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I was just thinking about diodes recently and I came across this video

https://youtu.be/bXEyCf1P0UU?t=35 Minute 0:32

He says the power disspitated by the diode will be the forward voltage drop Vf (0.7V usually) times the current through it. Now, that's my confusion. Why would the power drop only be 0.7V times the current. I mean if I applied 2V across it, it should be 2V*current not 0.7V right?

I thought the 0.7V=Vf is the minimum voltage across it we need (with correct polarity) for it to start conducting,

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  • \$\begingroup\$ You are right. I didn't watch the video, but typically diodes operate with something like 0.6 to 0.7V across them. With 1V across the diode, it may be conducting a lot of current and in danger of over-heating. With 2V across a diode, the current will be excessive and the diode will be in danger of quick death. But in all cases, the power dissipated by the diode is Vf * If. And you use the actual Vf when it is known. Or you can use Vf=0.7 when it has some moderate current through it. \$\endgroup\$ – mkeith Dec 9 '18 at 1:07
  • \$\begingroup\$ you use diodes in series with other components. if you replace a wire from Vin with a diode, the first component will see 0.7v less than Vin. If you replace a wire to gnd with a diode, the last leg of the circuit will be at 0.7v instead of 0v. \$\endgroup\$ – dandavis Dec 9 '18 at 2:16
  • \$\begingroup\$ Another place to skim a little would be here at EESE. \$\endgroup\$ – jonk Dec 10 '18 at 7:30
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As an approximation, the voltage drop across a silicon diode when it is conducting is always about 0.7V, so the power dissipation is current multiplied by 0.7V.

If you put 2V (forward bias) across the diode, for example a 1N4005, the current will be many times the maximum rated current and it will quickly burn out.

In practice the actual voltage will be less at very low currents and/or high temperatures and will be higher at very high currents and/or low temperatures but this is a good approximation for many everyday purposes.

Remember, the device and operating conditions (temperature, mainly) determines the forward drop at a given current. For any set of conditions you get to pick the current OR the voltage, not both.

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  • \$\begingroup\$ Ah. So basically under normal conditions when I see a circuit (properly built that is), I can assume that the silicon diode will have around 0.7V across it. I can however apply a higher voltage across it which will increase the current through it and risk burning it. \$\endgroup\$ – AlfroJang80 Dec 9 '18 at 3:09
  • \$\begingroup\$ Not risk burning it, you will burn it. Much more than the forward voltage and you're in the hundreds or thousands of amperes (though you'd be limited by parasitic resistance) \$\endgroup\$ – Hearth Dec 9 '18 at 4:28
  • \$\begingroup\$ +0.7 (or so) if it's conducting some reasonable fraction of the rated current, it could be a large negative voltage of course. There are some circuits where the exact voltage (and how it varies with current and temperature) is very important, even a few mV or uV matters and in those cases you have to use a more sophisticated model. \$\endgroup\$ – Spehro Pefhany Dec 9 '18 at 5:25
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The voltage across an abrupt-doping-profile diode changes by 0.058 volts per 10:1 change in current, by 0.026 volts for E^1 change in current (with E^1 being 2.718...), and by 0.018 volts for 2:1 change in current.

Now, if we assume the diode passes 1milliAmp at 0.06 volts, lets make a table

Current Voltage

10 milliAmps 0.600 + 0.058

1mA 0.600

0.1mA 0.600 - 0.058

0.01mA 0.600 - 2*0.058

0.001mA (1uA) 0.600 - 3*0.058

100nanoAmp 0.600 - 4*0.058

10nanoAmp 0.600 - 5*0.058

1nanoAmp 0.600 - 6*0.058 = 0.600 - 0.348 = 0.252volts

Notice the voltage across the diode, if doped in a certain profile, drops by 58 milliVolts per 10:1 reduction in current.

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