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Minimum input impedance (Op Amps Design Problem)

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mobius0

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Hi, I have a problem that asks me to find the minimum input impedance and maximum output impedance of an amplifier that amplifies a signal between 1k-ohm resistance source and a 5k-ohm load. The signal loss at each end from loading effects is not to exceed .5dB (total of 1dB).

I'm not really sure where to start here. How can I find the input impedance without knowing any of the voltages? I was trying to use the formula vi = (Ri/(Ri + Rs))*vs. I also do not see how the I can relate the signal loss to this.

Thanks
 

audioguru

Well-Known Member
Most Helpful Member
A non-inverting opamp has an input impedance of Megohms. It can have the 1k ohms source feed it and have a couple of high resistance resistors bias it. It can easily drive a 5k ohms load with a very small loss.
 

mobius0

New Member
The exact answers in the book are 16.88k-ohms and 296.3-ohms respectively. This should be a non-inverted amp since we just started the topic and haven't gotten really far yet. I understand that input should be high but how should I go about calculating it to get the exact value?

Thanks for the reply.
 

audioguru

Well-Known Member
Most Helpful Member
16.88k ohms is the minimum input impedance of the opamp circuit that will load the 1k ohms source and cause a 0.5dB loss. A higher impedance will be better because it will cause less loss. You could bias the opamp with a 47k or 100k resistor.

296.3 ohms is the maximum output impedance of the opamp circuit that will create a voltage divider with the 5k ohms load to cause a 0.5dB loss. A lower impedance is better because it will cause less loss. The output impedance of an opamp circuit is almost zero ohms. You could add 100 ohms in series with its output to prevent the opamp from oscillating if it drives the high capacitance of a shielded audio cable.
 

Electronworks

New Member
Here goes...

Your signal has an output impedance of 1k. Therefore if you hang a resistor to ground on the end of this, you will get a voltage drop. Example: your output impedance is 1k. Put a 1k load on this and your output voltage will now be half what you expect it to be.

Your input impedance to your amp looks like a resistor down to ground. Your 1k source resistor will cause a voltage drop at the input pin if the input has any input impedance.

Now for your output resistance. This will look like a series resistor in series with your load. if you have a 1k output resistance: If you have an o/p voltage of 1v and no load, you will measure 1v output. You put a load on the output and your voltage will drop to 0.5V.

Now for your question... which I have forgotten. Let me scroll down..:

Ah yes..

Your source impedance is 1k. your input is Rin. Your voltage drop is:

Rin / (Rin + 1k) and this has to equal 0.5dB (which is 0.944). I got this from:

dB = 20 log (Vin/Vout). therefore inverselog(-0.5/20) = 0.944. We use -0.5 cos it is a loss.

Working backwards, we can find Rin = 16857ohms

Likewise with the output, your load is 5k:

5k/(5k + Routput) = 0.944

Routput = 296.6ohms

Hope this helps! :D
 
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