The difference between charge amplifier and transimpedance amplifier in practical application

zhaoyanhao

The difference between charge amplifier and transimpedance amplifier in practical application [Copy link]

We all know in theory that op amps can construct different operational amplifier circuits, such as charge amplifiers, where capacitors are in the feedback loop of op amps, and transimpedance amplifiers, where resistors are in the feedback loop. However, in practical applications, in order to compensate, discharge charge, filter, or improve loop stability, resistors will be added to the feedback loop of charge amplifiers, and small capacitors will be added to the feedback loop of transimpedance amplifiers. This makes the actual form of the circuits consistent. So the question is, if the circuit form is consistent, what is the essential difference between the circuits? Six years ago, when I went to an interview, I told the other party that a charge amplifier is actually an integrator. The other party's eyes were white to the sky, and he kept making sarcastic laughs, which made me doubt my life. There are differences in name and essence, and they can't be understood together at all? I have always believed that the ratio of resistors and capacitors in the feedback loop and their respective magnitudes make it clear what kind of amplifier it is. I wonder if the teachers can give me some advice and answer my doubts for many years. Thank you for your advice and discussion!

zhaoyanhao

Manually summon teacher @gmchen @maychang

gmchen

The landlord's statement makes sense.

At least, the charge amplifier and the integrator are consistent in feedback circuit form. The interviewer who laughed at you may not have made a circuit himself.

As for charge amplifiers and transimpedance amplifiers, it depends on what the main components in the feedback loop are.

The transimpedance amplifier is a resistor feedback. For stability, a small capacitor can be connected in parallel with the resistor, but the capacitive reactance of this capacitor at the operating frequency will be much higher than the feedback resistor, so the main feedback element is the resistor.

The feedback element of the charge amplifier is a capacitor, and the charge is integrated across this capacitor to give the output voltage, which is (this is what makes it similar to an integrator: the output of an integrator is ),

However, in order to discharge the bias current at the input of the op amp, it is possible to connect a large resistor in parallel with the capacitor. The resistance of this resistor only needs to be sufficient to discharge the bias current. If the resistance is too small, it will not meet the integration requirements.

zhaoyanhao

gmchen posted on 2024-7-16 12:44 What the OP said makes sense. At least, the charge amplifier and the integrator have the same feedback circuit form. The interviewer who laughed at you may...

Thank you for your answer. I'm sorry that I have been a bit busy recently and have not replied to you in time.

Let me give you another example. We know that charge is the integral of current. Many detectors are actually current sources in essence, but many matched preamplifiers are charge amplifiers. I am confused. Does the charge amplifier really put out charge, or is it a current signal? For example, the detector outputs a pulse. This pulse may be a function of time on the horizontal axis and current on the vertical axis. So after passing through the charge amplifier, it should be the integral of the current. However, the output from my actual use is still a pulse, which is converted into voltage. So does the output voltage at time t correspond to the charge or the current at that moment? This may be a bit abstract and difficult to express. Some detectors say that they output charge signals. Then the charge is actually the total amount of current over a period of time, which is the result of integration. How can it correspond to time t? The essence of the problem may be how the detector outputs the charge signal, or how to choose the matched preamplifier. Why don’t they use transimpedance amplification but charge amplification when some are current sources? I have worked for many years, but I am still confused about this issue.

zhaoyanhao

gmchen posted on 2024-7-16 12:44 The original poster's statement makes sense. At least, the charge amplifier and the integrator are consistent in the feedback circuit form. The interviewer who laughed at you may...

Or can we understand it this way? The essence of charge amplifier is an integration circuit. So if the input current is a pulse, first increasing and then decreasing to 0, the charge amplifier output is always a charging slope. The reason why it is also a pulse is that the falling edge is actually the result of RF discharge. If the falling edge is not considered, the first half is a cumulative process, just like acceleration and speed. The current input is acceleration, and the charge amplifier output is speed. Even if the acceleration is a little bit, the speed is increasing. When the acceleration (current) reaches 0, the speed (output) also reaches the maximum. If you want to measure the actual current, you only need to measure the maximum output value. As for the falling edge of the output caused by Rf discharge, it is not considered.

I don't know if this understanding is correct, but it still doesn't solve the problem of why transimpedance is not used, or when to use charge amplification and when to use transimpedance amplification.

gmchen

The signal detected by the charge amplifier appears in the form of pulses, and we need to get the average value after accumulating it over a period of time (which can be sampled by ADC later), so we use a charge amplifier. If we use a transimpedance amplifier, we will get a pulse voltage signal, which cannot be processed later.

zhaoyanhao

gmchen published on 2024-7-19 12:06 The signal detected by the charge amplifier appears in the form of pulses, and it is necessary to obtain the average value after it accumulates over a period of time (which can be sampled by ADC later), ...

Teacher, since the signal detected by the charge amplifier appears in the form of pulses, let me ask the question in another way. That is, if I want to infer the total input charge through the output of the charge amplifier, should I just take the maximum value of the pulse, or should I add up the values of each time point on the output pulse to infer the total input charge? Since the charge amplifier has RF, the output must also be pulses, right?

maychang

zhaoyanhao posted on 2024-7-19 15:40 Teacher, since the signal detected by the charge amplifier appears in the form of pulses, let me change the way of asking the question. That is, if I want to use the charge amplifier...

The signal detected by the charge amplifier appears in the form of a pulse. Before the pulse arrives, the voltage across the integrating capacitor is zero. After the pulse arrives, the circuit starts integrating and the output voltage starts to rise. After the pulse, the circuit stops integrating and the output remains unchanged. As long as you measure this unchanged output voltage, you can deduce the input charge.

zhaoyanhao

maychang posted on 2024-7-19 16:29 The signal detected by the charge amplifier appears in the form of a pulse. Before the pulse arrives, the voltage across the integrating capacitor is zero. After the pulse arrives, the circuit begins to integrate...

"After the pulse, the circuit stops integrating and the output remains unchanged." In this case, it is not a charge amplifier, but a pure integrator. Because of the existence of RF, the output of the charge amplifier must also be a pulse. I have made a similar circuit as you said. Pulses come one by one, and the integrator outputs a step signal. The feedback loop adds a controllable switch to discharge when the output is saturated. However, it is impossible to have no leakage current in actual engineering. Even if a very small integral capacitor is used, if the input signal is weak, the input cannot offset the leakage, and the ideal voltage value that can be collected cannot be achieved in the end, and all is leaked.

maychang

zhaoyanhao posted on 2024-7-19 17:14 "After the pulse, the circuit stops integrating and the output remains unchanged." Teacher, in this case, it is not a charge amplifier, it is a pure...

[I have made a similar circuit as you mentioned. When pulses come one by one, the integrator outputs a step signal.]

Doesn’t the output step signal indicate that “after the pulse, the circuit stops integrating and the output remains unchanged”?

Indeed, it is impossible to have no leakage in actual engineering. When the input signal is weak, it cannot reach the voltage that can be collected, and it is all leaked. We can only choose an op amp with the smallest bias current and offset as possible, and a capacitor with the smallest leakage as possible to solve it.