# Amplifier negitive feedback capacitor and resistor in paralell

1. Jun 11, 2009

### chopficaro

ok so if we have an amplifier with direct negative feedback we have a voltage follower

if we put a resistor between that inverting input pin and ground, we can do these things:

if we put another resistor between the output and the inverting input we can adjust the gain

if, instead, we use a capacitor between the the output and the inverting input we can get an integrator

im looking at a schematic for a Mass Flow Controller right now that, instead of that resistor or that capacitor between the output and the inverting input, we have both a resistor and a capacitor, in parallel, between the output and the inverting input. what does that do and how can i measure what it does?

2. Jun 12, 2009

### Staff: Mentor

Welcome to the PF. What is the problem with having *just* a capacitor as the feedback? What would happen if there is an offset, and that keeps getting integrated?

3. Jun 12, 2009

### vk6kro

The capacitor has more effect at high frequencies.

Capacitors have less reactance as the frequency rises, so it is like having a smaller resistance there.

This means the amplifier would have less gain at high frequencies than at lower frequencies.

The setup you describe was not a voltage follower, though. It is an amplifier with reducing gain at higher frequencies.

The resistor sets a limit to the gain at low frequencies, otherwise the amplifier could become unstable.

Testing it is easy. Just apply a signal to the input. Measure the output.
Gain = Output / input.
Do this at different frequencies.

Last edited: Jun 12, 2009
4. Jun 12, 2009

### skeptic2

You might try plotting the phase and gain of the amplifier versus frequency using a polar plot. Obviously they are limiting the gain at higher frequencies. Can you find the roll off frequency? Why do you think they chose that frequency?

5. Jun 12, 2009

### Bob S

As you point out, a capacitor C from the output back to the inverting input is an integrator. If there is a resistor R1 in parallel with the capacitor, it is a leaky integrator with a decay time constant R1C. If you have an input resistor R2 to the inverting input (with the non inverting input tied to ground through a resistor R3**), then a short pulse of voltage V and time t will put charge Q = Vt/R2 on the capacitor, so the output voltage is Q/C, and the decay time constant is R1C.

**1/R3 = 1/R1 + 1/R2

6. Jun 12, 2009

### chopficaro

ah i think i see now. this amplifier is a small part of a mass flow controller. another amplifier in the schematic uses mechanical feedback to keep the valve open just enough to allow a certain amount of gas through that the user defines by setting a control voltage. this amplifier that we have been discussing is in the control voltage input network. i can see that if the user changed the control voltage to quickly, it could cause oscillations in the flow, rather than a steady desired flow value. i think that this capacitor is used to make sure that control voltage doesn't change too quickly. thank you all for helping me understand

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