1
\$\begingroup\$

I was wondering what the effect of an op amp like the one in the image below is for an input signal such as \$0.5\sin(2000\pi t) + 0.5\$ V? Are the DC and AC components amplified with the same gain? And also, what would be the phase relationship between the input and the output? enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ are you asking about ideal op-amps or real op-amps? \$\endgroup\$ – The Photon Oct 9 '14 at 3:12
  • \$\begingroup\$ Real op amps since the I have to build the circuit in an electrical measurements lab. \$\endgroup\$ – user3266738 Oct 9 '14 at 3:16
1
\$\begingroup\$

As long as the AC signal frequency is within the op amp's bandwidth the AC and DC gains are the same for the circuit you have drawn. The gain is

$$\frac{v_{O}}{v_{I}} = 1 + \frac{R_f}{R_i}$$

If your op amp datasheet specifies its gain-bandwidth product you can easily calculate the bandwidth if you know the closed loop gain you need.

I've drawn your circuit in Circuit Lab with your input signal \$v_{I}(t) = 0.5\sin(2000\pi t)+0.5\text{V}\$, and I'm using \$R_{i} = 1\text{k}\Omega\$ and \$R_{f} = 100\text{k}\Omega\$ for a gain of \$101\$:

schematic

simulate this circuit – Schematic created using CircuitLab

If you run CircuitLab's DC solver you will see that \$v_{O} \approx 50.5\text{V}\$. Unless you happen to be using supply voltages greater than \$50\text{V}\$ the op amp will not actually be able to force \$v_{O}\$ that high and it will saturate.

If you need a high gain like \$101\$ as I've simulated and you are not be able to get rid of an undesirable DC offset like \$0.5\text{V}\$, you will need to add AC coupling. For most op amp circuits you can simply add a capacitor in series with your input to block the input's DC offset (you just need to determine the appropriate capacitance for the frequencies of interest). However, for this circuit that would be a bad idea since the op amp's non-inverting input bias current (which is very low but non-zero) would have nowhere to flow except into the AC coupling capacitor. To avoid this you also need to add a resistor from the non-inverting input to ground. Think of this as a simple \$RC\$ high pass filter. The AC coupled non-inverting amplifier looks like this:

schematic

simulate this circuit

If you run Circuit Lab's DC solver on the AC coupled circuit you will see that \$v_{O} \approx 0\text{V}\$.

You can run a frequency domain simulation in Circuit Lab for the Bode plot of the AC coupled circuit. You can see that the gain is very low at DC and low frequencies, is \$101\$ in the midband (including your input frequency of \$1\text{kHz}\$), and then decreases at \$-20\text{dB/decade}\$ at high frequencies. I don't know what frequencies are important to you so you might need to choose different capacitor and resistor values for the \$RC\$ filter.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

Most datasheets have a gain vs. frequency plot. That should get you partway towards your answer. At each frequency of interest, find the gain of the opamp itself from the plot and analyse the circuit as if it was DC with that internal gain. Then add the components back together, keeping track of which result was for which frequency.

Note that the gain plot does not include any feedback. It is the gain of the opamp itself. Vout/(Vin+ - Vin-)

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

The closed-loop gain of the circuit is 1+Rf/Ri. That gain is flat until you start approaching the closed-loop bandwidth of the circuit, which is the gain-bandwidth of the amplifier/closed loop gain. At that point the gain is 3dB less, and starts to roll off at 20dB/decade. (For a dominant-pole compensated op-amp which applies to the majority of general-purpose op-amps.)

The AC and DC components of the signal are both multiplied by the closed-loop gain, and the output will reflect that as long as inputs and outputs are within the voltage range, current and power output, and slew rate capability of the amplifier and its power supplies.

That's a pretty good approximation of what goes on, if you need to be more exact you can write the open-loop transfer function from the graph in the data sheet and do the math to get the closed-loop transfer function expression.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.