[SOLVED] Questions about compensated attenuators

[cvinyas]

Hello,

I am a newbie and I have these 2 questions. If you are so kind, please help me to answer them because I don't know the complete answer.

1. Does the input impedance of a compensated attenuator depend on the frequency? Why?

Take into account in question 1 that there's no signal source connected to the compensated attenuator. (meaning, no voltage loading effect)

2. If a sine wave is applied to the input of a compensated voltage attenuator, what is the phase shift of the output voltage? Why?

Thanks in advance.

 

[chuckey]

A compensated attenuator is two resistors, to get your attenuation (say in the ratio of 9 : 1, giving you a 10 :1 reduction in level). With this arrangement the bottom resistor is shunted by the input capacity of the amplifier and its stray capacity in the wiring. Without a suitable capacitor across the top resistor, it in conjunction with the input capacity will form a low pass filter. A small capacitor is connected across the top resistor. This capacitor is often adjustable in order for the two resistors with their two capacitors have the same time constant ( C X R = c X r). So as the frequency is raised the effect of the capacitors become the main input impedance, and hence falls as the frequency is raised.
2. There is no phase shift if the attenuator is set up properly (see above).
Frank

 

[cvinyas]

Thanks for the answer. So there is no phaseshift because both time constants are the same, no? TC1 = 1/(2*pi*frequecy) = RC = TC2 = rc. Is that correct?

Carlos.

 

[LvW]

Hi carlos,

A quick look on the voltage divider ratio tells you everything:

Z2/(Z1+Z2)=N(s)/D(s) with

N(s)=R2/(1+sT2) with T2=R2C2
and
D(s)=R1/(1+sT1)+R2/(1+sT2) with T1=R1C1.

For T1=T2 this leads to

Z2/(Z1+Z2)=R2/(R1+R2) >>real, independent on frequency and without phase shift.

 

[cvinyas]

Ok!! Thanks so much for the answer!! You solved me the doubt!