• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Capacitor discharging.

Status
Not open for further replies.

alphacat

New Member
Mr. Ron has showed me this circuit:


I wanted to ask please, how come the diode is always in forward state?

Thank you for any help.
 

Attachments

crutschow

Well-Known Member
Most Helpful Member
The pulse output goes from +2V to +10V.
 

alphacat

New Member
Thank you.
I have 2 questions please:
1. Could you tell me please what is the meaning of the numbers 0,200n,200n,800n,2u?
2. Even if the lower limit of the voltage source is 2V, how come the diode stays in forward state all the time?
 

crutschow

Well-Known Member
Most Helpful Member
1. I not completely familiar with the syntax, but I believe the 200n and 200n are the 200ns rise and fall times, the 800n is the 800ns flat portion of the pulse, and the 2u is the 2us waveform period.

2. The diode always has a positive voltage applied to it's anode with a minimum of +2V. What's not to understand about that keeping it forward biased? As long a diode has a more positive voltage on its anode then its cathode, it will remain forward biased.
 
Last edited:
Status
Not open for further replies.

EE World Online Articles

Loading
Top