JPH061415B2 - AC power controller - Google Patents

Description

Detailed Description of the Invention

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an AC power control device that smoothly controls AC power supplied to a load.

[Prior art]

Generally, the window glass installed in the cockpit of an aircraft is a heater.
Is embedded, and current is supplied to it to add glass.
Due to the heat, the glass becomes cloudy and has ice drops.
I try not to wear it. Is this current an on-board AC source?
It is necessary to adjust the value according to the temperature of the outside air.
Therefore, conventionally, phase control by thyristor is performed.
Was there. However, if phase control is performed using a thyristor, an AC wave
Since the shape becomes discontinuous and generates noise, this noise
To prevent life, one cycle of alternating current is the minimum unit
The energization time is controlled, and the power with different maximum values is applied.
Multiple power supply circuits to generate, continuous desired energization control
After entering the state, the first power used until that time
The closest to the power supply circuit and the power that was being supplied up to that point.
Alternately select a second power supply circuit that supplies high power
The second power supply until the energization period reaches the desired control state.
It is conceivable to sequentially lengthen the energization period of the supply circuit. [Problems to be solved by the invention] However, the power supply circuits with different powers are switched.
Moment, due to the voltage / current phase mismatch during switching,
A large current may flow, causing fluctuations in the power supply voltage.
There is. Had a problem. [Means for Solving Problems] In order to solve such problems, the present invention is
The first one cycle of Dentsu start time for a predetermined time
A delay circuit for delaying is provided. [Operation] Generation of a large current generated during switching is suppressed. [Embodiment] FIG. 1 is a block diagram showing an embodiment of the present invention.
In the figure, 1 is a filter circuit, 2 is a power supply circuit, and 3 is
Zero level detection circuit, 4 heater heater time control circuit, 5
Is a gate signal generation time control circuit, 6 is a gate signal generation circuit
Path, 7 is a thyristor circuit, 8 is overcurrent detection
Circuit, 9 is heater current detection circuit, 10 is sensor sensor
Detection circuit, 11 is an overheat detection circuit, 12, 1
3 is a switching circuit, 15 is an AND circuit, and 16 is a relay circuit.
-, 16a is a contact of relay 16, 17 is a transformer, 18 ₁ , 18 ₂ Is
Interlocking type power switch, 19 is a switch for sensor check
Tsuchi, 20 is a switch for overheat check, 2
1 is a switch for power-on check, 22 is a heater,
23 is an operation indicator lamp, 24 is the temperature of the window glass
The thermistor 25 is a resistor. The filter circuit 1 is used for intrusion of noise from the outside and this device.
Prevents noise from leaking to external devices. Power supply
Circuit 2 converts the AC voltage into the DC voltage required for the operation of this device.
It is supposed to be converted. Zero level detection circuit 3
Zero level is detected every cycle of the flow waveform, and one cycle starts
A pulse signal is generated at a time point,
As shown in FIG. 2, the input terminal 3a, the output terminal 3b, the resistor
Anti-32a-32b, diodes 33a-33c, transistor 34,
It is composed of a capacitor 35 and an inverter 36. The heater energization time control circuit 4 displays the current supplied to the heater as a window.
Taking into account the temperature of the glass and the time elapsed since the power was turned on,
Conduct the energization time control with one cycle of the AC waveform as the minimum unit.
It is becoming like this. This circuit is added as shown in Figure 3.
Width circuit 40, output voltage for about 3 minutes after power is turned on
3 minutes timer 41, comparing section 42, 8Hz
An oscillator 43 for generating a triangular wave of
They are composed of resistors 40a-40f, 41a-41f, 42
a, 42b, 43a ~ 43m, 44a ~ 44g, differential amplifier 40q, 41q, 42q, 43q
~ 43s ,, 44q, 44r, diode 41t ~ 41v, 42t, 43t ~ 43w, 44
t, capacitors 40x, 40y, 41x, 42x, 43x, 44x, input terminal 4a ~
4c and output terminals 4d to 4f. The gate signal generation time control circuit 5 is one of the thyristor circuits 7.
Gate required to control the energization time for one cycle
It is designed to generate signals, as shown in Fig. 4.
, Resistors 50a-50e, capacitors 51a-51d, timers 52a-5
2c, buffers 53a to 53c, OR circuit 54, diode 55a
~ 55c, inverters 56a ~ 56b, set-set type drive
Up-flop 57, AND circuit 58, input terminals 5a-5d,
It is composed of output terminals 5e to 5h. The thyristor circuit 7 has a thyristor 70 as shown in FIG.
a to 70f, input terminals 7a to 7m, output terminals 7p to 7r
ing. As shown in FIG. 6, the gate signal generation circuit 6 operates as a NAND circuit.
Path 60a, 60b, resistors 61a-61k, capacitors 62a-62d, in
Bata 63a-63d, buffers 64a, 64b, AND circuits 65a-65
l, PS type flip-flop 66a, 66b, diode 67a
~ 67g, transistors 68a ~ 68d, pulse transformers 69a ~ 69
c, input terminals 6a to 6g, output terminals 6j to 6m, 6p to 6z
Has been. The overcurrent detection circuit 8 is, as shown in FIG.
Resistors 81a-81h, Capacitors 82a, 82b, Diodes 83a-83
c, differential amplifiers 84 and 85, input terminal 8a, output terminal 8b,
It is composed of 8c. The heater current detection circuit 9 is a transformer 9 as shown in FIG.
1, diode 92, differential amplifier 93, reference voltage 94
The rectified voltage generated by the heater current is the standard
When the voltage exceeds the value of 94, a "1" level signal is sent.
The reference voltage 94 is applied to the heater 22.
Rectification generated by the maximum specified value of the heater current supplied
Selected slightly larger than voltage. Therefore,
When the heater current is normally supplied, the heater current detection
Path 9 outputs a "1" level signal. The sensor short detection circuit 10 is a differential amplifier 10a, a reference
It consists of voltage 10b and inverters 10c and 10d
Is energizing the relay 16 but is not supplied to the input terminal 10e.
The sensor voltage is less than the reference voltage 10b
In this case, a "1" level signal is supplied to the input terminal 10f.
When over current is detected, the input terminal 10g is set to "0" level.
Relay signal is supplied when relay 1 is detected
6 is being deactivated. The overheat detection circuit 11 includes a differential amplifier 11a and a reference voltage.
Pressure 11b, inverter 11c, diode 11d,
A voltage greater than the reference voltage 11b is supplied to the input terminal 11e.
"0" level from output terminal 11g when overheat is detected
The signal of is sent. The operation of the device thus configured is as follows.
However, for easy understanding, the power supplied to the heater is
The description will be given assuming that there is one type. Switch in FIG.
18 ₁ , 18 ₂ When the power is on, the power supply circuit 2
, The voltage + V necessary for operation is supplied to the
Relay 16 is operating, the filter circuit 1
The generated AC waveform is supplied to the zero level detection circuit 3.
It This circuit, as shown in FIG.
The Zener diode 33c is inserted in series in the source circuit.
And the direct current voltage + V is supplied to the transistor 34.
There is. Therefore, if the Zener diode 33c is short-circuited,
If so, the transistor 34 has a DC voltage + V
Is supplied, the operation start level is shown in FIG.
As shown in (a), it is connected to + V, and the value of the AC waveform is
Transistor 34 is turned on when it becomes smaller than the value of
Become. Therefore, this time, the Zener diode 33c is effective.
, The breakdown voltage is
Equal to the DC voltage + V supplied to the transistor 34
Then, the transistor 34 has the original operation start level +
It turns on only when the level of the AC waveform is lower than V by V.
Then, the signal shown in FIG. 8 (b) is output. That is,
The transistor 34 turns on at the zero level of the AC waveform,
It turns off at the next zero level. The above explanation is the reverse between the base emitters of the transistor 34.
It is said that the voltage is zero, but this value is about 0.7 volt
This level is the characteristic behavior of the transistor 34.
It is an inevitable starting level. But straight
Current, breakdown voltage of Zener diode 33c
By selecting properly, the circuit operates as a whole.
The starting level can be adjusted to zero level. That is, when the DC bias voltage is supplied in the operating direction,
Operation of the transistor 34 with respect to the transistor 34
DC bias voltage so that the starting level is zero volts
Insert a Zener diode 33c for level shifting to cancel
Zero level detection can be performed by turning on. That
Then, the voltage generated in the resistor 32c as shown in FIG.
It is differentiated by the sensor 35 and the resistor 32d and is shown in FIG. 8 (c).
Signal, which is clamped by diode 33b and
It is inverted by the inverter 36 to produce the pulse shown in FIG. 8 (d).
Is output from the output terminal 3b. The pulse output from the output terminal 3b is the gate shown in FIG.
Is supplied to the input terminal 5a of the control signal generation time control circuit 5,
It is supplied to the terminal B of the timer 52b via the tour 53b. Ta
The imager 52b is supplied with a "1" level signal at the terminal CD.
When the signal supplied to terminal B falls,
Generates "1" level signals from terminals and "0" level signals from terminals
However, this state continues for the time determined by the resistor 50b and the capacitor 51b.
It is going to continue. And the terminal CD is "0"
When it becomes bell, the level of terminal Q will be reversed.
ing. Therefore, the zero level as shown in Fig. 8 (d)
When the pulse from the circuit detection circuit 3 is supplied to the input terminal 5a
A pulse having the period T shown in FIG. 8 (e) is output to
It During this period T, as described above, the resistor 50b and the capacitor 5
1b, longer than half cycle of AC power supply, shorter than 1 cycle
It is set so that it will be loud. Output signal of timer 52b
Via the buffer 53c, AND circuit 58, and output terminal 5g
It is supplied to the terminal 6d of the gate signal generating circuit 7 (described later).
As described above, the AND condition of the AND circuit 58 is satisfied).
Therefore, as shown in FIG. 6, the NAND circuit 60b, the inverter
The oscillation circuit consisting of the data 63b has a "1" level at the input terminal 6d.
Resistor 61d and capacitor 62b while the signal is being supplied.
Generates the gate signal shown in Fig. 8 (f) with a short cycle determined by
To do. On the other hand, when the power is turned on, the heater energization time control circuit shown in Fig. 3 is
3 minutes timer 41 sends out "0" level signal
However, this level should gradually rise for about 3 minutes.
It has become. Therefore, the differential amplifiers 44q and 44r are "1".
Level signals are sent through these terminals 4f and 4g.
Is output as the end of the gate signal generation circuit 6 shown in FIG.
AND circuits 65a and 65 through the children 6e and 6f and the buffers 64a and 64b.
It is supplied to one terminal of c. AND circuit 65a, 65c
The other terminal is the output of the zero level detection circuit 3 shown in FIG.
Via the inverter 56b of the gate signal generation time control circuit 5
Has been supplied. And the zero level detection circuit 3
Short time after detecting zero level as shown in Fig. 8 (d)
Only during the period, the level is "0", and during other periods
It is at "1" level. For this reason, the game shown in FIG.
The other input of the AND circuits 65a and 65c of the gate signal generation circuit 6
Is supplied with the signal shown in Fig. 8 (d).
The AND circuits 65a and 65c are
A "1" level signal is being sent. Because of this
Lip flops 66a, 66b are set and both
A signal of "1" level is generated from the output terminal Q of.
This signal is sent from the AND circuit 65e as "1" level.
The high frequency of the high frequency sent from the inverter 63b.
The oscillation waveform is the AND circuit 61i, diode 67b, resistor 61f
Is supplied to the transistor 68a via this transistor
Turn on. This allows the pulse transformer 69a to
Then, the gate signal is sent to the thyristors 70a and 70b of the thyristor circuit 7.
Signal is supplied and the thyristor is turned on. This rhino
Listers 70a, 70b supply transformer 17 as shown in FIG.
The signal supplied to the transformer 17 is transformed.
And is supplied to the heater 22. Thyristors are connected in antiparallel as shown in Fig. 5.
Therefore, one of the thyristors has a positive half of the AC waveform.
It turns on with the waves. Generally, the thyristor is the anode
When the gate voltage is supplied, the gate signal
If it is supplied for a short time, it turns on, but
In some cases, it may not always turn on. Good
If used under various environmental conditions, this will not occur.
However, it is not economical to always demand good environmental conditions.
Become. Even in such a case, the gate signal is not only once.
To ensure that it is turned on repeatedly
It can be in a state. Therefore, with this device, 20KH
Generate a gate signal with a frequency around z, and
Signal to the thyristor to ensure reliable operation.
It The thyristor that is in the ON state is an anode
To reverse the polarity of the power supply voltage supplied between
It turns off. Therefore, as shown in Fig. 5,
Keep the iristor in anti-parallel so that the AC waveform is a negative half-wave.
So that the high-frequency gate signal continues to be supplied even when
If this is set, the service that was in the off state at the time of the positive half-wave
Irista is on when the AC waveform reaches a negative half-wave
become. Here, the gate signal is a half cycle of the input AC waveform.
Be sure to turn it off at the time of over 1 cycle
For example, it will turn on at the end of one cycle of the input AC waveform.
The thyristor was off. With this device, as shown in Fig. 8 (g), the negative half-wave
I'm trying to turn on the iris. And negative
Gate signal after time t1 when the half wave of
Is supplied, the negative half
The thyristor, which had been off at the time of half-wave, turned on,
Positive half wave is output. At time t2, the gate signal is
When no power is supplied, but when the power supply polarity is positive half-wave
The thyristor that is on until t3 remains on.
Continue the state. At time t3, it becomes a negative half-wave
So, until now, i.e., the one that was on in the positive half-wave
The lister turns off, but from this point, as shown in (f)
, The gate signal starts to be supplied again, so the positive half-wave
The thyristor that had been turned off at turned on, and is shown in (g).
As the positive half-wave continues, the negative half-wave output
It is output from the star circuit 7. In this way, the AC waveform
Zero level in one direction, from positive to negative
Make sure to generate the gate signal every time you cross
, The AC waveform is continuously output. At time t4, the gate signal is no longer supplied, but before
As mentioned above, the thyristor that was on at this point
It remains on even if the gate signal is not supplied.
It However, at time t5, the polarity of the AC waveform changes
Then, the thyristor that was on until now is turned off. That
Then, as shown in (f), the trigger signal is supplied after time t4.
Since it is not supplied, the thyristor circuit 7 will also be output after this point.
No force signal is generated. The heater energization time control circuit 4 is constructed as shown in FIG.
The three-minute timer 41 is, as shown in FIG. 9 (a),
The output voltage monotonically increases from the moment the power is turned on and gets tired in about 3 minutes.
The oscillator 43 is shown in FIG. 9 (b).
So that a triangular wave of about 8Hz is generated.
It Therefore, the differential amplifier 42q is as shown in FIG. 9 (c).
The output signal level from the 3-minute timer 41 is higher than the triangular wave
Also outputs a "0" level signal for a large period. This signal
Is transmitted from the output terminal 4e, and the gate signal is generated as shown in FIG.
Terminal 5b of live time control circuit 5, inverter 56a, OR circuit
It is supplied to the terminal CD of the timer 52b via 54. this
Therefore, the signal shown in Fig. 9 (d) is supplied to the terminal CD of the timer 52b.
Be paid. On the other hand, the frequency of the AC power supply is about 400Hz.
Therefore, the cycle of the triangular wave is 50 times that of the AC power supply.
Therefore, during the period of one cycle of the triangular wave,
It corresponds to the cycle. And the signal shown in (d) is
The gate signal is supplied to the thyristor circuit 7 during the "1" level.
Is supplied, the thyristor circuit 7 will remain for 3 minutes after the power is turned on.
It is turned on intermittently, and the period during which it is on is time
As the time elapses, the timer shown in (a)
After the input voltage is saturated, it turns on continuously. For this reason,
As shown in Fig. 9 (e), the output from the thyristor circuit 7
The AC waveform that is generated is the cycle during the output period as time passes.
The number of le is increasing. The above is the explanation when the temperature of the glass is not added,
The thermistor attached to the glass is actually
The resistance value depends on the temperature, so when the power is turned on, the temperature
Is low and the resistance is low in the normal state. others
Therefore, the differential amplifier 40q of the amplifier circuit 40 has a non-inverting input terminal.
The voltage is higher than the voltage at the inverting input terminal
Then, this circuit uses the signal of "1" level, that is, the heater 2
2 is outputting a signal to heat it so that it becomes hot
It However, as mentioned above, when the power is turned on, the three-minute timer
Since the output voltage of 41 gradually increases, the differential amplifier 41q
Output level is lower than the output level of the differential amplifier 40q,
The output level of the differential amplifier 40q is
Clamped to the output level of the dynamic amplifier 41q,
Signal of the boosted level is the inverting input terminal of the differential amplifier 42q.
Is supplied to. Then, the heater is heated and the glass temperature rises.
And the resistance of the thermistor becomes high and the differential amplifier 40q
Since the voltage supplied to the transfer input terminal also increases, the difference will eventually occur.
The output level of the dynamic amplifier 40q decreases. And the difference
The output level of the dynamic amplifier 40q is the output level of the differential amplifier 41q.
When smaller, diode 41t is reverse biased
Is supplied to the inverting input terminal of the differential amplifier 42q.
Signal is dominated only by the output signal of the differential amplifier 40q,
Control is performed so that the lath temperature becomes the equilibrium temperature. The AC waveform output from the thyristor circuit 7 is the transformer 17
Is supplied to the heater and the voltage is changed from the standard of the heater 22 to the required voltage.
It is converted and supplied to the heater 22. Intermittent with a transformer
When performing continuous energization time control, intermittent time
If it is short, before the electromagnetic energy in the transformer is consumed
The next energization will be started, so energize again
Do not supply current of opposite polarity when
Then, the magnetic flux in the iron core is saturated and a large current flows. This
Therefore, the AC waveform as shown in Fig. 10 (a) is supplied.
(B) when intermittently controlling this AC waveform.
As shown in the figure, the power supply is restarted even though the power supply has ended with the positive half wave.
In this case, it is necessary to restart the power supply from the negative half wave. This
In order to realize that, this device is shown in Fig. 10 (c).
As shown in the figure, when the waveform shown in (a) changes to the negative direction,
A gate signal, and the gate signal has a negative AC waveform.
Change to a positive half-wave and stop before the half-wave ends.
The positive half-wave changed to a negative half-wave.
When choosing to ensure that the thyristor is turned off.
It FIG. 10 shows the configuration as described above.
Thus, energization control is performed with one cycle of the AC waveform as the minimum unit.
The power is turned on as shown in Fig. 9 (e).
It gradually becomes longer than the time of charging, as shown in Fig. 10 (b).
At the time when the zero level is crossed in the polarity direction with the AC waveform
Energization is started from and the AC waveform has the same polarity as when energization started.
The power supply is stopped at the time when it crosses the zero level in the direction. So
Then, continue this control until the glass temperature reaches the specified temperature.
To be Also, after reaching the specified temperature, there is no change in the outside air temperature.
When the glass temperature changes, the glass temperature is brought to the specified temperature.
Control for returning is performed. In the above description, the electric power supplied to the heater 22 is one type.
Came. However, at the top of the problem to be solved by the invention,
As solid, it controls the power required for the heater 22 on and off.
Is a problem in that the temperature of the heater 22 is controlled.
But the power supply of the aircraft is not very stable
, The voltage inside the machine drops when the heater is energized.
Fluorescent lamps used for lighting have flickering, which may be inconvenient for passengers.
Give a pleasant feeling. Therefore, in the present invention, first, low power is used.
Conduct the energization time control, gradually increase the energization period,
When the entire period is energized (this state has been explained so far)
The power of this value and its power
So that even greater power is alternately supplied to the heater
ing. At that time, the energization time of the one with larger power is short at first
Gradually increase over time. And power
Contrary to this, the small energization time of
I try to keep it short. Next, such control operation will be described. I mentioned above
As shown in FIG. 4, the timers 52a and 52b have a terminal CD of "1" level.
Output at the falling edge of the signal supplied to terminal B when
Q starts to shift from "0" level to "1" level
And a signal of "0" level is supplied to the terminal CD
Is reset and the terminal Q is unconditionally set to the "0" level.
It is configured as Therefore, is it the terminal Q of the timer 52b?
When a "1" level signal is generated from the timer 52a
Generates a signal of "0" level from the terminal Q. That
The timer 52b outputs the signal shown in FIG. 9 (e).
During this period, a signal of "1" level is sent from the terminal Q.
Therefore, the timer 52a outputs the signal shown in FIG. 9 (e).
The signal of "1" level is output from the terminal Q during the non-use period.
It is becoming like this. As shown in FIG. 4, the output of the timer 52a is the gate signal generation.
The output of the timer 52b is supplied to the NAND circuit 60a of the circuit 6.
Gate through flip-flop 57 and AND circuit 58
The sixth signal supplied to the NAND circuit 60b of the signal generation circuit 6
As can be seen, NAND circuit 60a and NAND circuit 6
0b constitutes an oscillation circuit (oscillation frequency of about 20 KHz),
These oscillator circuits have "1" level for each NAND circuit.
It oscillates during the period that the signal is supplied. As mentioned above
As described above, the timers 52a and 52b alternately generate outputs.
Therefore, oscillation circuits including NAND circuits 60a and 60b are output alternately.
Is occurring. These oscillation outputs are
It is supplied to the paths 65h and 65l, but other inputs of these AND circuits
The force is applied to the flip-flop 66 through the AND circuits 65f and 65g.
Output from a, 66b is supplied. This flip
The stop circuit is a buffer 64 from the heater energization time control circuit 4.
a, the signal supplied through the AND circuit 65a, and the buffer
64b, controlled by a signal supplied via the AND circuit 65c
It is being controlled. Heater energization time control circuit 4
While the output level of the operational amplifier 40q shown in Fig. 3 is low,
Outputs "1" level signal to both differential amplifiers 44q and 44r
However, the output of the differential amplifier 40q rises to a certain level.
The differential amplifier 44q sends out a "0" level signal.
Become so. And the output level of the differential amplifier 40q is
When it gets higher, both differential amplifiers 44q and 44r are at "0" level.
Signal will be sent. The output of the differential amplifier 40q is
Its input, ie the thermistor connected to terminal 4a
The thermistor 24 changes according to the resistance value of the
The window glass of the cockpit in which the heater 22 shown in FIG.
It changes with the temperature. The temperature of the window glass is the current flowing through the heater 22.
The temperature of the heater rises above the first temperature.
Then, the differential amplifier 44q outputs a signal of “0” level to the heater.
Temperature difference above the second temperature, which is higher than the first temperature
The dynamic amplifier 44r also outputs a "0" level signal.
Set a constant. For this reason, when the power is turned on, the third minute timer 41 operates so that the third
The differential amplifiers 44p and 44r in the figure both send out "1" level signals.
The gate signal generator circuit 6 of FIG.
It is added to one terminal of the drive circuits 65a and 65c. AND circuit 6
The other terminals of 5a and 65c are the outputs generated by the zero level detection circuit.
Since the pulse is supplied,
AND circuits 65a and 65c send out a signal of "1" level,
The flip-flops 66a and 66b are set and their Q output
Are both "1" level. Q of flip-flops 66a, 66b of the gate signal generation circuit 6
Outputs are diodes 43u and 43t of heater energization time control circuit 4
Since it is being supplied to the
A voltage is generated at the resistor 43i shown in. This voltage is differential
Since it is supplied to the inverting input terminal of the amplifier 43r, the differential
The amplifier 43r is the level of the triangular wave supplied to the non-inverting input terminal.
The triangular wave shown in Figure 11
To do. Note that the characteristic d in FIG.
Is the output. Therefore, as the level of the 3-minute timer 41 changes, FIG.
As described in (e), the inverter of the gate signal generation circuit 6
The duration of the 20KHZ signal output from the data 63b gradually increases.
Become Then, from the AND circuit 65e in FIG.
Since the bell signal is being sent, the inverter 63b
The output 20KHz signal is AND circuit 65i, diode 67
b, supplied to the transistor 68a via the resistor 61f,
Since the transistor is turned on and off at about 20 KHz, the gate
The signal is supplied to the thyristors 70a and 70b,
Turns on. As a result, the heater 2 is passed through the transformer 17.
Heater current is supplied to 2. The output of the inverter 63b is the ninth
As shown in Fig. (E), as time elapses,
Therefore, the duration is getting longer. This and
When the thyristor turns on and off, the number of turns on the primary side of the transformer 17 is
Since it is connected to the largest tap, the heater current is
Corresponding to the current of 7 amps and the current is shown in Fig. 12 (a).
Flow as shown in. The period during which the heater current of 7 amps flows is gradually reduced as described above.
It becomes longer and eventually becomes a continuous energized state, so
The temperature eventually reaches the above-mentioned first temperature. As a result,
The output level of the differential amplifier 40q shown in FIG.
It has risen to the corresponding value, and this is
"1" level signal detected by 44q and sent up to now
To "0" level. This signal is the flit shown in FIG.
It is supplied to the reset terminal R of
Reset the flap. For this reason, until now "1"
The AND circuit 65e that was sending the level met the AND condition.
And the AND circuit 65f meets the AND condition instead.
Since it stands, a signal of "1" level is sent from there,
It is supplied to the AND circuits 65j and 65h. On the other hand, the output of flip-flop 66a changes from "1" level.
By changing to the “0” level, as described above, the triangle
The wave level is shifted as shown in waveform b in FIG.
It Therefore, the output waveform of the inverter 63b shown in FIG.
Again outputs an intermittent signal as shown in Fig. 9 (d).
It On the other hand, the timers 52a and 52b shown in FIG.
As such, when one is producing output, the other is delivering output.
Since it is designed so that it does not come out, the inverter 63 in FIG.
a indicates that the signal shown in (e) has stopped as shown in Fig. 9 (f).
During this period, a signal of about 20 KHz is sent out.
Contrary to (e), issue (f) has a shorter duration over time.
I will continue to connect. Therefore, AND circuit 65h and AND circuit
A 20 KHz signal is output alternately from the path 65j, which causes
The set of thyristors 70a and 70b and the set of thyristors 70c and 70d intersect.
Turn on each other. Then, the set of thyristors 70c and 70d is connected.
What is being continued is the one with few primary windings of the transformer 17.
Therefore, the current supplied to the heater 22 is also
Is larger, the current is 14 amps
Flow into the heater 22. At this time,
The duration of the signal sent from the connection circuit 65j
The signal sent from the AND circuit 65h becomes longer with
On the contrary, the duration of the trust becomes shorter with the passage of time.
The current flowing through the heater 22 is as shown in Fig. 12 (b).
become. In Fig. 12 (b), the small amplitude part is 7
1 for the part where the current of the pair is flowing and the part where the amplitude is large.
This is the part where a current of 4 amps is flowing. When the current flowing through the heater increases in this way, window glass
As the temperature reaches the second temperature, it is shown in Fig. 6 this time.
The flip-flop 66b is also reset. For this reason,
In the AND circuit 65f, the AND condition is not satisfied,
Since the AND condition is satisfied in the AND circuit 65g, this time
The 20KHz signals are alternately sent from the hand circuits 65k and 65l,
The set of thyristors 70c and 70d and the set of thyristors 70e and 70f alternate
Turn on. These are connected to the heater 22 via the transformer 17.
Alternately supplying 14 amps and 20 amps
Therefore, the current flowing through the heater 22 is shown in Fig. 12 (c).
Will come to you. In (c), the part with large amplitude is 20
The amperes and the small amplitude part are 14 amps. In this way, even if phase control is performed for each wavelength
If the current to be changed is changed stepwise, the voltage fluctuation will decrease,
Flickering of the in-flight fluorescent lights also disappears. However, the amplitude of the current can be changed by switching between the two circuits and
The phases must be continuous, but the phase
Matching is likely to occur, at which time an excessive current flows,
The voltage fluctuation rate becomes large. So this inconsistency
In order to eliminate the noise, the flip-flop 57 shown in FIG.
Anti-50d, 50e, Capacitor 51d, Diodes 55a, 55b, Anne
A switching circuit 58 is provided, and when the current value is switched, the AC waveform
The first one cycle is followed by a slight delay.
That is, when the flip-flop 57 is set,
Output time is determined by resistor 50d and capacitor 51d
Is output after a delay, and the AND circuit 5
8 becomes active after a delay. In addition, a timer 52c is provided to
The waveform duration before switching at the time of shape switching is slightly longer.
is doing. This causes overcurrent due to phase mismatch.
It can be prevented. No delay is required after the first cycle.
Therefore, the diode 55b does not cause a delay after two cycles.
I am trying to stay. The AC waveform output from the thyristor circuit is sent to the transformer 17.
Is converted to the required voltage from the standard of the heater 22
However, at this time, a part of the winding of the transformer 17 is detected by the heater current.
It is configured by the transformer 91 of the circuit 9. For this reason,
The current supplied to the heater 22 is picked up by the transformer 91.
The differential amplifier 9 is up-converted and rectified by the diode 92.
3 is supplied to the inverting input terminal. The differential amplifier 93 is described above.
As described above, the maximum specification value of the heater current is applied to the non-inverting input terminal.
Causes a voltage slightly higher than the voltage supplied to the inverting input terminal.
Since the quasi voltage 94 is supplied, it is possible to
When the data current flows, it outputs the output signal of "0" level.
It On the other hand, the output terminal 5e of the gate signal generation time control circuit 5
Is most of the period when the heater current is supplied
Since the signal of "1" level is sent by
And the "0" level output from the heater current detection circuit 9
When both signals of
The lister is turned on and the heater current is supplied to the heater 22.
The operation display lamp 23 for indicating that the power is being supplied.
Turn on the light. The current output from the thyristor circuit 7 is the same as that of the transformer 17.
Input terminal 8 of the overcurrent detection circuit 8
is supplied to a. This current is applied to the resistor 81a shown in FIG.
Flow into the thyristor as shown in Fig. 13 (a).
Generated AC voltage of the magnitude corresponding to the current value
Is applied to the inverting input terminal of the differential amplifier 84 shown in FIG.
Supplied. V / 2 is applied to the non-inverting input terminal of the differential amplifier 84.
If a bias is supplied, its output is shown in FIG. 13b.
The signal shown in is output. However, if this is the case,
The input waveform must be rectified. However, differential
When the standard of the amplifier 84 is examined in detail, the input is negative.
Those whose operation is guaranteed for 0.3 V or more (Example:
For example, there is LM2904). Therefore, ground the non-inverting input terminal
In the circuit of Fig. 7, input and minus 0.6V
Applying a signal with amplitude up to degrees to the inverting input terminal,
A positive half-wave waveform with amplitude V is output. Sanawa
The differential amplifier 84 simultaneously performs rectification and amplification.
It has been done. The output of the differential amplifier 84 is smoothed by the resistor 81c and the capacitor 82b.
And its smoothed output is a reference voltage determined by resistors 81f and 81e.
When the pressure becomes larger than the pressure, the differential amplifier 85 becomes "1" level.
Send the output signal. This "1" level signal is output
It is output through the child 8b, 8c and the gate signal generation circuit 6 is input.
Input terminal of sensor terminal 6g and sensor short detection circuit 10
Supplied on 10f. Therefore, the gate signal generation circuit 6 is
Transistor 68d shown in Fig. 6 turns on,
Since the bases of the stars 68a to 68c are set to the ground potential,
The gate signal is no longer supplied to the thyristor,
Turn off. On the other hand, the sensor short detection circuit 10
The relay 16 is activated by the signal supplied to the input terminal 10f.
It is deenergized and the contact 16a is opened. If the window glass becomes overheated for some reason, the
The resistance value of the mister 24 increases. This thermistor is
The current is supplied from the data energization time control circuit 4
Then, when the window glass overheats, the overheat detection circuit 11
Voltage supplied to the inverting input terminal of the differential amplifier 11a at
The pressure increases. If this voltage exceeds the reference voltage 11b, the difference
The dynamic amplifier 11a generates a "1" level output signal,
The signal is inverted by the inverter 11c and detected by the sensor short
Since it is supplied to the inverter 10d of the path 10, also at this time
Relay 16 is de-energized. In addition, the thermistor 24 is short due to some cause.
And the differential amplifier 10a of the sensor short detection circuit 10
The voltage supplied to the inverting input terminal is the non-inverting input terminal
Since it is smaller than the supplied voltage, the differential amplifier 10a
It outputs a "1" level signal. Therefore, the relay 16
Is de-energized and power is not supplied to the thyristor circuit 7.
It Window glass is overheated by overcurrent and
It also overheats when exposed to warm outside air, so at this time as well
The overcurrent detection circuit detects overheating of the window glass.
Will end up. Therefore, in the present invention, the heater current is
Detected, when the heater current is normal value,
The rent detection circuit does not work. Sanawa
The heater current detection circuit 9 shown in FIG.
When it is normal, a "1" level signal is transmitted. This
Therefore, the differential amplifier 11a of the overheat detection circuit 11
This "1" level signal is supplied to the inverting input terminal of
There is. On the other hand, the non-inverting input terminal of this differential amplifier circuit is
-The generated current is supplied to the mister 24. Only
However, the voltage at the non-inverting input terminal corresponds to the temperature of the window glass.
Voltage that changes and is lower than the power supply voltage, but inverted
Since the voltage at the input terminal is at "1" level, almost no
It is at the same level as the source voltage. From this, the heater current
Is greater than the normal value, the differential amplifier 11a
The voltage at the output terminal is always higher and its output is "0" level.
It's a bell. Therefore, the inverter 11c
Outputs "1" level signal to detect sensor short
Affects the operation of the path 10, ie the operation of the relay 16
Absent. However, when an overcurrent flows through the heater 22, the heater current
Since the detection circuit 9 sends out a signal of "0" level,
The output signal should not affect the overheat detection circuit 11.
Yes. Therefore, the overheat detection circuit 11 does not
The voltage generated in the controller 24 is detected and the value is above the allowable value.
When the sensor short detection circuit 10 is driven,
Therefore, the relay 16 is de-energized, so that the heater 22 is supplied with current.
Is not supplied and prevents the window glass from overheating. Well
At this time, the AND condition of the AND circuit 15 is also satisfied.
Therefore, the lamp 23 lights up, and the alarm is notified.
It When the switch 20 is powered on, the timer 41 shown in FIG.
Create a state where the output voltage of (3 minutes timer) is saturated
As the power is turned on, the temperature of the thermistor 24
You can check the control status. The switch 21 artificially creates an overheated state.
It is a switch. In the above example, the heater of the glass window in the cockpit was overheated.
Although the circuit has been described, it is not limited to this, and AC power can be applied.
Those which perform variable control can be generally used. [Advantages of the Invention] As described above, the present invention is based on the selected power supply circuit.
The power sent from the road is delayed by a predetermined time from the time of switching.
Since it is the one that caused the overcurrent,
It has an effect that it can be reduced.

[Brief description of drawings]

FIG. 1 is a block diagram showing an embodiment of the present invention, FIG. 2 is a circuit diagram showing a zero level detection circuit, FIG. 3 is a circuit diagram showing a heater energization time control circuit, and FIG. 4 is a gate signal generation time. FIG. 5 is a circuit diagram showing a control circuit, FIG. 5 is a circuit diagram showing a thyristor circuit, FIG. 6 is a circuit diagram showing a gate signal generating circuit, FIG. 7 is a circuit diagram showing an overcurrent detection circuit, and FIG. 8 is a gate. Waveform diagrams of respective parts for explaining a signal generation state, FIG. 9 is a waveform diagram of respective parts for explaining an operating state of the thyristor when the power is turned on, and FIG. 10 is a waveform diagram showing an energized state of the thyristor, 11 FIG. 12 is a graph showing the generation state of a comparison signal with the temperature of the controlled object, FIG. 12 is a graph showing the waveform of the current supplied to the heater,
FIG. 13 is a waveform diagram for explaining the operation of the overcurrent detection circuit. 3 ... Zero level detection circuit, 4 ... Heater energization time control circuit, 5 ... Gate signal generation time control circuit, 6 ... Gate signal generation circuit, 7 ... Thyristor circuit , 8 ... Overcurrent detection circuit, 9
.... Heater current detection circuit, 10 ... Sensor short detection circuit, 11 ... Overheat detection circuit.

Claims (1)

[Claims] 1. An alternating-current power control device for supplying power to a load by alternately selecting power supply circuits having different maximum values,
An AC power control circuit comprising: a delay circuit that delays electric power sent from a selected electric power supply circuit for a predetermined time from a switching time point and outputs the delayed electric power.