丨〇 2. The gates of N5 to N8 are respectively coupled to WWL1 to 〇[4. The sources of N5 to (4) are respectively coupled to WBL1 to WBL 4. [0034] The register unit 1〇〇 also includes four n-channel M〇s elements, labeled N9 to N12. The N9 to N12 drains are coupled to the db" input/output of the storage element 1〇2 The gates of N9 to N1 2 are respectively coupled to WWL1 to 驿[4. 1272618 V. Description of the Invention (10) [0035] The register unit 1〇〇 also includes four n-channel M〇s elements, labeled as N1 3 to N1 6. The N1 3 to N1 6 gates are respectively coupled to the sources of n 9 to N1 2 . That is, N 9 to N 1 2 are respectively connected in series to N1 3 to N1 6. N1 3 to The gates of N1 6 are respectively coupled to WBL1 to WBL4. The sources of N1 3 to N1 6 are coupled to Vss 〇 [0036] N5 to N16 constitute a total of four write circuits 1 〇 8. One of the write circuits 108 It is composed of N5, N9 and N13. The other write circuit 108 is composed of N6, N10 and N14. Another write circuit is composed of N7, N11 and N15. The last write circuit is 1〇8. It consists of N8, N12 and N16. [ 0037 ] The foregoing splicing manners of N5 to N16, storage elements 1〇2, WBL1-4, and WWL1-4 enable the storage element 1 〇 2 to be written by any of the four functional blocks. For example, assume that the functional block coupled to write 想3 wants to write a high value (i.e., a binary one) to storage element 〇2. The function block will set WWL3 and WBL3 to a high value, that is, the function block will charge WWL3 and WBL3 to a value substantially close to Md. A high value on WBL3 will turn N15 on and a high value on WWL3 will turn Nl 1 on. Therefore, the grounded discharge path is supplied to the complement DB input/output of the storage element 1〇2, so that a high value is stored in the storage element 102. What is the N17 and the storage element 1〇2 of Figure 2? 1 will invert the low value received by the complement DB input/output, and the storage element 102 will store the inverted value, that is, the high value will be stored in the true value 〇 input / output f terminal. In other words, since WBL3 is high, the N15 will provide a low value on its drain. A high value on WWL3 turns Ni 1 on, while Ni 1 sends a low value on its source from the drain of N1 5 to the complement of storage element 丨〇 2

第15頁 1272618 五、發明說明(π) 輸入/輸出DB。 [ 0 038 ]反之,假設耦接至寫入埠3的功能方塊想要將 一低值(即二進位的〇 )寫入儲存元件丨〇 2。功能方塊會設 定WWL3上的高值及WBL3上的低值,亦即,功能方塊將WWL3 充電到實質上接近Vdd的值,且將WBL3放電到實質上接近 Vss的值。WWL3上的高值使N7導通。因此,N7提供放電路 徑給儲存元件1 〇2的真值d輸入/輸出,並使低值存入儲存 元件1 0 2中。 [0039]暫存器早元也包括一反相器。反相器 106的輸入搞接至儲存元件1〇2的真值輸入/輸出 [ 0040 ]暫存器單元1〇〇也包括二個N通道M〇s元件,標 示為N1及N2。N1的汲極係耦接sRBL1。N1的閘極係耦接至 RWL1。N1的源極係耦接至…的汲極。N2的源極係耦接 Vss ° N2的閘極則耦接至反相器106的輸出。 [004 1 ]暫存器單元丨〇〇也包括二個n通道元件,標 示為N3及N4。N3的汲極係耦接至RBl2。N3的閘極係耦接至 RWL2 ° N3的源極係耦接至…的汲極。N4的源極係耦接Page 15 1272618 V. Description of invention (π) Input/output DB. [0 038] Conversely, assume that the function block coupled to write 想要3 wants to write a low value (ie, binary 〇) to storage element 丨〇 2. The function block sets the high value on WWL3 and the low value on WBL3, that is, the function block charges WWL3 to a value substantially close to Vdd and discharges WBL3 to a value substantially close to Vss. A high value on WWL3 turns N7 on. Therefore, N7 provides a circuit path to the true value d input/output of the storage element 1 〇 2, and stores the low value in the storage element 102. [0039] The register early element also includes an inverter. The input of the inverter 106 is connected to the true value input/output of the storage element 1 [ 2 [0040] The register unit 1 〇〇 also includes two N-channel M 〇 s elements, denoted as N1 and N2. The drain of N1 is coupled to sRBL1. The gate of N1 is coupled to RWL1. The source of N1 is coupled to the drain of .... The gate of the N2 source coupled to Vss ° N2 is coupled to the output of the inverter 106. [0041] The register unit 丨〇〇 also includes two n-channel elements, designated N3 and N4. The N3's drain is coupled to RBl2. The gate of N3 is coupled to the drain of RWL2 ° N3 to the drain of .... Source connection of N4

Vss。N4的閘極則耦接至反相器1〇6的輸出。 [ 0 042 ]整體來看,N]^,jN4及反相器1〇6即稱為讀取電 路104 °其中一個讀取電路104係由反相器106、N1以及N2 所組成。另一個讀取電路104則由反相器106、N3以及N4所 組成。 [ 0043 ]在一具體實施例中,暫存器單元1〇〇係依據二 相位的時脈來運作。在一具體實施例中,RBL1及RBL2在Vss. The gate of N4 is coupled to the output of inverter 1〇6. [0 042] Overall, N]^, jN4 and inverter 1〇6 are called read circuits 104. One of the read circuits 104 is composed of inverters 106, N1 and N2. The other read circuit 104 is composed of inverters 106, N3 and N4. [0043] In one embodiment, the scratchpad unit 1 operates in accordance with the clock of the two phases. In a specific embodiment, RBL1 and RBL2 are

第16頁 1272618 五、發明說明(12) 相位1期間會保持浮接(f 1 0 a t i n g ),而在相位2期間,會 預先充電至高值。讀取電路丨〇 4則致能儲存元件丨〇 2由二個 功能方塊中的一或二個來讀取。 [0 044 ]例如,假設耦接至讀取埠2的功能方塊想要讀 取存於儲存元件102中的值。再進一步假設存於儲存元件 1 02中的值為一低值。功能方塊會設定RWL2上的高值。反 相态1 0 6則從儲存元件1 〇 2接收該低值,並於其輸出端產生 一咼值,由N4的閘極來接收。反相器1〇6輸出端之高值會 使N4導通,而RWL2的高值會使N3導通;因而,接地的放電 路徑會提供給節點RBL2。因此,RBL2上於相位2期間預先 充電的高值,會在相位1期間,經由通過N3及“的路徑而 放電成低值。因此,在讀取埠2上,功能方塊會從 件1 0 2讀取二進位的0。 [0045]反之,假设存於儲存元件1Q2中的值為一高 值。功能方塊會設定RWL2上的高值,以讀取存於儲存元件 102中的位元值。反相器1〇β從儲存元件1〇2接收其高值, 並於輸出端產生一低值,由N4的閘極來接收。反相器1〇6 輸出端的低值會使N4不導通。因此,雖然RWL2上的^值會 使N3導通,但是沒有放電路徑可提供給節點RBL2。=此: RBL2上於相位2期間預先充電的高值,在相位1期間合 在高值。因此,在讀取蟑2上,功能方塊會從儲存元㈢件 讀取二進位的1。 [ 0 046 ]為了更完整地了解本發明的優點, 圖三到圖六敘述四種習用的暫存器單元。圖三到圖六中的Page 16 1272618 V. INSTRUCTIONS (12) During phase 1, it will remain floating (f 1 0 a t i n g ), and during phase 2, it will be precharged to a high value. The read circuit 丨〇 4 enables the storage element 丨〇 2 to be read by one or two of the two functional blocks. [0 044] For example, assume that the functional block coupled to the read block 2 wants to read the value stored in the storage element 102. It is further assumed that the value stored in the storage element 102 is a low value. The function block sets the high value on RWL2. The inverse phase 1 0 6 receives the low value from storage element 1 〇 2 and produces a 咼 value at its output, which is received by the gate of N4. The high value of the output of inverter 1〇6 will turn N4 on, and the high value of RWL2 will turn N3 on; therefore, the grounded discharge path will be provided to node RBL2. Therefore, the high value pre-charged on phase R2 during phase 2 will be discharged to a low value via phase N3 and during the phase 1. Therefore, on reading 埠2, the function block will be from the device 1 0 2 Read the binary 0. [0045] Conversely, assume that the value stored in storage element 1Q2 is a high value. The function block sets the high value on RWL2 to read the bit value stored in storage element 102. The inverter 1 〇β receives its high value from the storage element 1〇2 and generates a low value at the output, which is received by the gate of N4. The low value of the output of the inverter 1〇6 makes N4 non-conductive. Therefore, although the value on RWL2 turns N3 on, no discharge path can be supplied to node RBL2. = This: The high value pre-charged during phase 2 on RBL2 is high at phase 1. Therefore, On reading 蟑2, the function block reads the binary one from the storage element (3). [0 046] To more fully understand the advantages of the present invention, Figures 3 through 6 describe four conventional register units. Figure 3 to Figure 6

1272618 五、發明說明(13) 每個單元包括四個寫入埠及二個读 1 0 0做比較。圖三到圖六的暫存項_ ,以/、圖一的單元 的暫存器單元⑽類些方面與圖-號,以求簡化與明晰。彳㈣凡件會使用相同的標 [0047]圖二係一習用夕勒六即口口 ^300 ^ # ^ ° ^ =單 1 〇。中·_Ν16 ;取而而代暫之存:不包括暫存器單元 no玫C:人办 代的疋,暫存器單元300包含了 四條反向寫入位元線(WBLX)的垂直 WBLX1、WBLX2、fBLX3 及WRT U 、屬接線軚不為 m則的源極,LX=二提其:二接至N9、㈣、 [^048 ]可觀察到,單元3〇〇的每個寫入埠具有3條金 錄。⑽一 9ΛΛμ ΰ早兀100的母個寫入埠只有2條金屬接 炊曰早70的缺點之一是其包含相當多的金屬接線。尤 ::’单心00比單元100多出四條用以將位元值寫入儲存 ^件102的金屬接線,這可能會使單元川的尺寸大於單元 100,並且增加金屬層的擁塞。 [,9]圖四係一習用之暫存器單元4〇〇的方塊圖。暫 Q包括一儲存元件1G2、二個讀取電路1〇4、 ML1-2、RBLl-2、,Ll-4及WBL1-4,如圖一的暫存哭單元 =〇。然而,暫存器單元400不包括暫存器單元1〇〇中的N5_ N16。取而代之的是,暫存器單元4〇〇包括四個通閘 1272618 五、發明說明(14) (passgate) ’ 或傳輸閘(transmissi〇J1 gate),標示 為PG1、PG2、PG3以及PG4。PG卜4係由一N通道電晶體丁及一 p通道電晶體平行耦接而成,其中兩個電晶體的源極耦接 在一起,而汲極亦耦接在一起。PG1—4的汲極係耦接至儲 存元件102的真值輸入/輸出D。pG1-4的源 WBLH。肌Η則分別麵接至pG1_4之“道電晶=接至 極〇 [ 0050 ]暫存器單元4〇〇也包括四條反向寫入字組線 (WWLX)的水平金屬接線,標示為釋^丨、wwlx2 及WWLX4,其分別輕接至pG卜4之?通道電晶體的閘 專送WBL1_4的補數。補數㈣ ^, [ΓΓ;3 "3 ^ 口 早疋1〇〇的母個寫入埠只有2條么屬 接線。單元400的缺點之一曰平頁Ζ條金屬 尤其是,單元400,;; 一:的金屬接線。 多出四條用以將位元值寫二入的 一 3 =似,會比單元_ 可能會使單元4。〇的尺寸寫大:單存二件广2:金屬接線’這 擁塞。 寸大於早兀100,並且增加金屬層的 [0 0 5 2 ]圖五係 一 ^ ^ . 存器單元500包括—儲存^存器單元500的方塊圖。暫 -2、RBU_2、二存;"件102、二個讀取電路…、 元100。然而,暫存号單二,WBU-4 ’如圖-的暫存器單 Ν5-Π6。取而代之的ΊI,暫存器單元100中的 疋暫存器單元400包括四個通閘,1272618 V. INSTRUCTIONS (13) Each unit consists of four writes and two reads of 100 for comparison. The temporary storage item _ of Figure 3 to Figure 6 is simplified and clarified by the aspects of the register unit (10) of the unit of Figure 1 and the figure-number.彳 (4) The same standard will be used for the pieces. [0047] Figure 2 is a syllabus used in the syllabus of the mouth ^300 ^ # ^ ° ^ = single 1 〇. In the middle of the _ Ν ; ; ; ; ; ; ; ; : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : WBLX2, fBLX3 and WRT U, the source is not the source of the m, LX = two mention: two connected to N9, (four), [^048] can be observed, each write 单元 of the unit 3〇〇 3 gold records. (10) A 9 ΛΛμ ΰ 兀 100 母 埠 埠 之一 埠 埠 埠 埠 埠 埠 埠 埠 埠 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一 之一In particular, the single cell 00 is four more than the cell 100 for writing the bit value to the metal wiring of the memory device 102. This may cause the cell size to be larger than the cell 100 and increase the congestion of the metal layer. [, 9] Figure 4 is a block diagram of a conventional register unit 4〇〇. The temporary Q includes a storage element 1G2, two read circuits 1〇4, ML1-2, RBL1-2, L1-4, and WBL1-4, such as the temporary crying unit = 如图. However, the register unit 400 does not include N5_N16 in the register unit 1A. Instead, the register unit 4 includes four pass gates 1272618, a description (14) (passgate) or a transmissi (J1 gate), designated PG1, PG2, PG3, and PG4. The PG 4 is formed by parallel coupling of an N-channel transistor and a p-channel transistor, wherein the sources of the two transistors are coupled together, and the drains are also coupled together. The drain of PG 1-4 is coupled to the true value input/output D of the storage element 102. Source of pG1-4 WBLH. The tendons are respectively connected to the pG1_4 "channel crystal = connected to the pole [0050] register unit 4" also includes the horizontal metal wiring of the four reverse write word lines (WWLX), marked as release , wwlx2 and WWLX4, which are respectively connected to pG Bu 4? The gate of the channel transistor is dedicated to the complement of WBL1_4. Complement (4) ^, [ΓΓ; 3 " 3 ^ 口早疋1〇〇的母写There are only two wires in the port. One of the disadvantages of the unit 400 is the flat metal strip, especially the unit 400,;; one: the metal wiring. There are four more ones for writing the bit value. = like, will be more than unit _ may make unit 4. 〇 size is written large: single storage two pieces wide 2: metal wiring 'this congestion. Inch is greater than early 兀 100, and increase the metal layer [0 0 5 2 ] map The memory unit 500 includes a block diagram of the memory unit 500. Temporary-2, RBU_2, and 2 memory; "block 102, two read circuits..., and 100. However, temporary storage No. 2, WBU-4 'Scratchpad unit 如图5-Π6 as shown in Fig.. Instead, 疋I, the 疋 register unit 400 in the register unit 100 includes four pass gates.

第19頁 1272618 五、發明說明(15) 與圖四的單元4〇〇類似,標示為PG1、PG2、PG3及PG4。暫 存器單元500也包括四個反相器,標示為512、514、516及 518。暫存器單元5〇〇不包括圖四單元4〇〇中的金屬接線 WWLX1 - 4。取而代之的是,PG1-4之P通道電晶體的閘極係 分別耦接至反相器512、514、516及518的輸出,而反相器 512、514、516及518的輸入則分別耦接至WWL1-4。 [ 0053 ]可觀察到,單元50 0的缺點之一是,雖然其包 含與圖一單元1〇〇相同數量的金屬接線,但卻比單元1〇〇包 含更多的電晶體。尤其是,假設反相器5丨2到5丨6至少包括 二個電晶體,則單元5 0 0至少會比單元1〇〇多出四個用以將 位元值寫入儲存元件102的電晶體,這可能會使單元5〇〇的 尺寸大於單元1〇〇。 [ 00 54 ]圖六係一習用之暫存器單元6〇〇的方塊圖。暫 存單元600包括一儲存元件1〇2、二個讀取電路1〇4、 RWU-2、RBU-2、WWL1-4、WBU-4 及N5-12,如圖一的暫 存器單元100。然而,暫存器單元6 〇〇不包括暫存器單元 100中的N13-N16。取而代之的是,暫存器單元6〇()包括四 個反相器,標示為612、614、616及618。N9-N12的源極係 分別耦接至反相器6 1 2、6 1 4、6 1 6及6 1 8的輸出,而反相器 612、614、616及618的輸入則分別耦接至WBU-4。 口 [ 00 55 ]可觀察到,單元6〇〇的缺點之一是,雖然其包 含與圖一單元100相同數量的金屬接線,但卻比單元丨'〇〇包 έ更多的電晶體。尤其是,假設反相器6丨2到6丨6至少包括 二個電晶H,單元600至少會比單元1〇〇多出四個用以將位Page 19 1272618 V. Description of the invention (15) Similar to the unit 4〇〇 of Figure 4, denoted as PG1, PG2, PG3 and PG4. The register unit 500 also includes four inverters, designated 512, 514, 516, and 518. The register unit 5 does not include the metal wiring WWLX1 - 4 in the unit 4 of Figure 4. Instead, the gates of the P-channel transistors of PG1-4 are coupled to the outputs of inverters 512, 514, 516, and 518, respectively, and the inputs of inverters 512, 514, 516, and 518 are coupled respectively. To WWL1-4. [0053] It can be observed that one of the disadvantages of cell 50 is that although it contains the same number of metal wires as cell 1 of Figure 1, it contains more transistors than cell 1 . In particular, assuming that the inverters 5丨2 to 5丨6 include at least two transistors, the cell 500 will have at least four more cells than the cell 1 to write the bit values to the storage element 102. Crystal, this may make the size of unit 5〇〇 larger than unit 1〇〇. [0054] Figure 6 is a block diagram of a conventional register unit 6〇〇. The temporary storage unit 600 includes a storage component 1〇2, two read circuits 1〇4, RWU-2, RBU-2, WWL1-4, WBU-4 and N5-12, such as the register unit 100 of FIG. . However, the register unit 6 does not include N13-N16 in the register unit 100. Instead, the register unit 6() includes four inverters, labeled 612, 614, 616, and 618. The sources of the N9-N12 are respectively coupled to the outputs of the inverters 6 1 2, 6 1 4, 6 16 and 61 18, and the inputs of the inverters 612, 614, 616 and 618 are respectively coupled to WBU-4. Port [0055] It can be observed that one of the disadvantages of cell 6〇〇 is that although it contains the same number of metal wires as cell 100 of Figure 1, it has more transistors than cell 丨'. In particular, it is assumed that the inverters 6丨2 to 6丨6 include at least two electro-crystals H, and the unit 600 has at least four more than the unit 1〇〇 for placing the bits.

1272618 五、發明說明(16) 元值寫入儲存元件102的電晶體,這可能會使單元6〇〇的尺 寸大於單元100。 、、[0 056 ]雖然本發明及其目的、特徵和優點已詳細敘 述,其它實施例亦可包含在本發明之範圍内。例如,雖然 本發明所述之實施例具有四個寫入埠及二個讀取埠,但^ t明亦可適用於具有不同數目寫入埠及讀取埠的暫存器單 ^〜之以上所述者,僅為本發明之較佳實施例而已, ::能以之限定本發明所實施之範圍。大凡依本發明 ::範圍所作之均等變化與修飾’皆應仍屬於 利 =之範圍内,謹請f審查委員明鑑,並祈惠准,是寻: 主轉 °1272618 V. INSTRUCTION DESCRIPTION (16) The element value is written to the transistor of the storage element 102, which may cause the cell 6〇〇 to be larger than the cell 100. [0056] While the invention and its objects, features and advantages have been described in detail, other embodiments may be included within the scope of the invention. For example, although the embodiment of the present invention has four write buffers and two read buffers, it can be applied to a temporary memory device having a different number of write buffers and read buffers. The above is only the preferred embodiment of the present invention, and the scope of the invention can be limited thereto. According to the invention, the equal changes and modifications made by the scope of the invention are still within the scope of profit = **, please be sure to review the committee, and pray for the right, is to find:

第21頁 1272618 圖式簡單說明 【圖式簡單說明】 [0 0 1 9 ]圖一係本發明之多埠暫存器單元的方塊圖。 [0 0 2 0 ]圖二係本發明之圖一儲存元件的方塊圖。 [0 0 2 1 ]圖三到六係習知技術之暫存器單元的方塊 圖號說明: 1 0 0 -暫存器單元 ’ 104 -讀取電路 108-寫入電路 400 -暫存器單元 512, 514, 516, 518-反 612, 614, 616, 618-反 102-儲存元件 1 0 6 -反相器 30 0-暫存器單元 5 0 0 -暫存器單元 器 6 0 0 -暫存器單元 器Page 21 1272618 Brief Description of the Drawings [Simple Description of the Drawings] [0 0 1 9] Figure 1 is a block diagram of a multi-turn register unit of the present invention. [0 0 2 0 ] Figure 2 is a block diagram of the storage element of Figure 1 of the present invention. [0 0 2 1 ] FIG. 3 to FIG. 6 are block diagrams of a conventional register unit of the prior art: 1 0 0 - register unit '104' - read circuit 108 - write circuit 400 - register unit 512, 514, 516, 518- counter 612, 614, 616, 618-reverse 102-storage component 1 0 6 - inverter 30 0-scratchpad unit 5 0 0 - register unit 6 0 0 - temporary Memory unit

第22頁Page 22

Claims (1)

% 六、申請專利範圍 1 · 一種暫存器單元,包括: 一儲存it件,具有真值輸入及補數、輸入;以及: N個寫入電路,每一値皆耦接至該儲存元件,每一寫 人電路包括: " 一第一輸入接線,用於傳送一二進位值以窝入該 ....... ..... . " .*· - .· . - . ·· - ........ ......... . ...... · 儲存元件中; -第一電晶體,將該第一輸入接線耦接至該真值 輪入; 第一輸 二電晶 體;以 電晶體 通,以 2. 第一及 線。 3. 第一、 電晶體 4. 第一、丨 電晶體 第上及第三電晶體,串接至該補數輸入,其中該 入接線亦耦接至該第三電晶體,以選擇性地使談第 體辱通,提供該二進位值之一補數至該第丄電晶 . --- ' .............. 及 一第二輸入接線,柄接至該第一電晶體及該第二 ,用於選擇性地使該第一電晶體及該第二電晶體導 選擇性地使該二進位值能寫入該儲存元件。1 如申請專利範圍第1項所述之暫存器單元,其中該 第二輸入接線係每一該Ν個寫入電路僅有之輸八接 如申請專利範圍第2項所述之暫存器單元,其中該 第二及第三電晶體係每一該Ν個寫入電路中僅有之 如申讀專利範圍第1項所述之暫存器單元,其中該 第裏及第三電晶體係每一該N個寫入電路中僅有之%. Patent application scope 1 · A register unit, comprising: a storage unit having a true value input and a complement, an input; and: N write circuits, each coupled to the storage element, Each writer circuit includes: " a first input wiring for transmitting a binary value to nest into the ..... . . . . . . . . . . . . . . . ·· - .......................... In the storage element; - a first transistor that couples the first input connection to the true value Wheeling; the first transmission of two transistors; with the transistor through, to 2. First and line. 3. First, the transistor 4. The first, the first and third transistors of the transistor are serially connected to the complement input, wherein the input wiring is also coupled to the third transistor to selectively enable Talk about the first insult, provide one of the binary values to the Dijon. --- ' .............. and a second input wiring, the handle is connected to The first transistor and the second are configured to selectively cause the first transistor and the second transistor to selectively enable the binary value to be written to the storage element. [1] The register unit of claim 1, wherein the second input wiring system has only one input of each of the write circuits, such as the temporary register described in claim 2 a unit, wherein the second and third electro-crystal systems are each of the ones of the write circuits, such as the register unit of claim 1, wherein the first and third electro-crystal systems Only one of each of the N write circuits 0608-A40741TWFl.ptc 第23頁 1272618 案號 921(W(U1 a jEje 六、申請專利範圍 5·如申請專利範圍第i項所述之暫存器單元 於2 〇 6·如申凊專利範圍第】項所述之暫存器單元 第二電晶體之一汲極係耦接至該補數輪冬。: :7 ·如申請專利範圍第6項所述之暫存丨器單元 事三電晶體之一源極係為接地。 丨8 ·如:申:請專利範圍第7項餘述之^^^^ 第二電晶體之一源極係耦接至該第三電晶體之一 9·如申請專利範圍第1:現 第一、第二及第三電晶體包括金氧半導體(M〇s 10 ·如申請專利範圍策^ 第一、第二及第三電晶體包含n通道M〇s元件。〜該 括· 11 ·如申凊專利範圍第]項所述之暫存器單元,更包 杜取電路’輕接至該儲存元件,用以從 件讀取該二進位值。 、 夂4埽存元 大於j 2:如申請專利範圍第n項所述之暫存器單元 士 13‘如申請專利範圍第仏項所述之暫存哭單一 母一該Μ個讀取電路包、括: ;w早兀,其 : 其中N大 ’其中該: ’:其中該 5 cML ll? 八γ該 及極。 ΐ 其'呼...該..:1 其中Μ 中 及 一讀取電路,將該真值輪人耦接 14 ·如申請專利範圍第13碩、/ γ出接線 ,所返之暫存器單元 其中0608-A40741TWFl.ptc Page 23 1272618 Case No. 921 (W (U1 a jEje VI. Patent Application No. 5) The register unit described in item i of the patent application scope is at 2 〇 6 · One of the second transistors of the register unit is coupled to the complement wheel winter.: :7 · The temporary storage unit of the sixth embodiment of the patent application scope One source is grounded. 丨8 · For example: please refer to the patent scope of item 7 ^^^^ One of the second transistor is coupled to one of the third transistors. Patent Application No. 1: The first, second and third transistors include gold oxide semiconductors (M〇s 10 • as claimed in the patent) First, second and third transistors comprise n-channel M〇s components ~ Included · 11 · The register unit described in the scope of the patent application section [7], the packaged circuit is 'lighted to the storage element for reading the binary value from the piece. 夂 4埽存元 is greater than j 2: as described in item n of the patent application, the temporary storage unit 13' as described in the scope of the patent application, the temporary crying single mother, one reading Take the circuit package, including: ;w early, which: where N big 'which should be: ': where 5 cML ll? eight γ and the pole. ΐ its 'call...the..:1 where Μ a read circuit, the true value wheel is coupled to the 14; as claimed in the patent range 13th, / γ output wiring, the register unit is returned 0608-A40741TWFl.ptc 第24貢 六、申請專利範圍 每一該Μ個讀取電路更包括 你吃二f二输入接線’輪接至該讀取電路,甩於選摆α 存元件讀取該二進位值Γ導通選擇性地致能從該錯 Λ ^2 ^' ^ ^ ^ ^ 11 ®; ^ 11 31 ^ ^ ^ ^ ^ ^ ^ ^ ^ 大於I6。·如’吻^ ' 17:^暫存器單元^包漱 N個寫入旱’母一個恰好咆 輅接至該儲^路’每一個將其所對應之其中一個寫入埠; 體。 70件’且每一寫入電路恰好包含三個電晶 該二屬專利範圍第17項所述之暫存器單元,其中 使該三%電S Ϊ中之第一條傳送一第—訊號,以選擇性地 新該儲存元Γΐ中之第一個與第二個導通,而選擇性地更 19如由▲中之該位元。二… :I- 該二條金遥\讀專利範圍第1 8項所述之暫存器單元,其电; 之一值 屬接線中之第二條傳送用以選聲性地更新該位元 • . ΛΛ ';'',:;. ^· ^ ;' .ΛΪ;·.Γ: ”...·. 'i··.- -· -:v ::.,. ," ...:, 20 ·如由▲士 在考一你甲,專利範圍第19項所述之暫存器單元,其中: 地金屬接線中之談第二條上所傳送的讓值亦選擇性: :以一個電晶體中之第三雜噂 0608-A40741TWFl.ptc 第25頁0608-A40741TWFl.ptc The 24th tributary, the scope of the patent application, each of the reading circuits further includes that you eat two f two input wirings' to the reading circuit, and select the two storage elements to read the two The carry value Γ is selectively enabled from the error Λ ^2 ^' ^ ^ ^ ^ 11 ®; ^ 11 31 ^ ^ ^ ^ ^ ^ ^ ^ ^ is greater than I6. · For example, 'Kiss ^' 17: ^ register unit ^ package 漱 N writes to the drought 'mother one just 咆 至 to the storage path' each one of which corresponds to one of them; 70 pieces 'and each writing circuit contains exactly three crystals. The second unit is a register unit according to item 17 of the patent scope, wherein the first one of the three parts of the electric S 传送 transmits a first signal, To selectively turn on the first one of the storage cells and the second one, and selectively to be 19 as the bit in ▲. Second... :I- The two Jin Yao \ read the register unit described in item 18 of the patent scope, its electricity; one of the values is the second transmission in the wiring to selectively update the bit. ΛΛ ';'',:;. ^· ^ ;' .ΛΪ;·.Γ: ”...·. 'i··.- -· -:v ::.,. ," ... :, 20 · If the singer is in the test case, the register unit mentioned in item 19 of the patent scope, wherein: the transfer value of the ground metal connection is also optional: The third choke in a transistor 0608-A40741TWFl.ptc Page 25 1272618 介 __案號 921Q0941_P 年 六、申請專利範圍 至該三個電晶體中之該第二個電晶艘。 21 ·如申請辜利範圍第20項所述之暫存器單元,其中 該儲存元件具有竊接至該三個電晶艨中之該第—個電晶體 之第一輸入,以及麵接至該三個電晶體中之該第二個電 .、 ^ .晶體之'一::::第二’輪 I ; . 、* ..... · · ' . ,........ 22 ·如申請專利範圍第21項所述之暫^存器果元、其·- 用以選探性地更新該位元。 f 23 ·如申·請專利範圍第丨8項所述之暫;存器翠元r其^隹^^ 該N個寫入埠之該二條金屬接線中又該第一條接線會共同 地決定談N偃寫入埠中的哪一假會^ 2 4 ·如申請專利範圍第丨7項所述之驚存器軍元,其申 .大於2·。十 25·如申請專利範圍第17項所述之暫存器單元,其中凡 括 26·如申請專利範圍第24項 所 ^ ί二個讀取電路,耦接矣該餚存本件,甩以從該儲 存凡件讀淑該位元值、 2 7 ·.獐多埠暫存器單元,包栝· 卜元件,甩以儲存一位元;Ϊ N組金:屬接; Ni電晶耀,:每組僅有三個電晶體,丨每組電晶體係 :接至該儲存元件v甩以將讓位炙寫獻存 每組僅有二條金屬接線崎1272618 _ __ Case No. 921Q0941_P Year VI. Patent application scope to the second electro-crystalline ship in the three transistors. The register unit of claim 20, wherein the storage element has a first input that is stolen to the first one of the three transistors, and is connected to the first The second of the three transistors, ^. The 'one of the crystal:::: the second 'round I; . , * ..... · · ' . ,........ 22 · The temporary storage unit as described in item 21 of the patent application scope, which is used to selectively update the bit. f 23 ·For example, please refer to the temporary range mentioned in item 8 of the patent scope; the memory of the Tsui Yuan r r ^ ^ ^ ^ ^ The N writes of the two metal wiring and the first wiring will be jointly determined Talk about which fake meeting in N偃 is written ^ 2 4 · If you apply for the stunner military yuan mentioned in item 7 of the patent application, its application is greater than 2·. 10.25. The register unit as described in claim 17 of the patent application, wherein the two reading circuits of the 24th item of the patent application scope are coupled to the food storage device, The storage unit reads the bit value, 2 7 ·. 獐 埠 埠 埠 单元 , , , 卜 卜 卜 卜 卜 卜 卜 卜 卜 卜 卜 Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ϊ Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni There are only three transistors in the group, and each group of electro-crystal systems: connected to the storage element v甩 to place the yielding bits in each group with only two metal wirings. 線係耦接至該N組電晶體中對應的〆組,用以將該位元寫 入至該儲存元件。 • 2 8 ·:如申請專利範圍第 大於2'。: 卜29 ·如申請專,利範圍第27項所述之驾存I '备4 • . ' · .·· . - V: , · . '·-··; - · - - … ::, . - … 1 3 0 ·:如球請專利儀 對丨每一該N組電晶體而言,該僅有的二個 個丨係耦接至該儲存元件之w赛督 儲;存元件。. 3 1 ·如申請專利範圍第3 〇項戶斤述之暫存益單元,其中 對每一該N組電晶體而言,該僅有的二個:電晶體中之第之 個係耦接至該儲存元件之一補數輸入’以將該位元之一補 數寫入該儲存元件。 • 32·如申請專利範圍第31頊所述f暫存器單元,其中 對每一該N組電晶體而言,該僅有的二個電晶體中之第三 傭係耦接至該僅有的三個電晶體中之該苐二假,用以將該 位;元之該補數送至該僅有的三個電晶體中之該第二假> 一 :33·如申請專利範圍第32項所述之暫存器單元.,其中 對於每一該N組電晶體與該N組金屬接線中之該對應的一組 而i言,該僅有的二條金屬接線中之第一條會傳遂該位元’ 以寫入該儲存元件。 34·如:參 對於每一該N組電晶體與該N組金屬接線申之讓對惠爾The wire system is coupled to a corresponding one of the N sets of transistors for writing the bit to the storage element. • 2 8 · If the scope of the patent application is greater than 2'. : 卜29 ·If you apply for a special, the driving situation mentioned in item 27 of the profit range I ' 备 4 • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - ... 1 3 0 ·: For the ball, for each of the N sets of transistors, the only two of the ties are coupled to the storage element. 3 1 · As for the temporary storage unit of the third paragraph of the patent application scope, for each of the N sets of transistors, the only two: the first one of the transistors is coupled To the storage element one of the complement inputs 'to write one of the bits to the storage element. 32. The f register unit according to claim 31, wherein for each of the N sets of transistors, a third of the only two transistors is coupled to the only one The second of the three transistors is used to send the bit; the complement of the element to the second of the only three transistors > a: 33 · as claimed in the patent scope 32. The register unit of claim 32, wherein for each of the N sets of transistors and the corresponding one of the N sets of metal wires, the first one of the only two metal wires is Pass the bit ' to write to the storage element. 34·如: Reference to each of the N sets of transistors and the N sets of metal wiring 第27買 0608-A40741TWFl.ptc 1272618 ^ . /, __案號92100941 年S月曰 修正_ 六、申請專利範圍 而言,該僅有的二條金屬接線中之該第一條係耦接至該僅 有的三個電晶體之該第三個之一閘極,其中該僅有的二條 金屬接線之該第一條會選擇性地使該僅有的三假電晶體之 該第三個導通,以選擇性地將該位元之該樵數送至該僅有 ' .. : .: ..... ·. :的三 '- - : :: 1 '-. 3 5.如申丨請專利範圍第27項所述之暫存器單元 > 更包專」 至:少二個讓取電路,耦接:支 - · ,. _ ( · 存元件讀取該位惠值。: 3 6.如申請專利範圍第35項所述之暫存器單元,其中 . * . 該至少二個讀取電路包含二個讀取電路A27th buy 0608-A40741TWFl.ptc 1272618 ^ . /, __ case number 92100941 S month 曰 amendment _ 6. In the scope of patent application, the first of the only two metal wires is coupled to the The third one of the three transistors, wherein the first strip of the only two metal wires selectively turns the third of the only three dummy transistors on, To selectively send the number of the bits to the three '-: ::: 1 '-. The register unit described in item 27 of the patent scope is further packaged to: less than two access circuits, coupled: branch - · , . _ ( · memory component reads the bit value.: 3 6 The register unit according to claim 35, wherein the at least two read circuits comprise two read circuits A 0608-A40741TWFl.ptc 第28頁0608-A40741TWFl.ptc第28页
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