the cuurent of diode clamper circuit

[EDA_hg81]

The current of clamper circuit

The circuit is as the attached image.

Since the diode is a practical diode, I am wondering how this circuit can supply 200mA for diode when diode is conducted.

Many thanks in advance.

 

[Prototyp_V1.0]

I'm not sure of your question.

Peak current depends on frequency and value of C and R.

 

[EDA_hg81]

I'm not sure of your question.

Peak current depends on frequency and value of C and R.

Please correct me if I am wrong:

peak current dI= (C x dV)/dt and dt = RC, so dI = dV/R ( R is the resistance of parallel resistor).

Thanks.

---------- Post added at 01:15 ---------- Previous post was at 01:12 ----------

I am confunsing with the dt in equation dI= (C x dV)/dt, is dt = RC or dt = something else?

 

[Prototyp_V1.0]

Since you have no values for any components, the general answer is yes. If the voltage rise is quick enough, it should be able to deliver 200mA and more.

 

[EDA_hg81]

Sorry for forgeting providing the values.

The cap is 100nF, the resistor is 100k and the diode is BAS16XV2T1G.

for cap to be quickly charged to dV, the equation is as:

dV = (I x dt)/C, and dt = RC, so dV = I x R

do you think this is the reason why the value of R is 100k?

Thanks.

 

[Prototyp_V1.0]

dt = RC

No, that's illegal because you should get rid of the delta:

 

[EDA_hg81]

Thank you.

I know why should use 100K resistor it is for "The magnitude of R and C are chosen so that is large enough to ensure that the voltage across the capacitor does not discharge significantly during the diode's "Non conducting" interval"

Clamper (electronics) - Wikipedia, the free encyclopedia