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I guess it's a fundamental question...

Let say I have a diode_1 that can handle 1W power and another diode_2 that can handle only 0.1W power. If these two didoes are connected in series (diode_1 first), what would be the power handling?

If I apply 1W to these series-connected didoes, would the 2nd diode be damaged?

Thanks!

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  • \$\begingroup\$ Assuming they have the same forward voltage at the same current ... 0.2W. \$\endgroup\$ – Brian Drummond May 16 '18 at 20:29
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If I apply 1W to these series-connected didoes ...

Didoes?

We don't talk about "applying power" to a diode. We pass current through it and power is dissipated in the process. Power dissipated in a device is given by \$ P = VI \$ where P is in watts (W), V is the voltage across the device in volts (V) and I is the current through the device in amps (A).

The voltage across a typical silicon diode is about 0.6 to 0.7 V over a wide range of currents. If we pass 1 A through your series connected diodes they will both dissipate \$ 0.6 \times 1 = 0.6 \ \mathrm W \$.

The 1 W diode will survive.

The 0.1 W diode will rapidly heat up and the internal junction will be destroyed. The diode will fail.

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Assuming they're both the same type diodes with close forward voltage drops, then it's limited by the lower power rating. Here's why.

The power dissipation in the diode is due to the current through it times the forward voltage drop. In a series circuit, the same current flows through both. Therefore, each diode (voltage drop) will consume power, and if the forward voltages are equal, then the power will be equal.

You wouldn't apply power to the diode. It's consumes power based on what I described above.

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