# Simple Amplifier Design

Thread Starter

#### Elevon

Joined Jun 7, 2020
18
What exactly is the purpose of the resistor and the capacitor in series entering the op-amp's input terminal, and then in parallel after the op-amp when the signal is being fed back into the op-amp's input terminal?

#### Audioguru again

Joined Oct 21, 2019
3,317
You forgot to post the schematic for us to see if the opamp is inverting or is non-inverting.
I do not see how a resistor and capacitor can be in series and also be in parallel. The schematic will show how.

Thread Starter

#### Elevon

Joined Jun 7, 2020
18
Here is the schematic:

I am confused about how to calculate the current flowing back into the inverting input, beause the design requirement states, "Choose so that 500 µA flows through it when () is applied across it." I dont know how to interpret it because it says the current through Rf must be 500 microamps when a maximum voltage of 10V is applied, but surely if the supply is AC then the value won't be fixed at 500 microamps due to the changing reactance from the parallel capacitor, and if it is DC then the current wont be able to pass beyond the first capacitor C1?

#### Audioguru again

Joined Oct 21, 2019
3,317
The inputs of opamps draw extremely low currents that can be ignored in this circuit.

C1 and R1 cut low frequencies and Cf and Rf cut high frequencies. The capacitors do not pass DC.
The AC gain for middle frequencies is Rf/R1.

Since the + input is grounded then the negative feedback will cause the - input also to be grounded.
Then Rf= 10VAC/500uA=20k ohms for middle frequencies.

#### WBahn

Joined Mar 31, 2012
26,398
Here is the schematic:
View attachment 213744
I am confused about how to calculate the current flowing back into the inverting input, beause the design requirement states, "Choose so that 500 µA flows through it when () is applied across it." I dont know how to interpret it because it says the current through Rf must be 500 microamps when a maximum voltage of 10V is applied, but surely if the supply is AC then the value won't be fixed at 500 microamps due to the changing reactance from the parallel capacitor, and if it is DC then the current wont be able to pass beyond the first capacitor C1?
"Choose so that 500 µA flows through it when () is applied across it."

What is "it"? Are the two "it"s referring to the same thing, or two different things?

What does "when ( ) is applied" mean?

A common design approach for simple amplifiers is that the capacitors are sized so that, at the frequencies of interest, each either acts very much like a short or very much like an open. At DC, all capacitors look like open circuits and at very high frequencies all capacitors look like short circuits. So you really only have two possibilities here, either C1 transitions from an open to a short before Cf or the other way around. Sketch out the amplifier gain for both possibilities and see which one makes the most sense and go with that.

Thread Starter

#### Elevon

Joined Jun 7, 2020
18
"Choose so that 500 µA flows through it when () is applied across it."

What is "it"? Are the two "it"s referring to the same thing, or two different things?

What does "when ( ) is applied" mean?

A common design approach for simple amplifiers is that the capacitors are sized so that, at the frequencies of interest, each either acts very much like a short or very much like an open. At DC, all capacitors look like open circuits and at very high frequencies all capacitors look like short circuits. So you really only have two possibilities here, either C1 transitions from an open to a short before Cf or the other way around. Sketch out the amplifier gain for both possibilities and see which one makes the most sense and go with that.
Apologies, the orignal text said "Choose the feedback resistor so that 500 microamps flows through it when a voltage equivalent to the maximum output voltage (10V in this case) is applied across it. The answer you provided explains nicely to me how it works, but I am yet to find a method of calculating the correct resistance and capacitance values for a current of 500uA. I suppose I can work them out using a bode plot and some nodal analysis.

#### Audioguru again

Joined Oct 21, 2019
3,317
The Rf resistor is 20k since it has 500uA in it and 10V across it. The capacitors affect the low and high frequency response of the opamp.
If Cf has a high capacitance then you would never get 500uA in Rf or 10V across it.

#### MrAl

Joined Jun 17, 2014
8,372
Apologies, the orignal text said "Choose the feedback resistor so that 500 microamps flows through it when a voltage equivalent to the maximum output voltage (10V in this case) is applied across it. The answer you provided explains nicely to me how it works, but I am yet to find a method of calculating the correct resistance and capacitance values for a current of 500uA. I suppose I can work them out using a bode plot and some nodal analysis.
Resistor current does not change with frequency therefore a fixed voltage of any frequency causes the same AC current.
That, and considering the virtual ground, you should be able to figure it out.

#### atferrari

Joined Jan 6, 2004
4,386
@WBahn @Audioguru again

I am puzzled; why a differentiator is discussed as amplifier?

Could you enlight the audience?

#### Audioguru again

Joined Oct 21, 2019
3,317
@WBahn @Audioguru again
I am puzzled; why a differentiator is discussed as amplifier?
It is not a differentiator that has no input resistor. The input RC acts like a simple highpass filter.

#### atferrari

Joined Jan 6, 2004
4,386
It is not a differentiator that has no input resistor. The input RC acts like a simple highpass filter.
Sorry AG, the double negative made this even more puzzling to me. At lost here.

#### Audioguru again

Joined Oct 21, 2019
3,317
This circuit in this thread has an input resistor in series with a capacitor feeding the virtual ground input of the inverting opamp. Then it is a highpass filter that cuts low frequencies but passes high frequencies. The opamp has a determined gain of Rf/Ri at high frequencies. The cutoff frequency is determined by the input and its series capacitor values.

A differentiator has no input resistor so its input capacitor passes only extremely high frequencies, It almost has infinite gain at an infinite frequency.

#### WBahn

Joined Mar 31, 2012
26,398
@WBahn @Audioguru again

I am puzzled; why a differentiator is discussed as amplifier?

Could you enlight the audience?
This is not really a differentiator, it's a high pass filter so the degree to which it functionally acts as a differentiator depends on the component values and the spectral content of the signal. Regardless, amplifier is a pretty generic term that can be applied to just about any device that transforms one signal into another.

#### atferrari

Joined Jan 6, 2004
4,386
This circuit in this thread has an input resistor in series with a capacitor feeding the virtual ground input of the inverting opamp. Then it is a highpass filter that cuts low frequencies but passes high frequencies. The opamp has a determined gain of Rf/Ri at high frequencies. The cutoff frequency is determined by the input and its series capacitor values.

A differentiator has no input resistor so its input capacitor passes only extremely high frequencies, It almost has infinite gain at an infinite frequency.
So, what about what I found in Wikipedia?

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#### MrAl

Joined Jun 17, 2014
8,372
This circuit in this thread has an input resistor in series with a capacitor feeding the virtual ground input of the inverting opamp. Then it is a highpass filter that cuts low frequencies but passes high frequencies. The opamp has a determined gain of Rf/Ri at high frequencies. The cutoff frequency is determined by the input and its series capacitor values.

A differentiator has no input resistor so its input capacitor passes only extremely high frequencies, It almost has infinite gain at an infinite frequency.
A differentiator does often have an input resistor in series with the cap. That is to keep noise down somewhat.
If we had a resistor that is somewhat larger then it acts more like a traditional high pass, but as that resistor value comes down the circuit becomes more and more like a true differentiator.
This is easy to imagine. Say we start out with a 1000 Ohm resistor. That is probably more like a traditional high pass than a true differentiator, although ti depends on the cap value and the application too. As we come down to 100 Ohms, then 10 Ohms, it becomes more like a true differentiator. As we get down to 1 Ohm, it is very much mroe a differentiator than a high pass, and as we come down to 0.1 Ohms the resistance starts to approach the ESR of the cap so we get hard pressed to call it anything other than a differentiator even though it has a series resistor with it.

Of course no matter what the resistor is, the real test is the application. Did the designer want a traditional high pass function or did he want a differentiator with attention to noise and less high frequency response to help reject out of band signals. So with a casual observation it gets hard to tell the difference between the two.

#### MrAl

Joined Jun 17, 2014
8,372
So, what about what I found in Wikipedia?
Hi,

I think some of that image got cut off.

#### atferrari

Joined Jan 6, 2004
4,386
Last edited:

#### atferrari

Joined Jan 6, 2004
4,386
Regardless, amplifier is a pretty generic term that can be applied to just about any device that transforms one signal into another.
Surprising claim (at least for me).

#### WBahn

Joined Mar 31, 2012
26,398
Surprising claim (at least for me).
I probably should have caveated it with saying that it almost certainly wouldn't be considered an amplifier if the output power doesn't come from something other than the input signal.

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