Predicting maximum current flow of diode in half/full-wave rectifier

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t0mbst0n3

Newbie level 3 Diode forward voltage drop = 0.6 V
R3 = 18 kΩ
Vs = 10sin(2π*10000t) V

Half wave rectifier circuit diagram: Full wave rectifier circuit diagram: How do you predict the maximum current flow for the D1 diode for both scenarios? I'm stuck at this part because the Vs value has an unknown variable.

KlausST

Super Moderator
Staff member Hi,

What do you mean with unpredictable variable?
In any case you know it's a sinwave and you know it's amplitude.
Thus the max voltage is clear.

And because there is an ohmic load .... just use Ohm's law.

Klaus

t0mbst0n3

Newbie level 3 Hi,

What do you mean with unpredictable variable?
In any case you know it's a sinwave and you know it's amplitude.
Thus the max voltage is clear.

And because there is an ohmic load .... just use Ohm's law.

Klaus

Sorry, I don't get you at the "it's a sinwave and you know its amplitude." Amplitude is 10, so therefore maximum voltage is 10? Then what is the point of the Vs equation?

KlausST

Super Moderator
Staff member Hi,

Amplitude is 10, so therefore maximum voltage is 10?
Correct. This is true for every pure sinewave.

Then what is the point of the Vs equation?
That´s not the question.
But maybe it´s just to check if you recognize that the max voltage is 10V.

Klaus

• t0mbst0n3

t0mbst0n3

Points: 2

FvM

Super Moderator
Staff member How do you predict the maximum current flow for the D1 diode for both scenarios?
Diode current is equal to I(R3) = Vo/R3. You want to calculate Vo, which is Vs - diode voltage drop (0.6 or 1.2 V respectively).

There's no D1 in circuit 2, b.t.w. Apparently you mean a different diode.

• t0mbst0n3

t0mbst0n3

Points: 2

t0mbst0n3

Newbie level 3 Diode current is equal to I(R3) = Vo/R3. You want to calculate Vo, which is Vs - diode voltage drop (0.6 or 1.2 V respectively).

There's no D1 in circuit 2, b.t.w. Apparently you mean a different diode.

Alright, understood.

For circuit 1, Vo = Vs - 0.6 because there is only one diode, yes? Just want to confirm.

For circuit 2, the diode current must be found for either D3/D4/D5/D6, since all four diodes share the same forward voltage drop = 0.6 V. So based on that logic, would the diode current be equal to I(R3,4,5,6) = Vo/R3, where Vo = Vs - 0.6 - 0.6 ? I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.

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FvM

Super Moderator
Staff member Voltage drop of the bridge rectifier is always 1.2V.

KlausST

Super Moderator
Staff member Hi,

I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.

Just to clarify: In an usual bridge recitifier circuit:
* for the positive halfwave diodes right_top and left_bottom are active and thus these two are connected in series.
* for the negative halfwave diodes left_top and right_bottom are active and thus these two are connected in series.
... although these diodes don´t have direct connection at the bridge rectifier circuit.

Klaus

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