Question about Vceo

Status
Not open for further replies.
The way I see it, Vcbo is the fundamental breakdown voltage where avalanche or punch through starts, and although specified at a particular current the current rise is very steep.
Vebo is more a mesure of the current increase due to the beta multiplication effects.
The associated I-V curves are shown using a common base stage for Vcbo (at various emitter currents) and with a common emitter stage for Veco.

If one is using a common base stage or some constant emitter current then the circuit is self compensating in that Vbe will reduce to maintain the constant emitter current and any collector to base current will be swept up by the base prior to reaching the emitter.
Whereas for a common emitter stage the collector to base charge gets to the emitter where beta multiplication takes effect.
So Vceo approaches Vcbo as the resistance from the base goes from open to short or as the charge from the collector into the base is removed before reaching the emitter respectively.

As I said, thats just from my point of view or way of viewing a BJT.

Thanks
-Antonio
 
"So Vceo approaches Vcbo as the resistance from the base goes from open to short or as the charge from the collector into the base is removed before reaching the emitter respectively."

I think you are absolutely right here, except to say "Vceo approaches ..." is confusing as Vceo is a function of the device under particular circumstances (B = open) and does not change. Rather "Vce approaches ...."

I am probably being pedantic, but my approach comes from years of device manufacture and test rather than application.

In my context

Vce breakdown voltages at Ic = ?

Vceo = base o/c
Vces = base short (tends to Vcbo)
Vce(sus) = low voltage Vce sustaining voltage
Vcev = reverse bias on base
Vcex = some impedance network on BE.


I suspect I am drifting way off topic so I'll shut up!
 
Status
Not open for further replies.