Hi guys I’m happy to join All about circuits community. This is my first post and I would like to talk about some points I didn’t understand concerning the positive clamper circuit (picture attached).
View attachment 220635
1- at the debut of the negative cycle the diode is forward biased but it also needs 0.3 V to conduct. Correct me if I’m wrong but when the negative cycle comes the diode anode is at O V but I can’t figure out what’s the potential at the cathode.
In other words I would like to know how the diode cathode potential varies during the half negative cycle.
2- since during the half negative cycle the capacitor is getting more and more charged until it reaches Vm how the diode could still conduct if the cathode which is connected to the positive side of capacitor more positive than the anode knowing that the anode is at 0V ?
3- I always used to draw the potential arrow from the positive side of the component to its negative one. From negative peak (-Vm) to 0 the diode doesn’t conduct anymore since the cathode is more positive than anode. Then the circuit is just Vi C and R. So for me I would draw the potential arrow from right to left for C, from top to bottom for R and from bottom to top for Vi since we are still in the negative cycle which means the bottom is most positive than the top. According to Kirchhoff law I will sum the arrow in the same direction and substract the arrow in opposite direction : Vout + Vi = Vc, where Vout is potential across R. This is how I would do. In others words I don’t get how we could say that from negative peak to 0 Vout is the sum of Vc and Vi instead of being Vc-Vi.
So these are the 3 points I didn’t really get from this circuit. Any help would be appreciated.
Thank you !

Moderator note : tried to revert edit. image is gone.

 

Hello,

This will explain the circuit:
https://www.theengineeringknowledge.com/diode-clamper-circuits/

Bertus

 

Hi guys. This subject seems interesting. Bertus you suggest this interesting link but there is still some points to clarify. As Mike said the diode (is it’s Si) needs 0.7 to work. Since the anode is at 0V the cathode needs to be negative by minimum 0.7 in order to work. How could then the cathode be negative since the right side capacitor becomes more and more positive by time. As Mike i don’t understand how could the diode work if its cathode is more positive than its anode during negative half cycle.
Thanks.
Jayden

 

Your attachment gives me an error saying it cannot be found.

 

I’m sorry for the attachment. Here is the circuit.

Any help would be appreciated.
Thank you.
Mike

 

2- since during the half negative cycle the capacitor is getting more and more charged until it reaches Vm how the diode could still conduct if the cathode which is connected to the positive side of capacitor more positive than the anode knowing that the anode is at 0V ?

The capacitor will charge positively until the negative peak is about 0.7V below ground (negative output) not 0V.
Then it will conduct at the negative peak only as needed to provide the load output current.

LTspice simulation is below for some arbitrary circuit values:
Note that the negative output voltage (yellow trace) is a little less than -0.7V.
Also note the diode current pulse (red trace) at the negative peak of the sinewave that replaces the capacitor charge loss from the load current due to R1 during the previous cycle.

 

The capacitor will charge positively until the negative peak is about 0.7V below ground (negative output) not 0V.
Then it will conduct at the negative peak only as needed to provide the load output current.

LTspice simulation is below for some arbitrary circuit values:
Note that the negative output voltage (yellow trace) is a little less than -0.7V.
Also note the diode current pulse (red trace) at the negative peak of the sinewave that replaces the capacitor charge loss from the load current due to R1 during the previous cycle.

View attachment 220679

Hi Crutschow,
Thank you for your reply.
1- I have simulated also the circuit on LTspice and here is the result :


As you can see at the first negative half cycle the Vout (blue) which is also the potential at the cathode diode is at -1V. And that's what I don't understand. I mean I agree that on negative cycle the diode is forward biased and that the anode is at 0 V (connected to ground) but from where the outpout and so the cathode is getting this -1 V ? I mean this -1 V comes from somewhere but I can't figure from where. Moreover I don't understand why from Vin = 0 V to Vin = -10 V the cathode potential varies from -1 V to -0.7 V.
2- If I did understand from -10 to 0 V (from negative peak to 0V for Vin) the capacitor potential is 9.3 V. Then when Vin begin to increase from -10 to 0 V the Vout and so the cathode has also to increase (the same amount of potential increased for Vin) in order to maintain the potential across the capacitor constant at 9.3 V. Right ?

Thank you.

Mike