Hello,
i have asked this question few weeks before in this forum [link below], but still i have doubt. I am modulating message signal using astable multivibrator circuit ,then passing through isolation using capacitive coupling technique in IC design. DC-level of modulated message signal is shifted, across coupling capacitor , although no circuit is connected on other side of capacitor. So, I would like to know why this DC-level is shifted?

In my previous thread , @Alec_t expalined me that this DC-level shift is due to amplifier biasing current (amplifier was connected with coupling capacitor on demodulator side)[shown]. But this time , nothing is connected on other side of capacitor but still DC-level is shifted?

Can someone explain me this?

Message signal : 1MHz
Oscillating Frequency: 27M Hz
Coupling Capaciotr (CC) :1pF
DC-level of Modulating signal before Passing CC : 0-5V
DC-level of Modulating signal after passing CC: +1.4 to -3.6

Link: https://forum.allaboutcircuits.com/...-passing-through-capacitive-isolation.170264/

Thanks.

 

A capacitor blocks DC so the average DC level on the output of the capacitor is determined by whatever the capacitor is connected to.

 

If there is absolutely nothing connected after the capacitor then the average DC voltage after the capacitor will depend on ehat voltage the capacitor is charged to. If the capacitor is discharged then the average voltage after the capacitor will be the same as the average voltage before the capacitor.

 

A capacitor blocks DC so the average DC level on the output of the capacitor is determined by whatever the capacitor is connected to.

Thanks for your response. But this time, i have not connected the capacitor with any circuit , but still its level is shifted ? I would like to know why and what causes this shift in voltage level?

 

From the thread you referenced, you wrote "Before it is between (0-5V) but on other side of capacitor it's now between (-3.4 and 1.6) as shown in figure."

What crutschow and AlbertHall said is the key.

Passing a signal through a coupling capacitor will remove the DC component of that signal. You signal goes from 0V to 5V and the DC offset will be 5 volts x the duty cycle of the pulse. By visual inspection of the waveforms you posted in the other thread your duty cycle is about 25% so the DC component of the pulse will be 25% x 5V = 1.3 volts.

Check it with a DVM to see the average (DC) voltage.

Since you seemed to have missed the point in earlier posts about blocking the DC, your signal had a 50% duty cycle it would go between some DC voltage -2.5V and some DC voltage + 2.5 volts. Because the signal assums a new DC component on the load side of the capacitor it is , and pretending there is only a resistor to ground the average (the DC component of the signal) will be 0V. So it will swing between those two voltages. Having a duty cycle of less than 50% will push the peaks up a little bit and your DC offset introduced by your amplifier will also be expected to shift the bottom of the signal (and the top being 5V higher).

The last 30 seconds of this 1 minute and 30 second video will make the effect clear:

and
https://www.youtube.com/watch?v=3h6tH592BEs

 

But this time, i have not connected the capacitor with any circuit ,

Strictty, if nothing is connected then you don't know what the voltage is.

 

A bit off :

But is it ok to bias the transistor like that?

 

Your scope probe will have a resistance to ground, and that will keep the average value of the signal at ground.

 

nothing is connected on other side of capacitor but still DC-level is shifted?

Whatever you are measuring the voltage with will provide a resistance to ground.
That's not "nothing".

 

From the thread you referenced, you wrote "Before it is between (0-5V) but on other side of capacitor it's now between (-3.4 and 1.6) as shown in figure."

What crutschow and AlbertHall said is the key.

Passing a signal through a coupling capacitor will remove the DC component of that signal. You signal goes from 0V to 5V and the DC offset will be 5 volts x the duty cycle of the pulse. By visual inspection of the waveforms you posted in the other thread your duty cycle is about 25% so the DC component of the pulse will be 25% x 5V = 1.3 volts.

Check it with a DVM to see the average (DC) voltage.

Since you seemed to have missed the point in earlier posts about blocking the DC, your signal had a 50% duty cycle it would go between some DC voltage -2.5V and some DC voltage + 2.5 volts. Because the signal assums a new DC component on the load side of the capacitor it is , and pretending there is only a resistor to ground the average (the DC component of the signal) will be 0V. So it will swing between those two voltages. Having a duty cycle of less than 50% will push the peaks up a little bit and your DC offset introduced by your amplifier will also be expected to shift the bottom of the signal (and the top being 5V higher).

The last 30 seconds of this 1 minute and 30 second video will make the effect clear:

and

Thanks, for your detailed explaination, which addressed my question.