Half-Wave Rectifier Average Diode Current

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Discussion Overview

The discussion revolves around the calculation of average diode current in a half-wave rectifier circuit, focusing on the theoretical and mathematical aspects of the problem. Participants explore various parameters such as voltage values and the behavior of the diode within the circuit.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant requests help with calculating the average diode current based on a provided circuit diagram.
  • Another participant emphasizes the importance of showing work and breaking down the problem analytically, posing guiding questions about voltage values and diode behavior.
  • A participant provides specific voltage values, stating that 120 volts AC corresponds to a peak voltage of 680 volts across the secondary coil and mentions a forward voltage drop of 0.7 volts across the diode.
  • There is a clarification regarding the ideal diode assumption, noting that even an ideal diode has a forward voltage drop that should be considered in calculations.
  • One participant suggests calculating the average current by determining the area of the waveform divided by its period, hinting at using a sinusoidal function for the calculation.
  • Another participant claims to have figured out the average diode current calculation independently, stating it is Vp / π divided by the load resistor.
  • A subsequent reply questions the use of peak values for calculating average values, suggesting a different approach involving V_dc and R_load.
  • One participant notes that both their calculations yield the same result, indicating some level of agreement on the outcome despite differing methods.

Areas of Agreement / Disagreement

Participants express differing methods for calculating average diode current, with some agreeing on the results while others challenge the assumptions and approaches used. The discussion remains unresolved regarding the best method to use.

Contextual Notes

Participants reference specific voltage values and assumptions about the diode's behavior, but there are unresolved aspects regarding the ideal diode model and the implications of using peak versus average values in calculations.

Petrucciowns
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http://img12.imageshack.us/img12/7084/helpphg.jpg


In the image shown above, how is the average diode current calculated? This is an old homework assignment and I'm taking notes. I would appreciate any help.

Thank you.
 
Last edited by a moderator:
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You should still show your work / thoughts, even if it's an old assignment / homework.

Learn to break down the situation and look at the problem analytically. A brilliant / lucky engineer is sometimes able to spot something immediately and/or produce an answer from seemingly nowhere. A good engineer should be able to break down the situation logically, and work their way through a problem. Here are some guiding questions:

1) What's 120VAC mean in V_RMS and V_Peak?
2) What's V_peak on the output side?
3) How much voltage does the (assume ideal) diode drop?
 
Last edited:
MATHLABdude those are all questions that I already answered. The 120 volts AC is the AC voltage from the source (outlet). Primary Peak current is 120 volts RMS / .707.

The peak load voltage is that across the secondary coil. 680 volts peak.

In a forward biasing rectifier such as this ,the voltage drop across the diode is .7 volts.

Assuming it is ideal it has no resistance, and therefore, 0v across its terminals.
 
Petrucciowns said:
MATHLABdude those are all questions that I already answered. The 120 volts AC is the AC voltage from the source (outlet). Primary Peak current is 120 volts RMS / .707.

The peak load voltage is that across the secondary coil. 680 volts peak.

In a forward biasing rectifier such as this ,the voltage drop across the diode is .7 volts.

Assuming it is ideal it has no resistance, and therefore, 0v across its terminals.

Sorry, this wasn't made clear. I'm unclear of what has 0V across its terminals. The diode (even an ideal diode, with 0 resistance) has a forward voltage drop (the internal resistance just models the shift in the forward diode voltage one sees with changing current). Regardless, since 680 >> 0.5 V, the impact of that is little more than a rounding error.

Good, now that you have your voltage across the secondary, you can calculate the peak current, which I assume that you've done. And you know that the current must be sinusoidal, except for when the diode blocks out the negative going portion of the waveform.

So now what's the area of that waveform divided by its period? That's the average value. (Hint, you don't actually need to get an expression for the full waveform with frequency and what not, just use a generic sin function, and scale the results up by your peak current).
 
Thanks for the help, but I figured it out on my own. The average Diode current is Vp / pi which is then / the load resistor.
 
Petrucciowns said:
Thanks for the help, but I figured it out on my own. The average Diode current is Vp / pi which is then / the load resistor.

You mean V_dc / R_load? Using the peak values won't give you the average values you're looking for.
 
Both of them, yours and mine come up with the same result.
 

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