Can anyone explain this for me.

I'm trying to use MCP604 as an buffer-amplifier.
But it it will not work properly.
Instead of having a gain of 1 i get <0,5.

Is 38kHz to high in this application?
When i read the specifications it don't seem to be a problem.

/Urban

 

Direct from the manufacturer -

[SIZE=-1]The MCP604 quad operational amplifier (op amp) has a gain bandwidth product of 2.8 MHz[/SIZE]

You need a faster op amp.

 

This is something i don't understand.
If the bandwidth is 2.8MHz how do i calulate that 38kHz is to much for this device?

Regards
/urban

 

yes I was also puzzled by that comment. I guess Beenthere get it wrong.

 

Duh - read MHz for Khz.

 

Duh - read MHz for Khz.

shocked.

 

Hello,

I think the two 1k2 resistors make a to hard load for the 38 kHz source.
Try changing them to 47K each.
Also put a capacitor between the source and the input of the buffer.

Bertus

 

What is the slew rate for this sucker, anyone have a link for the datasheet?

 

Slewrate 2,3V/uS

http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en010447

I'm actually measuring 1V at input at OP so there is no problem with the load
(and getting < 0,5V out).

Regards/
/urban

 

You need a cap (0.1uF-1uF suggested) between the voltage divider and the source to center the noninverting input at Vcc/2. Use 10k resistors for the divider.

If you want the output centered around 0v, you will also need to use a cap in series with the output, and a resistor to ground.

 

I think your measurement are correct. And by using the superposition theorem your result can be explained. But i am going to bed now. So if your problem is not solved. I will come back to you. And of course anybody is free to pitch in on this one

 

Ok so let us use some superposition. First we look at the contribution from the "bias" network. We short circuit Vin. But by doing this we also tie the opamp input to ground (A). So it will not have anything to say for output. Let us short circuit the bias input. As we see the result is that source has to drive a load equal to the two resistors in parallel(B). But Vout=Vin
I suggest you download the LM324 datasheet. Not because you should use this opamp. But is has a good collection of circuits examples. Find one of the example and let us what that will bring us
EDIT: I guess I was somewhat tiered yesterday. So I was perhaps wrong in my assumption that your measurement is correct. Or at least it will depend on your input test voltages. But the circuit behavior is now explained.