A diode with a resistor in parallel and in series.

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SUMMARY

The discussion focuses on a circuit involving a diode with a series resistor (R1 = 3kΩ) and a parallel resistor (R2 = 470Ω) connected to a voltage source (Vs = 20V). The diode has a voltage drop (Vd) of 2V, leading to a voltage of 18V across R1. Participants clarify that the voltage drop across the diode is not equivalent to the voltage across the resistors, enabling the calculation of current through R2. The insights gained emphasize the importance of understanding voltage drops in circuits with diodes and resistors.

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  • Basic understanding of electrical circuits
  • Familiarity with Ohm's Law
  • Knowledge of diode characteristics and behavior
  • Experience with series and parallel resistor configurations
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  • Learn about calculating current in series and parallel resistor circuits
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Electronics enthusiasts, electrical engineering students, and professionals working with diode-resistor circuits will benefit from this discussion.

Trespaser5
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I have a diode with a resistor in parallel(R2), but also diode resistor circuit has a resistor in series with it(R1). I have uploaded a drawing as well but I think it takes a bit of time for them to be authorised.
The values of the circuit are
Vs = 20v
R1 = 3kΩ
R2 = 470Ω
Diode = Has a Volt drop of 2v

I've found the formula Vo-Vd/Rl = I for a diode in series in a resistor but am not sure how this relates if the diode also has resistor in parallel with it. I'm guessing from what I've read that the diode does not have a resistance as such but only 'consumes' 2v, so can I use 18v along with the resistance in the remainder of the circuit to calculate It ? Or am I heading down the wrong path entirely. This is my second night on this question so could really do with some help. Thankyou
 

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My .doc viewer isn't working, but I think I can see the circuit. That would be a zener diode, and, yes, it leaves 18V across R1.

Knowing the voltage across R2, you can calculate the current through R2.
 
Thank you so much. I think I was mis understanding that the voltage drop stated was the same as the voltage shown across, when that twigged today I got it, what a feeling :) Thank you for your help and time :)
 

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