it doesn't seem to make much difference.

high gain, high values of resistance and changes in temperature make a difference, which can be estimated as worst case from the data sheet. Try running your tests cold and hot, using your freezer and your oven. You might also like to try a different circuit more susceptible to bias currents like an integrator.

 

Okay, I got a reliable setup for a gain of 1,000 and I get the same output with or without the R14 resistor.
What am I missing? The resistor seems to have no notable effect.
Can somebody suggest a setup that demonstrates the effect of this resistor?

 

high gain, high values of resistance and changes in temperature make a difference, which can be estimated as worst case from the data sheet. Try running your tests cold and hot, using your freezer and your oven. You might also like to try a different circuit more susceptible to bias currents like an integrator.

An integrator with DC tests? I think my output would just max out. I don't have a reliable way to make an AC measurement. Yes, heating would / should make offset more obvious, but I don't think I could get a consistent test on my bench. Not a bad idea if I had a good shop.

 

An integrator with DC tests? I think my output would just max out. I don't have a reliable way to make an AC measurement. Yes, heating would / should make offset more obvious, but I don't think I could get a consistent test on my bench. Not a bad idea if I had a good shop.

you could use a Vdc=0 as an input to your integrator, then you see only the effect of the bias currents.

I have used my freezer and oven at home to test circuits. My wife was a bit sceptical, but she gave me permission!

 

Hello,

To test the circuit at a cold temperature, you could use a coldspray as well:
https://www.circuitspecialists.com/super-cold-sprays
To test the circuit at high temperatures a hairdryer will do the job.

Bertus

 

You will never understand this until you understand what the bias current is and how it affects circuit operation. Understand that before you try to understand the magic behind a particular compensation technique.

You will never understand this until you understand what the bias current is and how it affects circuit operation. Understand that before you try to understand the magic behind a particular compensation technique.

Yes, understanding how it effects the circuit is where I am stuck. How would you demonstrate the effect of this resistor to a student. Quoting text of theories doesn't help. I can't get the theory to have a result in a circuit. The stories explain what it should do. Maybe I am just expecting more of a degree of influence. But if it has no measurable influence what good is it?

 

Hello,

To test the circuit at a cold temperature, you could use a coldspray as well:
https://www.circuitspecialists.com/super-cold-sprays
To test the circuit at high temperatures a hairdryer will do the job.

Bertus

I don't doubt these things would have an influence on bias. I need consistent conditions to test a circuit with and without the resistor. The resistor does not prevent bias drift with temperature does it?
Is that what I am missing?

 

I don't doubt these things would have an influence on bias. I need consistent conditions to test a circuit with and without the resistor. The resistor does not prevent bias drift with temperature does it?
Is that what I am missing?

no it does not prevent it, but it does demonstrate it if you want to show students. The integrator will demonstrate it too. If you want to prevent bias currents, you can't, but you can minimise the effects by your choice of amplifier.

Setting balanced resistors in a simple low gain low bandwidth amplifier is not my biggest concern in a design, so in that respect you're correct, it does not matter that much. However in some circuits it is a concern, that is the skill of a designer, to know when to take parametric data into account.

 

How much of an influence should the resistor have? Is it a matter of the ratio between input current and bias current? 1.6 mV through 1,000 ohms is 1.6 uA. If my bias current is 25 nA would I see only a 1.5% change in output voltage with or without the resistor? My test setup drifts more than that as I breath on it.

(edited to add ...)
Ah, ha!!!
Input current 100 nA compared to bias current of 25 nA (as measured V and calculated A) and yes it makes a difference!
-0.1 V in, 0.12 V out with a wire. 0.090 V out with a resistor.

 

Okay, now that I have a circuit where it makes a difference it all makes sense. It is a question of how much input current you have with respect to the bias current. So ... op amps with bipolar junction transistors for the input and high input resistances (around 1 M ohm) ,,, use the resistor. If input current is more than 20 times bias current would it make a difference?

 

but I find it difficult to believe my $0.25 LM741's are lower than the data sheet specs with a bias of 25 nA..

I don't. Error terms do not have guaranteed minimum values. Plus, measuring 25 nA accurately is difficult. It might be that your measurement system is masking the true values with its own errors.

If input current is more than 20 times bias current would it make a difference?

It depends, and this is the real answer to your main question. If your circuit error budget is only 5%, then a 20:1 ratio of input current to error current is ok. Example - audio tone controls. The output is AC coupled to the next stage and the gain is low enough that the output won't saturate because of the input offsets, so don't care.

If you need 0.01% accuracy, it isn't. Example - active filter conditioning the input to a 16-bit A/D.

How do you know when to use the resistor? When the circuit's accuracy requirements exceed the effects of the uncorrected error terms.

ak

 

Wish I had read this suggestion when I started to explore real integrators.

you could use a Vdc=0 as an input to your integrator, then you see only the effect of the bias currents.

It well could be, I guess, that a leaky cap could play against in this demonstration?

 

True, but on this planet a near-perfect capacitor is much easier to find than a near-perfect opamp.

ak

 

yes it makes a difference!

just to lower your excitement: there are plenty of times that balancing out the input resistance does NOT make a difference, or you don't care if it makes a difference: like for opamps with low input bias current (fets for example), or for ac applications.

the "balancing helps" theory is based on the assumption that the two input bias current is roughly equal for the two input pins and are at the same direction (both in or both out). Those assumptions are not always true.

Many times in applications where you care about input bias current, the better approach is to use an opamp with low input bias current, auto-zero opamps, or to null the input bias current.

Relying on the "balancing out" is a low cost but not terribly effective way to solve the problem.

 

you could use a Vdc=0 as an input to your integrator, then you see only the effect of the bias currents.
.........................

Only if the op amp adjusted for zero input offset voltage.
If you ground the input resistor to the integrator, the input offset voltage will also cause integrator drift.
What you want is an open circuit on the integrator input. That way the voltage offset has no effect and only the bias current will cause output drift.
You can then calculate the bias current based upon the integrator capacitor size and the drift rate.

 

I don't. Error terms do not have guaranteed minimum values. Plus, measuring 25 nA accurately is difficult. It might be that your measurement system is masking the true values with its own errors.



It depends, and this is the real answer to your main question. If your circuit error budget is only 5%, then a 20:1 ratio of input current to error current is ok. Example - audio tone controls. The output is AC coupled to the next stage and the gain is low enough that the output won't saturate because of the input offsets, so don't care.

So with FET input op amps that have bias down in the pA is there ever a need for the resistor?
If you need 0.01% accuracy, it isn't. Example - active filter conditioning the input to a 16-bit A/D.

How do you know when to use the resistor? When the circuit's accuracy requirements exceed the effects of the uncorrected error terms.

ak

 

Okay, now that I have a circuit where it makes a difference it all makes sense. It is a question of how much input current you have with respect to the bias current. So ... op amps with bipolar junction transistors for the input and high input resistances (around 1 M ohm) ,,, use the resistor. If input current is more than 20 times bias current would it make a difference?

It is not that simple. You might zero the op-amp out and 30 minutes later it is no longer zero. Why, because the temperature changed, and you decided not to use a compensating resistor between + input and ground. Look at " Average Input Offset Voltage Drift" of a LM741. It is 15uV/C (max). Without the input resistances being balanced, this parameter could eat your lunch.

I don't think you get it. I don't think you understand how input bias current works. Tell you what, I will indulge your stubbornness and give you a hint. Look at those two pins like you know nothing about about them. You connect a few resistors from them to ground and realize that they are constant current sources (or sinks). The next thing you realize is that they track pretty well over temperature. (as I said in an earlier post, manufacturers work hard to make this tracking happen.) So, now you have a blinding revelation where you see that if each pin sees the same resistance, they will also see the same voltage over temperature. Pretty cool stuff. OK, now take off the blinders and connect the dots.

 

It is not that simple. You might zero the op-amp out and 30 minutes later it is no longer zero. Why, because the temperature changed, and you decided not to use a compensating resistor between + input and ground. Look at " Average Input Offset Voltage Drift" of a LM741. It is 15uV/C (max). Without the input resistances being balanced, this parameter could eat your lunch.

I don't think you get it. I don't think you understand how input bias current works. Tell you what, I will indulge your stubbornness and give you a hint. Look at those two pins like you know nothing about about them. You connect a few resistors from them to ground and realize that they are constant current sources (or sinks). The next thing you realize is that they track pretty well over temperature. (as I said in an earlier post, manufacturers work hard to make this tracking happen.) So, now you have a blinding revelation where you see that if each pin sees the same resistance, they will also see the same voltage over temperature. Pretty cool stuff. OK, now take off the blinders and connect the dots.

Yes, drift is a given.
Re: "I don't get it."
Apparently not.
No, ,your hint does not give me a blinding revelation.
I can't relate anything you say to bias.

 

@hp1729
In the future, I will not be surprised to see you tell others about what you learned about op-amp input bias current in this thread. You can say you don't get it, but you do.

Have a nice day.

 

@hp1729
In the future, I will not be surprised to see you tell others about what you learned about op-amp input bias current in this thread. You can say you don't get it, but you do.

Have a nice day.

I am open to a test setup that demonstrates the purpose of that resistor, especially if it differs from mine.
.