Question over Coupling capacitors and Op-Amps

In summary: So even though the amplifier sees a changing DC voltage, it will only cancel out a fraction of that change due to the low reactance of the coupling caps.
  • #1
KrNx2Oh7
19
0
So this is the op amp I am looking at and I have a few questions, not entirely sure how to ask them, but I'll ask more as I find out more

http://postimage.org/image/i228xrkj7/
http://postimage.org/image/i228xrkj7/

So the capacitor C1 and C3 are used to block DC, but when you add these do you not
get voltage drops in the AC signal? Like when the op amp outputs a voltage does it not drop some voltage across the capacitor before it reaches Vout.

My second question is over C2. Apparently it is suppose to eliminate power-supply noise from reachin ghte op amps noninverting input. I don't see how this works. How does a capacitor in parallel with R4 stop noise, a non-DC signal?

Finally my last question, which might sound silly, but why do capacitors not end up in the Gain. Instead of Gain= -R2/R1, why don't you do the current equations and include -jwC in the transfer function to have capacitors show up in the gain.

silly questions, but I want to know why everything is in this configuration and why it works

thanks
 
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  • #2
The idea is to use coupling capacitors which are so large that even at the lowest frequency of interest their reactance is still comparatively much lower than other associated impedances in the circuit.
 
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  • #3
You haven't told us what frequency range the circuit is designed for, but the idea is the C2 provides a much smaller impedance from the op amp + input to ground than R3 and R4 in parallel.

You could get the same effect by reducing the values of R3 and R4, but then you would be wasting power because of the higher current flowing through them.

Similarly C1 will have a low impedance compared with R1, and C3 compared with RL, so for practical purposes they can be ignored when understanding how the circuit works.

Of course if you plotted the response of the circuit down to DC, the caps will affect the gain and phase shift of the amp at low frequencies.
 
  • #4
AlephZero said:
You haven't told us what frequency range the circuit is designed for, but the idea is the C2 provides a much smaller impedance from the op amp + input to ground than R3 and R4 in parallel.

You could get the same effect by reducing the values of R3 and R4, but then you would be wasting power because of the higher current flowing through them.

Similarly C1 will have a low impedance compared with R1, and C3 compared with RL, so for practical purposes they can be ignored when understanding how the circuit works.

Of course if you plotted the response of the circuit down to DC, the caps will affect the gain and phase shift of the amp at low frequencies.

Cool that was VERY helpful. Let me see if I understand everything.

C1 and C3 block the DC signal and only drop AC voltage slightly since the impedance is low for the capacitors at higher frequencies (Xc= 1/-jwC)

For C2 it lowers the voltage seen at V+ because it's impedance is also small. And +V is a DC voltage which I assuming is superimposed with a small noisy AC signal. So for DC the capacitor does nothing, but for the noisy AC it reduces what is seen in the V+.

For gain I'm not sure about what you said when u mentioned "plotted the response of the circuit down to DC". When is the gain just -R2/R1 vs when I ...

V+ = V*(R4 || Xc)/ [(R4 || Xc)+R3]
V- = V+
-[Vo - (V-)]/R2 = Vin - (V-)/(Xc+R1)
solve for Vo/Vin


One more thing. Why disnt we just ground the pos terminal
thanks a million
 
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  • #5
You would normally expect an op amp to have 2 symmetrical power supplies with a common 'ground' or 0Volts...usually +/-15V.
The + input is usually connected to the common ground to establish 0 volts.
Your amplifier is using a single supply which needs to be 'split', the 2 56k resistors act as a potential divider to 'split' the supply. The junction of the 56k resistors is connected to the + input of the amplifier.
This means that the output of the amplifier can increase and decrease symmetrically about an 'artificial' zero. The coupling capacitors play a role in this because
only AC (+/-) are transmitted by capacitors.
 

1. What is the role of a coupling capacitor in an op-amp circuit?

A coupling capacitor is used to connect two electronic components, such as an op-amp and a load, while blocking the DC component of the signal. This allows the AC signal to pass through, while preventing any DC offset from affecting the circuit. In op-amp circuits, coupling capacitors are typically used at the input and output stages to remove any DC biasing and to maintain proper voltage levels.

2. How do coupling capacitors affect the frequency response of an op-amp circuit?

Coupling capacitors have a capacitance value that determines the frequency at which they start to attenuate the signal. This means that for high-frequency signals, the coupling capacitor will have a smaller impedance and allow the signal to pass through with minimal attenuation. However, for low-frequency signals, the coupling capacitor will have a higher impedance and effectively block the signal. This can affect the frequency response of an op-amp circuit by limiting the bandwidth of the signal.

3. What are the considerations for choosing a coupling capacitor in an op-amp circuit?

When choosing a coupling capacitor for an op-amp circuit, there are a few factors to consider. The capacitance value should be large enough to allow the desired frequency range to pass through without significant attenuation. The capacitor should also have a high enough voltage rating to withstand any potential spikes in the circuit. Additionally, the capacitor should have low leakage and ESR (Equivalent Series Resistance) to minimize any unwanted effects on the signal.

4. Can coupling capacitors be replaced with other components in an op-amp circuit?

In some cases, coupling capacitors can be replaced with other components, such as transformers or resistors. However, these replacements may have different effects on the circuit's performance and should be carefully chosen and tested. For example, using a resistor instead of a coupling capacitor can cause a voltage divider effect and impact the signal's amplitude.

5. How can coupling capacitors be tested in an op-amp circuit?

Coupling capacitors can be tested in an op-amp circuit by using a multimeter to measure the capacitance value and checking for any significant deviations from the expected value. Additionally, the frequency response of the circuit can be tested by applying a signal generator and observing the output signal on an oscilloscope. Any unexpected changes in the frequency response may indicate an issue with the coupling capacitor.

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