CN102332869B - Highly integrated BTL audio power amplifier - Google Patents
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Abstract
本发明公布了一种高集成度BTL音频功率放大器,包括4个放大器、2个跨导放大器以及第一至第九电阻、内部补偿电容、基准电流模块,其中第一、第二放大器提供主通路并作为音频功率放大器的上下驱动单元,第三、第四放大器,以及第一跨导放大器、第五至第八电阻共同组成输出失调补偿模块,第一跨导放大器负责检测BTL两端直流共模失调,反馈到电流控制端微调整电流,进而减小失调,还在输出削波时微调增益使输出两端增益完全相同,使输出功率达到最大、效率最高,基准电流模块负责为第一跨导放大器输出端提供准确的输出中点;第二跨导放大器负责提供除米勒补偿外的系统的频率补偿,为系统提供足够的相位余量。
The invention discloses a highly integrated BTL audio power amplifier, including 4 amplifiers, 2 transconductance amplifiers, first to ninth resistors, internal compensation capacitors, and a reference current module, wherein the first and second amplifiers provide the main path And as the upper and lower drive units of the audio power amplifier, the third and fourth amplifiers, the first transconductance amplifier, and the fifth to eighth resistors together form an output offset compensation module. The first transconductance amplifier is responsible for detecting the DC common mode at both ends of the BTL Offset, feed back to the current control terminal to fine-tune the current, thereby reducing the offset, and fine-tune the gain during output clipping to make the gains at both ends of the output exactly the same, so that the output power reaches the maximum and the efficiency is the highest. The reference current module is responsible for the first transconductance The output of the amplifier provides an accurate output midpoint; the second transconductance amplifier is responsible for providing frequency compensation of the system except for Miller compensation, providing sufficient phase margin for the system.
Description
技术领域 technical field
本发明涉及一种音频功率放大电路,特别涉及BTL应用音频功放中的失调补偿、增益补偿、及相位补偿。 The invention relates to an audio frequency power amplifying circuit, in particular to offset compensation, gain compensation and phase compensation in a BTL application audio power amplifier.
背景技术 Background technique
在同样电压下,为了得到更高的音频输出功率,都会采用BTL方式应用,如图1所示是一种比较流行的用法,本用法省去了输出耦合电容,也降低外部分压消耗,但是本用法有两个缺点:没有失调补偿电路,输出外部需要补偿电容,BTL应用对电路输出端的失调电压要求特别高,如果失调处理不当,反而会影响静态功耗,还会产生很大的电流声,在大信号输出削波时,如果两端增益有差别,会影响整体的输出效率;另外BTL应用时电路较单声道应用使电路更加不稳定,相位余量变小,一般解决方式是外接100nF的电容,但是本发明改进内部结构,无需外部任何补偿电容。 Under the same voltage, in order to obtain higher audio output power, the BTL method will be used. As shown in Figure 1, it is a more popular usage. This usage saves the output coupling capacitor and reduces the consumption of external voltage division. However, This usage has two disadvantages: there is no offset compensation circuit, and compensation capacitors are required outside the output. BTL applications have particularly high requirements for the offset voltage at the output of the circuit. If the offset is not handled properly, it will affect the static power consumption and generate a large current noise. , when the large signal output clips, if there is a difference in the gain at both ends, it will affect the overall output efficiency; in addition, the circuit in BTL application is more unstable than that in monophonic application, and the phase margin becomes smaller. The general solution is to connect an external 100nF capacitor, but the present invention improves the internal structure without any external compensation capacitor.
发明内容 Contents of the invention
技术问题:本发明主要解决BTL失调及增益不一致问题,及电路稳定性问题。 Technical problem : the present invention mainly solves the problem of BTL imbalance and gain inconsistency, and the problem of circuit stability.
技术方案: Technical solution :
一种高集成度BTL音频功率放大器,包括4个放大器AMP、2个跨导放大器以及第一至第九电阻、内部补偿电容、基准电流模块,其中第一放大器、第二放大器为主要放大器,提供主通路,所述第三放大器、第四放大器、第一跨导放大器、第五至第八电阻共同组成输出失调补偿模块,同时第三放大器、第四放大器分别是音频功率放大器的上下驱动单元; A highly integrated BTL audio power amplifier, including 4 amplifiers AMP, 2 transconductance amplifiers, first to ninth resistors, internal compensation capacitors, and a reference current module, wherein the first amplifier and the second amplifier are the main amplifiers, providing The main channel, the third amplifier, the fourth amplifier, the first transconductance amplifier, and the fifth to eighth resistors together form an output offset compensation module, and the third amplifier and the fourth amplifier are respectively the upper and lower drive units of the audio power amplifier;
第一跨导放大器,负责检测BTL两端直流共模失调,反馈到电流控制端微调整电流,进而减小失调,同时第一跨导放大器还在输出削波时微调增益,使输出两端增益完全相同,使输出功率达到最大,效率最高,基准电流模块负责为第一跨导放大器输出端提供准确的输出中点;第二跨导放大器负责提供除米勒补偿外的系统的频率补偿,其等效的电容容量为100nF,为系统提供足够的相位余量; The first transconductance amplifier is responsible for detecting the DC common-mode offset at both ends of the BTL, and feeding back to the current control terminal to fine-tune the current, thereby reducing the offset. It is exactly the same, so that the output power reaches the maximum and the efficiency is the highest. The reference current module is responsible for providing an accurate output midpoint for the output terminal of the first transconductance amplifier; the second transconductance amplifier is responsible for providing frequency compensation of the system except for Miller compensation. The equivalent capacitance is 100nF, which provides enough phase margin for the system;
其中:第一放大器的正端接参考电压信号,第一放大器的负端分别与第一电阻的一端、第二电阻的一端连接;第一放大器的输出端接第三放大器的正端; Wherein: the positive end of the first amplifier is connected to the reference voltage signal, the negative end of the first amplifier is respectively connected to one end of the first resistor and one end of the second resistor; the output end of the first amplifier is connected to the positive end of the third amplifier;
所述第一电阻的另一端分别与第三电阻、第四电阻的一端连接,第三电阻的另一端与第二放大器的负端连接,第二放大器的正端接参考电压信号,第二放大器的输出端接第四放大器的正端; The other end of the first resistor is connected to one end of the third resistor and the fourth resistor respectively, the other end of the third resistor is connected to the negative terminal of the second amplifier, the positive terminal of the second amplifier is connected to the reference voltage signal, and the second amplifier The output terminal of the fourth amplifier is connected to the positive terminal;
所述第三放大器的输出端分别与第五电阻的一端、第二电阻的另一端、连接;第五电阻的另一端分别与第三放大器的负端、第六电阻的一端连接;所述第六电阻的另一端分别与第八电阻的一端、第一跨导放大器的正端连接; The output end of the third amplifier is respectively connected to one end of the fifth resistor and the other end of the second resistor; the other end of the fifth resistor is respectively connected to the negative end of the third amplifier and one end of the sixth resistor; The other ends of the six resistors are respectively connected to one end of the eighth resistor and the positive end of the first transconductance amplifier;
所述第八电阻的另一端分别与第四放大器的负端、第七电阻的一端连接;所述第四放大器的输出端分别与第七电阻的另一端、第四电阻的另一端连接; The other end of the eighth resistor is respectively connected to the negative end of the fourth amplifier and one end of the seventh resistor; the output end of the fourth amplifier is respectively connected to the other end of the seventh resistor and the other end of the fourth resistor;
所述第一跨导放大器的负端接中点电位信号,第一跨导放大器的输出端连接第九电阻的一端、第二跨导放大器的负端、第二跨导放大器的输出端;所述第二跨导放大器的正端分别连接第九电阻的另一端、内部补偿电容的一端,内部补偿电容的另一端接地。 The negative terminal of the first transconductance amplifier is connected to the midpoint potential signal, and the output terminal of the first transconductance amplifier is connected to one end of the ninth resistance, the negative terminal of the second transconductance amplifier, and the output terminal of the second transconductance amplifier; The positive end of the second transconductance amplifier is respectively connected to the other end of the ninth resistor and one end of the internal compensation capacitor, and the other end of the internal compensation capacitor is grounded.
有益效果:由于采用了失调补偿网络,电路随工艺的波动,失调明显减小,离散程度也大大降低,较没有此补偿网络的电路,在输出效率上面也有一定的提高,由于集成了BTL必须的104电容,使外围简化,应用更加简单,系统更加稳定。 Beneficial effects : due to the use of the offset compensation network, the circuit fluctuates with the process, the offset is significantly reduced, and the degree of dispersion is also greatly reduced. Compared with the circuit without this compensation network, the output efficiency is also improved to a certain extent. Due to the integration of BTL necessary 104 capacitors simplify the peripherals, make the application easier and the system more stable.
附图说明 Description of drawings
图1是一般BTL功能框图。图中包括:第一运算放大器 A、第二运算放大器 B。 Figure 1 is a general BTL functional block diagram. The figure includes: a first operational amplifier A and a second operational amplifier B.
图2是为带失调补偿及频率补偿的框图。 Figure 2 is a block diagram with offset compensation and frequency compensation.
图中包括:第一运算放大器 A、第二运算放大器 B、第三运算放大器 C、第四运算放大器 D、第五跨导放大器 E、第六跨导放大器 F。 The figure includes: the first operational amplifier A, the second operational amplifier B, the third operational amplifier C, the fourth operational amplifier D, the fifth transconductance amplifier E, and the sixth transconductance amplifier F.
具体实施方式 Detailed ways
本发明中利用了六个放大器,其中第一运算放大器A、第二运算放大器B确定主通路的增益,BTL的闭环增益 Utilized six amplifiers in the present invention, wherein the first operational amplifier A, the second operational amplifier B determine the gain of the main path, the closed-loop gain of BTL
GV={1+R2/(R1+R3)}+R4/(R1+R3)}+ΣGV(修正);,为了弥补工艺的波动,必须设置成其中R2=R4,R1=R3;如果在两运放增益不一致时,就会出现共模失调,偏离中点,补偿电路会启动,调整增益补偿电流,适当的降低先削波的运放,使两路增益完全一致。 GV={1+R2/(R1+R3)}+R4/(R1+R3)}+ΣGV (correction); In order to compensate for the fluctuation of the process, it must be set as R2=R4, R1=R3; When the gain of the op amp is inconsistent, there will be a common mode offset, and the compensation circuit will start when it deviates from the midpoint, adjust the gain compensation current, and properly reduce the op amp that clips first, so that the gains of the two channels are exactly the same.
共模失调,电路本身设置了一个基准电流i0=(VCC-2Vref)/R2,假设完全对称,没有补偿的电流的前提下,基准电流被第一运算放大器A、第二运算放大器B平分,在第一运算放大器A中流经电阻R2电流等于第二运算放大器B中流经电阻R4的电流,所以第一运算放大器A输出中点的电压为: Common mode offset, the circuit itself sets a reference current i0=(VCC-2Vref)/R2, assuming complete symmetry, under the premise of no compensation current, the reference current is divided equally by the first operational amplifier A and the second operational amplifier B, in The current flowing through the resistor R2 in the first operational amplifier A is equal to the current flowing through the resistor R4 in the second operational amplifier B, so the voltage at the midpoint of the output of the first operational amplifier A is:
i0*R2/2+Vref=Vref+VCC/2-Vref=VCC/2 i0*R2/2+Vref=Vref+VCC/2-Vref=VCC/2
设置Vref的目的是为了提高输入端承受的最大输入电平。 The purpose of setting Vref is to increase the maximum input level that the input end can withstand.
同理第二运算放大器B的输出中点的电压也为VCC/2。 Similarly, the voltage at the midpoint of the output of the second operational amplifier B is also VCC/2.
实际中第一运算放大器A、第二运算放大器B存在共模失调,第三运算放大器C、第四运算放大器D负责检测失调,当输出增益不匹配时,vo电压偏离中点,启动失调补偿电路,输出补偿电流,进而调整增益偏大的通道,使两路输出的增益相当;当工艺波动引起输出共模失调时,失调补偿模块启动,输出矫正电流,调整输出中点,达到矫正静态失调的作用。第三运算放大器C、第四运算放大器D是一简单的闭环应用运放,增益设置为3;由于设置在输出驱动部分,还起到减小输出的THD,进而减小整体的失真度。 In practice, the first operational amplifier A and the second operational amplifier B have a common-mode offset, and the third operational amplifier C and the fourth operational amplifier D are responsible for detecting the offset. When the output gain does not match, the vo voltage deviates from the midpoint, and the offset compensation circuit is started , output compensation current, and then adjust the channel with higher gain, so that the gains of the two outputs are equal; when the process fluctuation causes the output common mode offset, the offset compensation module starts, outputs the correction current, adjusts the output midpoint, and achieves the effect of correcting the static offset effect. The third operational amplifier C and the fourth operational amplifier D are simple closed-loop application operational amplifiers with a gain of 3; because they are set in the output drive part, they also reduce the THD of the output, thereby reducing the overall distortion.
由于采用的是BTL应用,两个单独工作的系统A、B,组合在一起,所需要的频率补偿更加复杂,本发明选择的补偿点,从整个系统来看,假设输入端接地,补偿点是系统的交流地,如果系统在上电、或其它干扰使第一运算放大器A、第二运算放大器B输出存在高频失调,那么补偿点存在干扰脉冲,由于是BTL应用,此干扰经过第一运算放大器A、第二运算放大器B的反馈一直持续下去,形成震荡,此震荡将一直持续下去,形成自激。此现象不是一般的相位补偿理论, Because of the BTL application, two systems A and B working independently are combined together, and the required frequency compensation is more complicated. The compensation point selected by the present invention, from the perspective of the entire system, assumes that the input terminal is grounded, and the compensation point is For the AC ground of the system, if the system is powered on or other interference causes high-frequency offsets in the output of the first operational amplifier A and the second operational amplifier B, then there is an interference pulse at the compensation point. Since it is a BTL application, this interference is passed through the first operation The feedback from the amplifier A and the second operational amplifier B continues to form an oscillation, and this oscillation will continue to form a self-excitation. This phenomenon is not the general phase compensation theory,
因为第一运算放大器A、第二运算放大器B自身的频率补偿没有问题,单独工作非常稳定,但是组合成BTL应用,为了系统的稳定,需要重新额外设定补偿网络。本发明在补偿点设置一到地电容,作为高频交流地,使第一运算放大器A、第二运算放大器B输出的高频波经反馈后直接经补偿电容到地,不流经主通路,形不成正反馈。防止高频自激。下面推导如何实现100nF电容。 Because there is no problem with the frequency compensation of the first operational amplifier A and the second operational amplifier B, they work very stably alone, but when they are combined into a BTL application, in order to stabilize the system, an additional compensation network needs to be re-set. In the present invention, a capacitor to the ground is set at the compensation point as a high-frequency AC ground, so that the high-frequency waves output by the first operational amplifier A and the second operational amplifier B are directly fed back to the ground through the compensation capacitor, and do not flow through the main path to form a Positive feedback. Prevent high frequency self-excitation. The following deduces how to realize the 100nF capacitor.
假设流入的电流为i,端口电压为v,电阻两端电压为VR;到地电容为c,等效阻抗为rin;等效等效电容为cin;gm跨导放大器的跨导为第一运算放大器A;则满足如下关系式: Suppose the inflowing current is i, the port voltage is v, the voltage across the resistor is VR; the capacitance to ground is c, the equivalent impedance is rin; the equivalent equivalent capacitance is cin; the transconductance of the gm transconductance amplifier is the first operation Amplifier A; then satisfy the following relationship:
i=VR*A+VR/R(1) i=VR*A+VR/R(1)
v=VR/jωc*R +VR(2) v=VR/jωc*R +VR(2)
由(1)、(2)可以得出rin=ri+cin=1/A*R+1/ jωc*R*A(3),等效图如图3所示。 From (1) and (2), it can be concluded that rin=ri+cin=1/A*R+1/ jωc*R*A (3), and the equivalent diagram is shown in Figure 3.
其中ri=1/A*R;cin=1/jωc*R*A;ri≈0,等效电容cin= (A*R) 1/jωc,所以等效电容放大了(A*R)倍;假设跨导为A=10,R=10K,内部集成电容c=1pF,所以cin=(10*10k*1p)F=100nF;利用了较小电容,实现大电容的功能,消除了高频自激。 Where ri=1/A*R; cin=1/jωc*R*A; ri≈0, the equivalent capacitance cin= (A*R) 1/jωc, so the equivalent capacitance is enlarged by (A*R) times; Assume that the transconductance is A=10, R=10K, and the internal integrated capacitor c=1pF, so cin=(10*10k*1p)F=100nF; a small capacitor is used to realize the function of a large capacitor and eliminate the high-frequency self- Excited.
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