Best option to attenuate the signal in this simple cct

[HankMcSpank]

Hi,

Here's my problem, in this diagram...



I have an output that is slightly too hot for the device that uses this circuit's output. The signal is about 2V peak to peak.

The first opamp is just a unity gain voltage follower. Therefore the input signal must also be about 2v peak to peak. The input is a piezo pickup, so the opamp must provide a Hi input Z.

What would be the best option to bring get that output signal down a little without affecting the following opamp path? (that said, if needs must & I have no other option, I can accept the attenuated signal impacting the following opamp circuit too!)

The device that uses the circuit's output is expecting the output impedance of this circuit to be 100 ohms.

I had thought of say a '10k preset' between the output & ground with the preset wiper being the new output...that will obviously reduce the signal, but I'm not sure how that affects the output impedance (or if it'll mess with the later opamp's config.

[Frank Berry]

You can get 10 dB of attenuation if you change the value of R2 to 1000 ohms and then put a 100 ohm resistor from the output of R2 to ground.

[HankMcSpank]

Hi Frank,

Many thanks,

Forgive my ignorance (never did get to grips with dB)....

What kind of drop in voltage terms can I expect to see?

...eg if the signal was 2V peak to peak before your suggested attenuation mod, what (approx) will it be afterwards?

Also, I'm guessing(!) that the 100 ohm resistor after R2 to gorund maintains the 100 Ohm impedance that the unit this signal feeds is expecting?

[Frank Berry]

With 2.0 volts going into the pad, the output wuill be .63 volts.
If you need less attenuation, you can change the value of R2.
The output impedance is controlled by the combination of the 1000 ohm resistor and the 100 ohm resistor.
With this combination, the actual output impedance is 90 ohms.

[HankMcSpank]

Thanks Frank - very helpful.

That amount of attenuation is a lot more than I was wanting - I'd be shooting for about 1.5V, therefore what is the formula can I use to establish the attenuation (so I can have less!)

Also, what formula are you using to establish the output impedance?!!!

Very basic stuff I'm sure - but nevertheless, we've all got to learn!

[Frank Berry]

My math was wrong. Sorry, I't early in the morning here.
With R2 at 1000 ohms and an additional 100 ohms to ground the actual output voltage will be about .2 volts. I don't know what I was thinking (or perhaps not thinking).

If you keep the resistance of R2 at 100 ohms and place a 330 ohm resistor between the output of R2 and ground, you will get about 1.5 volts out and keep the output impedance low.

This is a good link to help you calculate just about anything.

http://www.calculatoredge.com/

[richie00boy]

Level of attenuation is simple potential divider action - you can find this easily so I won't repeat it here.

Output impedance is just R2 in parallel with R3, 100 and 330 in the last example giving 76.7 ohms.

[Frank Berry]

Thanks richie00boy.

76.2 ohms is correct. If you actually need an output impedance of 100 ohms, you can simply add 23.8 ohms in series with the output.
I suspect that the output impedance isn't too critical but I may be wrong.

[traderbam]

If the series resistor is R2 (as in your diagram) and the shunt resistor from o/p to ground is R3, then

attenuation factor A = R3 / (R2 + R3)

output impedance Z = (R2 x R3)/(R2 + R3)


To help you out I have rearranged these so you can get the values of R2 and R3 you want straight from the attenuation and impedance you want:

R2 = Z / A

R3 = Z / (1 - A)


Eg: if you want an attenuation of 0.8 and an impedance of 100 ohms, you need to make R2 = 125 ohms and R3 = 500 ohms.

[mlloyd1]

assuming your opamp is happy driving 600 ohm loads.
some are not ....

mlloyd1