What is foward breakover voltage as applied to a scr?
- Ray;mondLv 78 years agoBest Answer
Breakover maybe a better term than than break down as the scr typically is not damaged by breakover, but the load may be damaged. Typically it occurs at about 1100 volts for an scr rated 1000 volts. The leakage microamps simulate a gate pulse. turning the scr on without a turn on signal. If the anode voltage is ac, the scr typically resets to off during the next negative half cycle, so typically the load is not damaged by a single pulse. NeilSource(s): I worked at scr testing long ago
- 8 years ago
well as know that SCR is a THREE pin device ANODE,CATHODE,GATE. when the SCR is forward biased i.e when ANODE is connected to POSITIVE voltage and CATHODE to negative voltage then the SCR is in forward biased state. in such a condition without applying GATE signal to SCR we can turn on the SCR, the method so called FORWARD VOLTAGE TRIGGERING. you know that SCR is 4 layer ,3 junction device. the junctions are J1,J2,J3. when the SCR is forward biased then the junction J1 and J3 disappears and the whole voltage that is supplied when SCR is forward biased appears across the reversed biased junction J2. if we keep-on increasing the voltage to SCR( which is nothing but the voltage across the reversed biased junction J2) this junction breakdown. which is the phenomenon that is present in PN diode when it is reverse biased. this voltage supplied to SCR when the junction J2 breaks is called the forward breakover voltge
- DavidLv 68 years ago
Do you mean "breakdown" voltage?
It's been a while since I messed with these things, but wouldn't that voltage vary according to the particular device?
There's a "depletion zone" right where the n-doped material meets the p-doped material, and to get current to flow from n to p, you've got to have 7/10ths of a volt for silicon doped material, and 3/10ths for germanium doped material.
I don't know what it would be if they used germanium on one side and silicon on the other, but there it is.
If you ran the current the other way, I guess it would depend on the way the device was made.Source(s): A foggy memory . . .