Half-Wave Rectifier: Understand & Calculate Output Voltage

In summary, when you turn on the circuit, the diode will not conduct until the capacitor is fully charged. Then, when the capacitor begins to discharge, the diode begins to conduct and the current will continue to flow in the opposite direction until the voltage on the capacitor becomes higher than the voltage on the source.
  • #1
Bassalisk
947
2
Hello,

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

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

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.)
 
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  • #2
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
 
  • #3
Then why would be subtract the CVD from Vmax? Isn't there a presence of the SAME Vmax on the output?
 
  • #4
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.
 
  • #5
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?
 
Last edited:
  • #6
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.
 

1. What is a half-wave rectifier and how does it work?

A half-wave rectifier is an electronic circuit that converts an alternating current (AC) input into a pulsating direct current (DC) output. It uses a single diode to only allow the positive half of the AC input signal to pass through, while blocking the negative half. This results in a unidirectional output voltage.

2. What is the output voltage of a half-wave rectifier?

The output voltage of a half-wave rectifier is equal to the peak value of the input AC voltage minus the voltage drop across the diode. It can be calculated using the equation Vout = Vpeak - Vd, where Vpeak is the peak value of the input voltage and Vd is the voltage drop across the diode.

3. How do I calculate the output voltage of a half-wave rectifier?

To calculate the output voltage of a half-wave rectifier, you will need to know the peak value of the input voltage and the voltage drop across the diode. You can then use the equation Vout = Vpeak - Vd to determine the output voltage.

4. What is the efficiency of a half-wave rectifier?

The efficiency of a half-wave rectifier is the ratio of the DC output power to the AC input power, expressed as a percentage. It is typically around 40-60%, as it only utilizes half of the AC input signal.

5. What are the applications of a half-wave rectifier?

A half-wave rectifier is commonly used in low-power applications where a unidirectional DC output is required, such as in battery chargers and small electronic devices. It is also used in signal processing circuits, where the negative half of the input signal needs to be eliminated.

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