# Voltage level shifting of Signal After passing through capacitive isolation.

#### Alex_Khan

Joined May 27, 2020
51
This may help to explain things in your setup.

View attachment 208935

The modulator output is measured between a and Ref1. The demodulator input is measured between b and Ground (Ground is Ref2.
Note that the voltage at b is affected by the current through R1, R2 and the transistor base junction. Hence the demodulator input waveform is offset from the modulator output waveform.
Thank you so much for your explaination. This explaination addressed the issue.

#### crutschow

Joined Mar 14, 2008
27,470
that means any excursions below ground will be useless.
It's a digital signal, so that's not a significant factor.
Are the grounds also isolated
I would expect them to be if it is an isolator circuit.
does the "demodulator" circuit make any sense?
Does to me.
It generates a low signal at the transistor output when a digital signal is received.

#### Papabravo

Joined Feb 24, 2006
16,512
It's a digital signal, so that's not a significant factor.
I would expect them to be if it is an isolator circuit.
Does to me.
It generates a low signal at the transistor output when a digital signal is received.
None of that was even remotely clear in the original post.

#### Papabravo

Joined Feb 24, 2006
16,512
This may help to explain things in your setup.

View attachment 208935

The modulator output is measured between a and Ref1. The demodulator input is measured between b and Ground (Ground is Ref2.
Note that the voltage at b is affected by the current through R1, R2 and the transistor base junction. Hence the demodulator input waveform is offset from the modulator output waveform.
Nice diagram -- explains a ton.
I was on the right track with asking about where the voltages were being measured from and to. Twice I got BS answers -- sorry I could not imagine what was going on.

#### Alex_Khan

Joined May 27, 2020
51
It's a digital signal, so that's not a significant factor.
I would expect them to be if it is an isolator circuit.
Does to me.
It generates a low signal at the transistor output when a digital signal is received.
Thank @crutschow for response. Its a digital control signal , with seprated grounds at each side of isolated cap. At demodulator side the recieved signal is amplified, filtered (using LPF) and orginal signal is retrieved using schmitt trigger.

#### Alex_Khan

Joined May 27, 2020
51
Nice diagram -- explains a ton.
I was on the right track with asking about where the voltages were being measured from and to. Twice I got BS answers -- sorry I could not imagine what was going on.
Thanks @Papabravo for your time. I think the things would be easily understood , if i would have explained it in better way .

#### Papabravo

Joined Feb 24, 2006
16,512
Thanks @Papabravo for your time. I think the things would be easily understood , if i would have explained it in better way .
We try to extract information to help explain what is presented. It doesn't always work, but all's well that ends well.

#### U-TurnSystems

Joined Jan 12, 2019
5
Read up on "dc restoration" using a diode and capacitor, to re-establish the peak dc voltage, despite changes in duty time (dc average) of a digital signal.

#### U-TurnSystems

Joined Jan 12, 2019
5
PS: The base-emitter junction is your "diode" in your case, clamping the positive excursions, and allowing negative peaks.

#### andrewmm

Joined Feb 25, 2011
1,642
As a thought experiment,
Have a capacitor, on one lead you put a pulse generator, going 0v to +5v, on the other side you have a scope.

The pulse gen and the scopes other lead are joined together , this is your referance.

So put a square wave into the capacitor 0 to 5v, what comes out the other side of the capacitor ?
what happens when you put 5v dc into the capacitor ?