Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

percentage of diode conducting in rectifier

Status
Not open for further replies.

ArFa

Junior Member level 2
Joined
Dec 27, 2010
Messages
23
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,438
Hi everyone...
Can anyone tell me how to find the % of the diode conducting in half-wave rectifier?
 

Well, I don't know to be honest.

If you can define "diode conducting" then I shall be able to help.
 

I mean to find when diode is conducting
 

It depends on the circuit and application but in general I would say a little under 50% of the time.

Keith
 

in half wave rectifier circuit ideally diode conducts 50% of time but practically it is less than 50% of time.
for conduction time, consider the duration in which diode ON or conducts.
 

Yes, considering forward bias voltage fo diode, it will be less than 50%.
 

If you want to estimate the percentage of time without simulation then draw your sine wave and mark the points on the half wave where the diode volts drop is exceeded.

Keith
 

I fear, the question is meaningless (and phrased vague, anyway) without giving additional parameters. A half-wave rectifier may e.g. have a capacitor or battery load ... In so far, the laconic answer from Protyp_V1.0 seems appropriate.
 

If the load as seen by the diodes is inductive or resistive then the diodes conduct for the complete half cycle. If the load has part capacitive load then the angle of conduction depends on the amount of ripple on the output. The lower the ripple the smaller the angle of conduction. Draw out your waveform, put a line across at peak voltage and another at peak volts- ripple volts. Look at the area of your waveform between theses lines, this is the conduction angle. Remember, peak volts = transformer volts - volt drop across diode.
Frank
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top