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

I can give you a straight answer but at this stage, you need to figure out things like this on your own, given that I have already given you a complete explanation.

 

Would 1 micro amp be too large ?

Define "too large".

 

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

Yes, that is correct.

But why don’t you tell us what you are trying to do.

What are you trying to drive with yor signal generator.

At what frequency?

Why do you think you need impedance matching?

Why do you think a ‘741 will do that?

 

N

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

Ohm's Law

View attachment 330988

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 2v signal into the input impendance of an LM741 buffer with impedance 2Mohm will draw 1uA?

 

Yes, that is correct.

But why don’t you tell us what you are trying to do.

What are you trying to drive with yor signal generator.

At what frequency?

Why do you think you need impedance matching?

Why do you think a ‘741 will do that?

 

Yes, that is correct.

But why don’t you tell us what you are trying to do.

What are you trying to drive with yor signal generator.

At what frequency?

Why do you think you need impedance matching?

Why do you think a ‘741 will do that?

I am trying to buffer a 2V signal from an oscilloscope into an RLC BSF. I need to know the maximum current that can be input into the non inverting terminal of a LM741

 

Please see a squick jotted diagram

 

I am trying to buffer a 2V signal from an oscilloscope into an RLC BSF. I need to know the maximum current that can be input into the non inverting terminal of a LM741

No, you need to know what voltage you can put on the input.

What is an RLC BSF?

 

Okay, thank you for the circuit.

Your signal generator can drive a 50Ω load. Your RLC circuit had 300Ω in series. So the signal generator can drive it directly without a buffer.

I would still suggest that you try to answer my questions because they will demonstrate your misconceptions that we can help you correct.

 

Hello,

What is an RLC BSF?

I think the ts means a band stop filter using a resistor, inductor and a capacitor.

Bertus

 

No, you need to know what voltage you can put on the input.

What is an RLC BSF?

RLC Band stop filter

 

I can give you a straight answer but at this stage, you need to figure out things like this on your own, given that I have already given you a complete explanation.

I'm confused, can you give me the figures in one message please

 

Did you look at the schematic I posted?

A Widlar current mirror sets the current in the differential amplifier to about 20uA.
View attachment 330986

So the maximum current that can be input is 20uF?

 

You don't input current, which is in Amps by the way. The device uses current to create power. How much that it uses is dependent on the component and the load on the component, not the power supply.

 

You don't input current, which is in Amps by the way. The device uses current to create power. How much that it uses is dependent on the component and the load on the component, not the power supply.

I'm not understanding. Do I need a resistor before the buffer amplifier then? This is my diagram

 

The PDF states that the absolute MAX input voltage is +/-15V. Is your AC source greater than that limit? If so, then a resistor is needed to drop the voltage to below the MAX input limit.

 

The PDF states that the absolute MAX input voltage is +/-15V. Is your AC source greater than that limit? If so, then a resistor is needed to drop the voltage to below the MAX input limit.

Hello, the input signal generator outputs 2V directly into an LM741 buffer. I just wanted to know if it would work? Or would the current be too large ?

 

Just get the idea that you have to control current into the opamp input out of your head. You control the voltage. If the voltage is within range, the current will not do any harm.

 

Hello, the input signal generator outputs 2V directly into an LM741 buffer. I just wanted to know if it would work? Or would the current be too large ?

Why? The opamp is superfluous anyway. It serves no purpose.

Why do you think you need it?

Hint: If your signal generator could only output 1mA, you would need a buffer.

 

Just get the idea that you have to control current into the opamp input out if your head. You control the voltage. If the voltage is within range, thw current will not do any harm.

So I don't need to worry about controlling the current? So if my voltage is within the range the buffer will work?

Looking at my diagram, the 2V input signal from the signal generator will be correctly buffered ? In the configuration I have drawn?