Conduction State Of An Ideal Diode
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Conduction State Of An Ideal Diode

The circuit of figure 121.7 (a) consists of an ideal diode (the black triangle), voltages and resistors (R1, R2, R3). The values of the voltages are as indicated in the diagram. R1 = 5 Ω, R2 = 10 Ω, R3 = 10 Ω.

(a) Is the diode conducting current?

(b) If resistor R2 is removed from the circuit, will the diode conduct current?

The strings: S7P4A41 (Linear motion).

The math:
Pj Problem of Interest is of type motion (linear motion). Diode current is linear.

(a) Assume the diode does not conduct current
So, diode is replaced by an open circle (figure 121.7 (b)).
So, voltage across R2 = v1 = 12[R2/R1 + R3] --------voltage divider rule.
So, v1 = 12(10/15) = 8 V.
Now apply KVL to right-mesh of figure 121.7 (b):
So, v1 = vD + 11
So, diode voltage, vD = 8 - 11 = -3.
So, diode is reverse biased and therefore not conducting.

If initial assumption is that the diode is conducting
Then, diode is replaced by a short circuit (figure 121.7 (c)).
So, v1 = v2
So, (12 - v1)/R1 = v1/R2 + (v1 - 11)/ R3
So, v1 = 8.75 V
So, v2 = v1 = 8.75 < 11
So, current is flowing in the reverse direction
So, diode is reverse baised and is not conducting.

(b) Yes. Diode will conduct if R2 is removed from the circuit.

Blessed are they that have not seen, and yet have believed. John 20:29

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