>  Hello ListMembers,
> 
> I have a very off-topic subject, maybe the correct destination is [EE]. I 
> hope You could bring some light on this. 
Hi Paul,

Your question is, in my opinion, not OT, but in fact here in Europe 
the answer to such questions belong to the education in electronics. 

Two spec items are important:

- Which frequency range has to be transmitted.
- What is the total resistance of the source plus the destination. 

Inserting a capacitor will limit the transmitting of low frequency 
signals. So what is the lowest frequency you will transmit and what 
attenuation is alowed for this frequency?

When you know these facts, you can calculate the capacitor value.

For instance, if -3 dB is allowed and R is the total resistance of 
source and destination together C=1/(2.pi.f.R). For other 
attenuations use the formula

attenuation= R/(SQRT(SQR(R)+SQR(1/2.pi.f.C))

Be aware that the attenuation caused by the internal resistance is 
freqency independent and is not contained in the frequency dependence 
of the specs. Frequency independent attenuation can easely be 
compensated by augmenting the amplification of an opamp circuit.

Things are a little bit more complicated if the transducer has a 
frequency dependent inner impedance. Than you must draw a model 
circuit and calculate from there.

I hope this will help you.

Regards, Harry

> 
> The question is: How can I determine the value of a capacitor between 
> amplifier stages, or between a transdutor and an operational amplifier? Of 
> course, the capacitor would permit only the AC component of the output of the 
> transducer to be amplified, but what should be the best value?
> 
> In the case of the transdutor, its resistance is about 500 ohms, and the 
> center frequency is 40 KHz.
> 
>  
> 
> Best regards,
> 
>  
> 
>                                  Paul
> 

-- 
Author: H.C. Croon
  INET: [EMAIL PROTECTED]

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