Frequency dependency of an inverting amplifier?

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sanstrian

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Title says it all...I would think that a plain inverting amplifier has no frequency dependence - is this true?

**broken link removed**
 
Hello there,

That is true in theory only using an ideal perfect op amp, but we never have that kind of device to work with in the real world so there is always going to be some frequency dependency with real world circuits like this. There will be gain loss and phase shift with increasing frequency once you pass a certain point, and even distortion depending on the slew rate.
 
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If you check the op amp data sheet you will find a spec called gain bandwidth product. (Gain X Bandwidth) For any given op amp this is the maximum frequency the op amp can run at with a gain of 1. At a gain of 2 it will be 1/2 as high.
 
The bandwidth of the closed-loop amp is the gain bandwidth product (GBW) divided by (Rin + Rf)/Rin. Thus for a inverting gain of 1 the bandwidth is 1/2 the GWB and for a non-inverting gain of one (Rf = 0) the bandwidth is equal to the GWB.
 
Hello again,


Guys, the bandwidth can not be calculated based on the gain bandwidth product (GBW) alone. We also have to take into account the op amps slew rate.
The max frequency can be calculated:
MaxFrequency=SlewRate*1e6/(2*pi*Vpeak)

where the SlewRate is the slew rate in volts per microsecond (V/us) and Vpeak is the peak amplitude of a sine wave at that frequency.
We then compare this to the GBW/G and see if we can really get the op amp to work at that high of a frequency. If the max frequency is higher than the GBW/G frequency then the op amp will work at F=GBW/G, but if the max frequency is less than GBW/G then it will not work very well.

For example, lets say we have an op amp with slew rate equal to 0.5 V/us and GBW=1MHz and we want to input a sine wave with a peak of 0.1v and we need a gain of 2. With a gain of 2 the output will presumably be equal to 0.2v, so the max frequency is:
MaxFrequency=SlewRate*1e6/(2*pi*Vpeak)
MaxFrequency=0.5*1e6/(2*pi*0.2)=398kHz.

Now since the GBW/2 is 500kHz and the MaxFrequency is 398kHz, this design should work at frequencies less than 398kHz.
Note however that it will not work at 500kHz because the slew rate limits the usable range.

For another example say we have the same op amp but an input peak voltage of 1v instead of 0.1v. Now the output would be 2v with a gain of 2
We again calculate the max frequency:
MaxFrequency=0.5*1e6/(2*pi*2)=39.8kHz.
but note now that even though we only want a gain of 2 we can only use this op amp up to about 40kHz. This is significantly different than before. The GBW/G is still 500kHz, but we can not use it even close to that frequency but are limited to about 40kHz.

Now you can see why the slew rate is so important.

Another thing to note is that the GBW/G number means the amplitude is 3db down, which could be a problem too.
 
So for a basic little inverting amplifier using a 741 op amp, measured at frequencies of 10 Hz, 1000 Hz, and 100 kHz...

we might expect the gain to behave like the theoretical at 10 Hz and 1000 Hz but begin to be measured less than the theoretical at 100 kHz?
 
The 741 opamp is 43 years old and has very poor performance.
Its datasheet shows that at 100kHz its open-loop gain is only 7 and its max output voltage is only 1.5V peak with severe distortion (a sine-wave or square-wave turns into a triangle-wave).
 

Hello again,

Yes, that is just about right.

There are four things we have to consider when we want to evaluate the possibility of using the amp at a certain frequency:
1. The gain bandwidth product of the op amp (GBW)
2. The slew rate of the op amp (usually expressed in V/us)
3. The gain (G) we want for the circuit
4. The input voltage peak

We can then calculate VoutPeak from VinPeak*G and the band width BW from GBW/G and the max frequency from MaxFreq=SlewRate*1e6/(2*pi*VoutPeak)

and this information will tell us if it will work or not.

For the 741 the GBW can be as low as 500kHz and the slew rate around 0.5 V/us, so for an input peak of 1v and gain of 1 we would have:

BW=GBW/G=500000/1=500kHz

and

VoutPeak=VinPeak*G=1*1=1

and then

MaxFreq=SlewRate*1e6/(2*pi*VoutPeak)=0.5*1e6/(2*pi*1)=79.6kHz roughly

This means with a an input of 1v peak we can use this up to about 80kHz, but with an input of 2v peak it would divide in half to about 40khz.


Another interesting view is what if we want to use this amp with a gain of 100 with the same input of 1v peak?

Now we have G=100 so that changes things quite a bit...

BW=GBW/G=500kHz/100=5kHz

and MaxFreq will be the same as before at about 80kHz, so now what limits the usable range of the op amp is actually the GBW/G not the max frequency due to the slew rate and input peak. Here the slew rate is good enough to follow a 1v peak input sine but the GBW/G will not allow any frequency above 5kHz to work properly.

As others have pointed out, there are better op amps out there including the low cost LM358 which work much better than the 741. The slew rate and GBW are roughly the same, but the input voltage requirements are less demanding. The 741 has a nasty bias requirement that might not work with voltages closer than about 4v to either power supply rail. That's really bad for something that has to run off of 5v for example.
 
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For the 741 the GBW can be as low as 500kHz and the slew rate around 0.5 V/us, so for an input peak of 1v and gain of 1 we would have:

BW=GBW/G=500000/1=500kHz
That's true for a gain of plus one. But since this thread started with a question about the frequency dependence of an inverting amp it should be noted that the BW = 500000/2 =250kHz for an inverting amp with a gain of minus one (as I noted in my previous post).
 
As others have pointed out, there are better op amps out there including the low cost LM358 which work much better than the 741. The slew rate and GBW are roughly the same ....
No.
The LM358 dual opamp has exactly the same opamps as the LM324 quad opamp. They are low power so in addition to their horrible crossover distortion their slew rate is terrible.
Their datasheets don't even show how bad is their slew rate when they have an output near their max rail voltage.
 

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This is from a lab I took an incomplete in almost two years ago as a physics student - 741 is what I used then - at least I guess it was a 741 - there are were a few different op amps but all contained 741 as part of the model number and the way the storage thing was labeled I was under the impression that they were all pretty similar.

The gist of the lab was to build a little inverting amp and measure the gain at 10 Hz, 1000 Hz, and 100 kHz, varying R_f for gain of .1, 10, and 100. The data I got at the time is no good - must have had something wired wrong. So instead of driving 200 miles to my university I'm just writing this part of the lab by talking about what I would have expected to happen.
 

Yeah, that's just not correct nor complete. You can never calculate the usable frequency on just the GBW alone.
 

The LM358 is better based on the input requirements. The 741 doesnt even work well at 5v.
 
I use MC34071 single, MC34072 dual and MC34074 quad opamps to replace LM324 and LM358 opamps when I want the input to operate as low as the negative supply which can be ground when there is no negative supply. Their max supply voltage is much higher than the others at 44V and they also work perfectly when the supply is as low as 3.0V. Their outputs also go down almost to ground in a single supply system.
They have no crossover distortion and have a good slew rate so the output can deliver 28V p-p up to 100kHz.

I use MC33171, MC33172 and MC33174 opamps when I need the low supply current of the LM324 or LM358. They have all the other features but have no crossover distortion and have a fairly good slew rate so the output can deliver 20V p-p up to 35kHz. Their max output current is a little less than normal.
 
Yeah, that's just not correct nor complete. You can never calculate the usable frequency on just the GBW alone.
I never implied that you do. I was just noting that, in your calculation for the BW you used a value of 1 for a gain of 1 and that's only true for a gain of plus 1 not minus 1. OK?
 
I never implied that you do. I was just noting that, in your calculation for the BW you used a value of 1 for a gain of 1 and that's only true for a gain of plus 1 not minus 1. OK?


Hello there Carl,

Well i cant figure out where you got that idea from. It may be a little bit better for a gain of +1 with the output shorted to the inverting terminal (voltage follower) but it doesnt change THAT much when we set it up as an inverting amplifier with a gain of 1.

Where did you get that information from as it sounds like one of those vague approximations we run into now and then?
 
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No, it's not a vague approximation. Read this.
 
Hi again Carl,


Ok sure

I think that site is interesting and informative. There's a slight problem though. When they talk about bandwidth they talk about a theoretical AC analysis, which does not show the effect of slew rate well enough to follow it as a guide line. That's what i meant about 'vague' approximation, but maybe the word 'vague' is not the best choice of words as it is still a little interesting. However, including the slew rate into the calculation is much more informative as i was trying to point out (that's all i was trying to do here really) and if you scroll down the linked page near the bottom you'll find this interesting little note:

START_LINK_QUOTE

MORE TO BANDWIDTH
Although fu (GBP) and gain play a major role in determining bandwidth, there's more to it. Slew rate and heavy loading can cut into your bandwidth. Slew rate would not only make an interesting design topic, but a challenging simulation topic as well. Its on the list of future topics.

END_LINK_QUOTE

An once again we can see that slew rate must play an important part in determining the usefulness of an op amp.
 
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