What is the maximum current I can input into the non inverting pin of a LM741 op amp? Is there a limit?

 

It's going to be quite low.

From National Semiconductor:

Why do you think you need a large current into an input?

 

An op amp input is a high impedance that only has a low bias current.

 

I just worked out the input to my non inverting pin of the amplifier is 20mA. Is this too large to input into a LM741.

 

What is the maximum current I can input into the LM741

 

I just worked out the input to my non inverting pin of the amplifier is 20mA. Is this too large to input into a LM741.

Yes, but it is extremely unlikely that you can force an opamp input to sink 20 mA, despite an apparently insane belief on your part that you can.

 

I just worked out the input to my non inverting pin of the amplifier is 20mA. Is this too large to input into a LM741.

Give us the schematic and we will then debate.

 

LM741 does not accept 20mA input.
The input resistance is 2 MΩ (typical).

 

What is the maximum current I can input into the LM741

Did you look at the schematic I posted?

A Widlar current mirror sets the current in the differential amplifier to about 20uA.

 

What is the maximum current I can input into the LM741

Hello,

Not sure how to draw the circuit. I basically need to impedance match a signal generator with a RLC BSF.

I need to use a LM741 op amp to do this but I realise the current may be too large to be input directly.

Could I have help on how you would draw this circuit please. Sorry I'm a amateur

 

If you apply 10 V into 2 MΩ you get only 5 μA, not 20 mA.

 

OHHHHHHH right am I being stupid then?
My input is 20mA

 

N

OHHHHHHH right am I being stupid then?
My input is 20mA

No. Not stupid, just not knowledgeable.
Your input is not 20 mA. Where did you get that information?

Ohm's Law



Here is how Ohm's Law works:

I = V / R

If V = 10 V, R must be 500 Ω in order to have a current in the circuit loop of 20 mA.

 

So my signal generator outputs 2V and 20mA could I feed this signal directly into the non inverting pin of the op amp? Or is the current too high,

 

I am using a Rhode and Schwartz apn62 signal generator. I just got told it was 20mA so should probably check the value

 

I am using a Rhode and Schwartz apn62 signal generator. I just got told it was 20mA so should probably check the value

The generator puts out 2 V with a maximum current of 20 mA.
If you put 2V into a 1000 Ω load, the generator puts out 2 mA.
If you put 2 V into a 100 Ω load, the generator puts out 20 mA.

Do you see how the output current depends on the load resistance, not on the maximum capability of the generator?

It also means that don't try to use a load resistance lower than 100 Ω. You will exceed the maximum capability if the generator.

 

The generator puts out 2 V with a maximum current of 20 mA.
If you put 2V into a 1000 Ω load, the generator puts out 2 mA.
If you put 2 V into a 100 Ω load, the generator puts out 20 mA.

Do you see how the output current depends on the load resistance, not on the maximum capability of the generator?

It also means that don't try to use a load resistance lower than 100 Ω. You will exceed the maximum capability if the generator.

Yes so my maximum current will be dependant on the impedance of the LM741 op amp then? As I input the 2V straight into the buffer amplifier?

 

This is a common mistake made by beginners and it applies to all voltage sources, for example, batteries and power adapters.

Suppose you have a 9 V battery capable of outputting 200 mA. It does not deliver 200 mA unless you ask for it.

Suppose you have a USB 5 V charger capable of outputting 2 A. It does not deliver 2 A unless you demand it.

 

So to clarify, if my 2V signal generator is fed directly into an LM741 with a 2MOhm impedance. The current drawn will be 2v÷2mOhm=0.000001A

 

Would 1 micro amp be too large ?