How to properly use an op-amp?

Question

ever since posting here I never been so lost of using op-amps before, hearing new things that I have never heard before (Vom, Vcm etc). I always thought OP AMPS is just plug it in and it will work everytime... Very wrong.

I have a couple of questions which would be most appreciated if anyone of you could answer them, before I asked them, yes I have been looking for the past 2 hours in this forums for previous questions that were asked. Still a little confused but it did clear some things up.

To keep things consistent I would be using This OP AMP throughout this whole example. MCP601

VCM: Common Mode Input Range

Here is what I understand -Its the range of which the MCP601 can happily accept with nothing going wrong, if one were to go over or under these ranges the op you will see unexpected error.

Example: Input = Audio Signal (1.2V pk-pk) VDD = 4.8V VSS = GND

VCM - Upper limit = 4.8-1.2 = 3.6

VCM - Lower limit = 0-0.3 = -0.3

VCM - \$V_{CM_{PP}}\$ = 3.6-(-0.3) = 3.9V

\$V_{IN}\$ - Positive Cycle of input = 600mV + (VDD/2) = 3

\$V_{IN}\$ - Negative Cycle of input = -600mV + (VDD/2) = 1.8

\$V_{IN}\$ = 1.2Vpk-pk

Meaning the input Vpk-pk is suitable?

VOM: Output Voltage Swing

Here is what I understand - Its the range of which the MCP601 is capable of outputting before clipping.

Example: Input = Audio Signal (1.2V pk-pk) VDD = 4.8V VSS = GND GAIN = 3.2

Input Bias = VDD/2 RL = 5k

VOM - Upper limit = 0+100mV = 100mV

VOM - Lower limit = 4.8-100mV = 4.7V

VOM - \$V_{OM_{PP}}\$ = 4.7-100mV = 4.6V

\$V_o\$ - Positive Cycle of input = (3.2*600mV)+(VDD/2) = 4.32V

\$V_o\$ - Negative Cycle of input = (3.2*-600mV)+(VDD/2) = 0.48V

\$V_o\$ - \$V_{o_{PP}}\$ = (4.32-0.48) = 3.84V (Before decoupling Cap).

This is how I understood to calculate for both \$V_{CM}\$ and \$V_{OM}\$. To me this OP-AMP shouldn't have a problem with the Vin as well as happily amplifying the Vin as well, however the opposite happened as it clips at 2.84Vpp. This doesn't make much sense to me from the calculation above. The VCM should be satisfied as well as the VOM. As the VOM has a Vpp of 4.6V which is > then my Vo of ideally 3.84Vpp and my VDD being 4.8V it should amplify to 3.84Vpp no problem?

If anyone can show me how to actually calculate VCM and VOM that would be amazing, I believe this method is missing something or I am not understanding some fundamental logic. I would like to gain the ability to understand input and output limitations through this method.

This configuration works if I increase VDD to ~6.1V if anyone can explain why through the VCM and VOM calculations I can probably correlate the two and probably will clear up any confusions I had.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Your 2nd datasheet snip is in mV not volts, and the output range is relative to the supply voltages. So with a 4.8V supply and 5K load (to 0V) the linear output range is 0.1 to 4.7V. If you bias the input and output at 2.4V you can get 4.6Vp-p. The op-amp output cannot exceed (or even meet) the supply voltages.

If the input is biased at 2.4V, your input range is -0.3 to 3.6V so you can only handle an input voltage of 2.4Vp-p = (3.6-2.4V)*2, based on the input range, however you also need to make sure the output does not saturate.

Your circuit has a gain of +3.2 so input voltage has to be within the range of +/-0.71875V or 1.4375Vp-p, which will yield the full output range, so the input range is not limiting.

You can use pretty much any op-amp on a single power supply provided you have enough supply voltage and provided you bias the input within the working range and provided you keep in mind the available output range.

In general for a low power circuit you would want to use higher value resistors than you show. You are loading the output with 5K||(2.2K+1K) which is lower than 5K, obviously, so the output swing is not guaranteed. Normally you can go at least 10x higher for the feedback resistors, maybe considerably more. If you can increase the load to 25K or 100K, and increase the feedback resistors by 100:1 it would be better. You may have to add a small capacitor across R3 to assure stability if you go very high with the resistors.

Answer 2

I believe I figured out the conundrum.

Taking up a project such as this and using op-amps to this extent, such as looking for characteristics you wouldn't usually look for coming out of university as input bias current, Vom, Vcm, etc.

Trying to joggle all these terms tend to confuse me and kinda overwrite the basic things I knew about op-amps.

When I was first introduced \$V_{OM}\$ and \$V_{CM}\$ for some reason I made myself believe as if the \$V_{in}\$ and \$V_{out}\$ didn't violate the \$V_{OM}\$ and \$V_{CM}\$ then clipping shouldn't exist. THIS IS completely WRONG

What I didn't account for is the voltage drop that the op-amp has internally due to it's architecture of the op-amp.

Meaning no op-amp can go rail-to-rail unless its perfect (no voltage drop in the internals).

For the problem above is a single power supply non-inverting amplifier, which means it requires a bias in order to swing "negative"

For Reference:

thus, its 4.576V -- 2.288V -- 0V

\$V_{DD_{pp}}\$ = 4.576V \$V_{DD_{p}}\$ = 2.288V

Through experiments, I found the voltage drop of the amplifier to be around ~1.616Vpp

We will do 2 case scenarios Where,

input_1 = 860mVpp

Input_2 = 1.14Vpp

Gain = 3.2

---

Input_1: 860mVpp

VCM:

-0.3 < \$V_{IN}\$ < 3.376

Vin:

1.858 < \$V_{IN}\$ < 2.718

Vin is within range of Vcm

VOM:

-15.424 < \$V_{OUT}\$ < 15

0.912 < \$V_{OUT}\$ < 3.664

Vo is within range of Vcm

You would expect your signal to behave as you predicted.

---

Input_2: 1.14Vpp

VCM:

-0.3 < \$V_{IN}\$ < 3.376

Vin:

1.658 < \$V_{IN}\$ < 2.798

Vin is within range of Vcm

VOM:

-15.424 < \$V_{OUT}\$ < 15

0.404 < \$V_{OUT}\$ < 4.052

Vo is within range of Vcm

You would expect your signal to behave as you predicted, however it isn't.

On my oscilloscope it clips at 2.96Vpp but we expected the output to be 1.14Vpp * 3.2 = 3.648Vpp? What is happening is the Voltage drop of the op-amp.

As mentioned above the voltage drop of the op-amp was ~1.616Vpp so doing the math in-tales

VDD -Vod = 4.576 - 1.616 = 2.96Vpp !! This essentially is telling us what our op-amp can actually drive up to. Which all makes sense now.

Essentially what an op-amp says rail - to - rail it means at least what I can see is that your Vin and Vout will never usually violate the op-amps VOM and VCM's

This is why when I increase VDD ~6.1V it works as the op amp can actually drive up to the expected output of 3.648Vpp as follow:

Vdd - Vod = 6.1 - 1.616 = 4.484 as the op-amp new limit is now 4.484Vpp and since 3.648Vpp < 4.484Vpp you can see it on the output.

Answer 3

Vpk-pk = 3.6-(-0.3) = 3.9V.
Meaning the input Vpk-pk is suitable?

Possibly. The mid point of CM range is not Vdd/2 here but 3.9/2=1.95V. This would then permit an input signal up to 3.9Vpp. . However your gain would clip the output.

The output stays in the linear range if the output is not clipping. It is defined for symmetrical clipping @100mV from both supply rails depending on loads >5k connected to VL=2.5V. This is because CMOS rail-to-rail Op Amps have a resistance at clipping on the order of 250 Ohms on either Nch or Pch driver. If the load goes to Vss=0 then there is less dropout above Vss but more dropout below Vdd as there is now twice the current compared to spec with [email protected]

Vin{pp}*Av=1.2*2.4=3.84Vpp will fit in the linear output range when the input and difference reference are both common (zero differential) near the middle of the CM range. (Remember near 2V for your supply) It also works for Vdd/2=Vcm bias in this example.

Advice: use R min values of 25k for feedback and load combined

All Op Amps output resistance is lowered by negative feedback gain. But clipping results in total loss of negative feedback. Since FET rise in RdsOn when Vgs reduces which here is Vdd , it is known to rise rapidly below 5V just like CD4000 family logic towards 1kOhm and higher at Vdd min.