Anybody out there know what the term ‘generator peak inverse voltage’ refers to? I need to find out what the peak inverse voltage of a particular generator is for a job and I have no idea what it means
PIV (sometimes PRV) is a specification for a solid state diode & a vacuum tube. I have never heard it applied to a generator. What kind of generator is this?
the generator is on a small oil vessel. it is a 25 kw 120 VDC machine. there are two generators that can be run together in parallel. There is a diode for each gen. to prevent reverse current, from one gen. to the other.
to clarify…I posted the original question for jbudd, as he did not have access to a PC at the time. He is really the originator of the post, hence his response to the “what kind of generator is it?” question instead of me.
Power supplies are supposed to have extremely low output impedance, so that they may supply their maximum rated output current with a minimal power (I²R) loss. This can be a problem if the power supply’s output is presented with another voltage source. The low impedance of the generagtor will attempt to draw high current from the external source, and burn itself up in the process. It sounds like the generator has a diode in its output line to prevent current from flowing back into it. My WAG is that the PIV rating is the maximum externally applied voltage that can be applied to the generator’s output without breaking down the protection diode and destroying the generator. But I’d like to see a wiring diagram to be sure.
I scratch my head over how this can be an even minor factor in cost consideration. 1KV PIV rated diodes are a dime each.
A 200 amp diode for 10 cents? 
The term “peak inverse voltage” applied to a generator makes no sense to me. However, applied to the diode in series with the output to prevent reverse current it does make sense. I assume the diodes are silicon semiconductors. In the forward direction they conduct large current with a low voltage across the diode. In the reverse direction they conduct small current with large voltage across the diode, up to a point. This point is the reverse breakdown or zener voltage. For reverse voltages exceeding the breakdown voltage the current through the diode is essentially unlimited (by the diode).
You need to find out what kind of diode it is and look up the characteristics, or ask the manufacturer the value the reverse breakdown voltage is.
Whoops. I missed the “KW” :o
Make that 10 dollars. Maybe 20.
The thing I’d be most concerned about is overheating the diode(s). Assuming each generator is rated for 25 kW / 120 VDC (continuous), and assuming the diode’s voltage drop is around 0.7 V, the diode would have to dissipate a maximum continuous power of 146 W. If it’s convection cooled, make sure the cooling mechanism is in good working order (inspect fan, dust in heat sink, etc.)
Why should the diode have a 0.7V voltage drop ??
That’s the typical forward voltage drop for a silicon PN junction, although it can go up to a volt under heavy current flow.
The graph is a little misleading until you realize that the full X axis is not linear. If it were, the blue line would be nearly parallel to I[sub]f[/sub].
That’s the typical forward voltage drop for a silicon PN junction, although it can go up to a volt under heavy current flow.
The graph is a little misleading until you realize that the full X axis is not linear. If it were, the blue line would be nearly parallel to I[sub]f[/sub].
That’s the typical forward voltage drop for a silicon PN junction, although it can go up to a volt under heavy current flow.
The graph is a little misleading until you realize that the full X axis is not linear. If it were, the blue line would be nearly parallel to I[sub]f[/sub] .
whoops
No Attrayant, it is not. The 0.7V is the cut in voltage. That is if you impose a voltage of less than 0.7V, you’ll have no current flow through it. If you impose a voltage of 1 volt, the output will be 1V.
Lot of people misunderstand it, just like you did.
I’m pretty sure Attrayant and Crafter_Man are correct on this one. It’s actually meaningless to speak of “voltage in” and “voltage out” with a device like a diode. More accurate to say that a voltage of at least 0.7V (or whatever the voltage drop is for that diode) has to be maintained across the diode for current to flow. If you measure the voltage potential across the diode while current is flowing through it, it will always be equal or greater to the minimum forward voltage drop for that material - it won’t magically drop to zero once current starts flowing. And while the current is flowing the diode will have to dissapate energy equal to the voltage drop times the current flowing through it.
Conventional diodes are not voltage regulators. As Andrew said, there is no “output” voltage to speak of. In this thread we suspect the aforementioned diode to be in series with the output current. Something like D1 in this circuit, except it should be on the output side of VR1.
The output voltage is that which is provided by the generator, minus the drop across the diode. The diode is being used as a polarity sensitive switch. If it weren’t for the fact that the current flow is so great, we could ignore the .7 volt drop altogether.
And the conduction curve is not linear. Silicon diodes begin conducting as early as .5V (see the graph I linked to).