Half-Wave Rectifier: Understand & Calculate Output Voltage

[Bassalisk]

Hello,

I want to come clean with rectifiers. Mainly half wave rectifiers with filter.

[PLAIN]http://www.sciencelobby.com/diodes/images/reservoir-capacitor.gif

Assume these conditions. Diode CVD is 1 V. At t=0 s, or initially, capacitor does not have starting charge. So everywhere else potentials are 0.

When we turn on the circuit, at first, diode won't conduct until 1 V is reached. That is small compared to those 12 volts. Diode will conduct until capacitor is fully charged, or until sinusoid reaches its peak(maximum).

After that, sinusoid starts to drop, but we have that voltage on the capacitor which is now higher than the voltage of the source. So diode doesn't let the current run in other direction and it is in reverse bias. Diode won't let any current until voltage of the source becomes higher than that on the capacitor right? And then capacitor recharges etc.

My question here is:

Did I got this right? How do we calculate output voltage? Its not steady because of that ripple.

If you calculate Vdc=Vmax-Vdiode, Why? Why would you subtract the diode voltage of the Max value because, sinusoid still reaches it maximum.

Does diode gets voltage drop like resistors? Or it just behaves like an element that needs certain voltage to let current through?

(and thanks PF and EE fellows I just passed all my exams with decent marks.)

 

[aks786]

it just behaves like an element that needs certain voltage to let current through

you got it right. diode has PN junction and PN junctions have minimum voltages to operate. and on load diode will offer voltage drop due to internal resistance

 

[Bassalisk]

Then why would be subtract the CVD from Vmax? Isn't there a presence of the SAME Vmax on the output?

 

[Jiggy-Ninja]

Because the diode takes 0.7 volts to be forward biased, and it takes 0.7V to keep it forward biased.

In that respect, you can think of it as a constant voltage resistor. Whenever you send something through a diode, 0.7 volts is lost.

 

[Bassalisk]

Hmmm so I did got that wrong. I thought that AFTER diode gets on the 0,7 Volts, it "transfers" that voltage on that parallel capacitor. So basically I thought like diode does nothing for 0-0,7 V then let's the current through, and likewise voltage is transferred...

So in a nutshell, when I get to those 0,7 V, those volts are lost at the diode, and the potential "after the diode in the circuit" is going to be 0 and then rise to Umax-0,7V (or 1V as I said in problem)

EDIT:

Can i get the oscilloscope the measure the current instead of voltage in National Instruments multisim?

 

[skeptic2]

Yes there are current sensors for oscilloscopes. They clamp around a conductor.

One thing to think about is that where the red line in your diagram is rising is the only time the rectifier is conducting, while the load is generally drawing current all the time. That means that the diode must be capable of conducting many times the current required by the load.