Diode and Ohm's Law question

Thread Starter


Joined Jan 5, 2011
I have this diode that is rated at 1000Volts @ 1 amp. This means that it can handle 1000 watts of power. With use of Ohm's law, W = VI, does that mean that I can pass 83 Amps through this diode @ 12 volts? Or does that law not apply?



Joined Apr 20, 2004
Not at all. The PIV rating (peak inverse voltage) will let it block up to 1000 volts. The junction can pass 1 amp of current at any voltage under 1000. Diodes are not resistors.

That's probably a 1N4007?
For an ideal diode, the forward voltage across the diode when it is conducting current will be .7 volts. In this condition it can conduct up to 1 amp of current. The 1000 volts is when the voltage is placed across the diode in the opposite direction. When the diode is reverse biased in this way the current through the diode is essentially zero. Above 1000 volts in the reverse direction the diode will breakdown or get destroyed.


Joined Dec 20, 2007
The maximum peak forward current of a 1N4007 rectifier diode is 30A for a duration no longer than 8.3ms (a half-cycle of a 60Hz sinewave).


Joined Nov 30, 2010
ecg125 is another number for 1n4007, and no, a part labeled 1 amp will not survive 83 amps, and the 1000 reverse volts that you don't apply to it times 1 amp maximum continuous forward current does not equal 1000 watts.

I think you need to read about diodes.


Joined Sep 30, 2009
For an ideal diode, the forward voltage across the diode when it is conducting current will be .7 volts. In this condition it can conduct up to 1 amp of current.
I have a question about this statement. In almost every text and post this is said. But when you look at a data sheet of diodes there is usually a graph which shows it different.

The .7V drop is(the way I see it) just a threshold voltage drop. As the amperage through the diode increases, the voltage drop also increases. So why do people say that the voltage drop is only .7V when it can be much more in a circuit?

This for me is one of the reasons that electronics is so hard to learn.


Joined Nov 30, 2010
It's laziness. Anyone that designs circuits has seen the graphs of forward voltage compared to forward current. Most people here say ".7" because it works for hobby stuff.

Most people also say .6 volts for a forward biased bipolar transistor. I did the experiments and found that Amps (collector) compared to Volts (base to emitter) is a straight line on logarhythmic graph paper. In other words, all the voltages from really tiny to about a whole volt can show up as Vbe depending on Ic. Are you going to explain that to every beginner?


Joined Jul 17, 2007
The 0.7v reference is just for convenience's sake. A 1N400x diode with 1A current flowing through it will have close to a 1V drop across itself (around 0.95v).

I did an informal test of a 1N4002 diode a few years ago; with 20uA current flow the Vf was 0.4v; .2mA was 0.5v, 2mA was 0.6v, 20mA was 0.7v. I didn't go to 200mA, but extrapolating from the plot I made, it would be close to 0.8v.

StayAtHomeElectronics mentioned an "ideal diode" that are frequently characterized with a Vf of 0.7v - and it doesn't matter how much current you put through them, the Vf will stay the same. That shouldn't be confused with a MOSFET as an "ideal diode" in synchronous switching supplies or synchronous rectifiers; as in that situation the MOSFET will not have a Vf to speak of (except for the intrinsic body diode), only the Rds(on) would be considered.
The 0.7 volt number for diodes is an approximation. The initial comments seemed to show a little misunderstanding of the functionality of a diode. A simplified model is the best place to start, especially given that there are no real circuit associated with the question.

Considering the 0.7 volt forward drop across a diode as a threshold voltage is also simplification the diode. It is not true that the diode does not conduct below 0.7 volts. It is true that once you get below that approximate voltage that the current that passes through the diode gets small, or even very small. That is what makes that simplification valid in many cases.

For the 1n400x, the forward drop is typically 1.4 volts at 20 Amps. That comes from the Fairchild Semiconductor datasheet. So, 20 uA gives 0.4v drop across the diode, from Sgt Wookie's informal test above, and 20 A gives 1.4v. 0.7 volts is not a bad starting point when you do not know the operating region of the parts.

Also, without some initial calculations you do not know what parameters are required in the part, so you can not always start with a part number or characteristic curves.