Please help with some questions about AGC

真正的小菜鸟

Please help with some questions about AGC [Copy link]

 

The circuit diagram is as above, which is used to achieve automatic gain. However, the simulation is not satisfactory. When the input signal amplitude changes, the output will also fluctuate greatly and cannot be stabilized within a certain range. I would like to ask how to modify it. This is the circuit diagram given by the teacher, but the overall principle is a bit difficult to understand. I hope the master can explain it a little. Thank you!

gmchen

There are two main problems with the AGC circuit of the original poster. The first problem is that the AGC control voltage is improperly set. Usually, the AGC amplifier circuit should not start control when the input signal is very small, so that the amplifier has a sufficiently large gain. When the input signal becomes larger and the output voltage reaches a certain threshold, the AGC circuit starts to start control. At this time, the gain of the amplifier begins to decrease, so that the output is maintained within a smaller range of variation. However, the original poster's circuit does not set a suitable threshold. The precision full-wave rectifier circuit will have a DC output regardless of the size of the input signal. After amplification and feedback to the multiplier, the input signal is attenuated. Therefore, the control threshold of this circuit is actually 0 level, which is inappropriate. The second problem is that the time constant is improperly set. In order to achieve the purpose of AGC, it is usually hoped that the AGC circuit will not work when the input signal changes in normal signal amplitude (such as amplitude modulation signal). Only when the carrier amplitude changes slowly due to some reason (such as fading, distance change, etc.) will the AGC function be used to maintain the output of the amplifier basically unchanged. Therefore, the time constant of AGC is usually much larger than the period of the modulation signal in the normal signal. The time constant determined by R6C3 in the original poster's circuit is only about 10 microseconds, which is obviously too small.

gmchen

The diagram of the teacher provided by the host uses a comparator as the starting threshold control, and the level of its in-phase end is 4mV. However, since a silicon diode is used in the previous rectifier link, the actual starting threshold is basically determined by the conduction threshold of the silicon diode, which is about 0.5~0.7V. The time constant of this circuit is (R2||R3)*C2, which is about 80 milliseconds, which is quite suitable. However, it should be pointed out that there are some inappropriate aspects of this circuit: First, it is not appropriate to use 4007 as the rectifier diode. The frequency response of that diode is very poor and cannot be used for high frequencies. It is recommended to use 1n4148 instead. Second, it is not appropriate to use a comparator as AGC control, because the output of the comparator has only two states, high and low, and it will actually cause low-frequency oscillation when the circuit starts.

真正的小菜鸟

gmchen posted on 2018-4-13 16:12 The teacher's diagram given by the OP uses a comparator as the starting threshold control, and the level of its in-phase end is 4mV, but due to the previous rectification link...

Thank you very much for your reply

真正的小菜鸟

gmchen posted on 2018-4-13 16:12 The teacher's diagram given by the OP uses a comparator as the starting threshold control, and the level of its in-phase end is 4mV, but due to the previous rectification link...

Sorry to bother you, "So the actual starting threshold is basically determined by the conduction threshold of this silicon diode, which is about 0.5~0.7V." Can you elaborate on this sentence? Doesn't R2 act as a pull-up resistor? Thank you

gmchen

When the output voltage is lower than the conduction threshold of the diode, which is about 0.5 to 0.7 volts, the diode will not conduct, so no AGC control voltage will be generated and the amplifier will not enter the controlled state. So the control threshold of this circuit is determined by the diode

gmchen

Your second sentence involves R2. I guess you want to ask why it is included in the calculation of the time constant. R2 is indeed a pull-up resistor, but it is also part of the RC charging and discharging circuit, so it must be included in the calculation of the time constant.

真正的小菜鸟

gmchen posted on 2018-4-17 07:39 Your second sentence involves R2. I guess you want to ask why it is included in the calculation of the time constant. R2 is indeed a pull-up resistor, but it is also an RC charge and discharge circuit...

Thank you for your answer. I will think about it again.

真正的小菜鸟

gmchen posted on 2018-4-17 07:39 Your second sentence involves R2. I guess you want to ask why it is included in the calculation of the time constant. R2 is indeed a pull-up resistor, but it is also an RC charge and discharge circuit...

Sorry to bother you. I would like to ask about the output of the comparator. Shouldn't it be close to the power supply voltage? But the simulation shows only 130mV. What is the role of the RC circuit behind it? Thank you

gmchen

A real rookie posted on 2018-4-24 21:41 Sorry to bother you. I would like to ask you about the output of the comparator. Shouldn’t it be close to the power supply voltage? But the simulation shows only 130mV. The rear R...

The actual working state of this circuit is as follows: when the negative input level of the comparator is lower than the positive input level, the comparator output is open, and the power supply charges C2 after being divided by R2R3, so the output voltage of the comparator will not immediately reach the positive power supply voltage. The output voltage of the comparator is fed back to the multiplier, so as this voltage (actually it is the control voltage of the AGC circuit) increases, the output voltage of the multiplier also increases. When the output voltage of the multiplier increases and the level of the negative input terminal of the comparator is higher than the positive input terminal, the comparator outputs a low level, which is close to the negative power supply (about -12V). However, due to the large capacity of C2 and the certain output resistance of the comparator, the output of the comparator will not drop to -12V immediately, but there is a short drop process. During this process, due to the drop in control voltage, the output of the multiplier will also drop, and finally a dynamic balance will be reached. This is the result of the AGC action. It is best to connect a resistor in series at the output end of the comparator in this circuit, so that when the comparator outputs a low level, the control voltage will not drop too fast, which can make the AGC action more stable.

gmchen

The simulation shows that 130mV is possible, because it is possible that the entire AGC circuit reaches dynamic equilibrium at this voltage. The specific value depends on the gain of the multiplier.

gmchen

As for the role of the RC circuit, the output of the comparator has been smoothed due to the role of RC in the previous answer. In fact, the feedback control of the AGC circuit cannot be too fast. Imagine that if the AGC reacts immediately when the output of the amplifier changes, then the output of the AGC amplifier will remain basically unchanged. In that case, the amplitude change information contained in the input signal (such as the amplitude modulated wave) will be obliterated. This situation is called anti-modulation, which should be avoided by the AGC circuit. Therefore, the time constant of this RC circuit should be much larger than the period of the lowest spectral component in the useful signal (such as the modulation signal of the amplitude modulated wave).

gmchen

Let me say a little more. I have replied to a post before saying that it is not appropriate to use a comparator. The main problem is: if the output resistance of the comparator is very small, then when the comparator outputs a low level, the control level (the level at the top of capacitor C2, which is the output level of the comparator in this circuit) will drop rapidly, which may cause a cliff-like drop in the output of the multiplier. Due to the existence of C1, this cliff-like drop will take a period of time to be reflected at the input of the comparator, allowing the system to return to dynamic balance. If the design is not appropriate, this situation will cause the system to oscillate. The previous reply suggested connecting a resistor in series at the output of the comparator in order to avoid this kind of cliff-like drop.

gmchen

The comparator in this circuit has an open collector output, so only a cliff-like drop will occur. If a push-pull output is used, very steep edges may appear in both the rising and falling directions, which may cause oscillation. Therefore, it is necessary to connect a resistor in series with the comparator output. A more common method is to use an op amp to build an integrator circuit or a low-pass filter.

真正的小菜鸟

gmchen posted on 2018-4-25 17:30 The comparator in this circuit is an open collector output, so only a cliff-like drop will occur. If a push-pull output is used, then in both the rising and falling directions...

Thank you for your patient reply

真正的小菜鸟

gmchen posted on 2018-4-25 17:30 The comparator in this circuit is an open collector output, so there will only be a cliff-like drop. If a push-pull output is used, there will be no drop in both the rising and falling directions...

I have to trouble you again, I'm really sorry I would like to ask how to quantitatively estimate the input and output of this circuit. I really don't know how to calculate it. Sorry to bother you, I really trouble you.

真正的小菜鸟

A real rookie posted on 2018-6-3 16:37 I am sorry to bother you again. I would like to ask how to quantitatively estimate the input and output of this circuit. I really don’t know...

Sorry for the trouble

gmchen

I don't know which circuit you want to analyze. There are two circuits in the first post, but both circuits have some problems. I have actually analyzed the circuits in several previous replies. I suggest you correct the problem first, and then determine what you want to ask. You can also search AGC on this forum. I once posted a post specifically about AGC circuits.

gmchen

A real rookie posted on 2018-6-3 16:37 Sorry for the trouble

In order to answer your question, I looked at the circuit diagram in the first post again. It seems that your circuit is slightly more reliable than your teacher's. In your circuit, the time constant problem still exists, but I misread the reference level last time. When I looked at it last time, my eyes were blurry and I mistook an arrow for grounding, so the previous answer about the reference level was wrong. The reference level in your circuit is 5V. The circuit analysis is actually very simple. The output is rectified by precision + amplifier and output to U4. This output voltage should be a quasi-DC, and the value is equal to the average value of U2 output. When this voltage is lower than the reference voltage, the output of differential amplifier U5 is a value close to the positive power supply voltage, so the gain of the multiplier is the largest. When the output voltage of U4 is lower than the reference value, the output of U5 is close to the negative power supply voltage, and the gain of the multiplier is close to 0. When the output voltage of U4 is near the reference voltage, U5 works in a linear amplification state, and the output voltage drops from close to the positive power supply to close to the negative power supply, so the gain of the multiplier has a decline process, which is the process of AGC control. Therefore, during the AGC control process, the output voltage should be maintained at an average value of about 5V, that is, the peak value is about 8V.

gmchen

A real rookie posted on 2018-6-3 16:37 Sorry for the trouble

You said in your first post that the AGC control is not ideal. How bad is it? Can you post the test data or something? That will improve the efficiency of the analysis.