Transistor Base to Emitter Voltage

[alexhong81]

just wondering, what is the difference between Base-Emitter Saturation Voltage and Base-Emitter on Voltage ?
VBE (sat) = 0.5V
VBE (on) = 0.7V

 

[cabraham]

The sat value should be higher than the on value. The first value of 0.5V sounds more like "Vcesat" instead of "Vbesat". Can you double check.?
Anyway, when a bjt is driven fully into saturation, Vce drops to a low value, 0.1 to maybe 0.5V, or even higher/lower in some cases. The on voltage for moderate current values at room temp are around 0.65V. If driven harder, the bjt saturates, & Vbe increases a little.

Claude

 

[alexhong81]

I am sorry, you are right, I got the wrong value here.
I am looking at this NPN transistor ZXTD6717E6, Diodes Inc.

VCE(sat) = 0.245V max (havent seen such low VCEsat before)
VBE(sat) = 0.93V typical
VBE(on) = 0.865V typical

If i am not wrong, When VCE > 0.245V and VBE > 0.93V the transistor will "on" and current ID will flow.
Then what is VBE(on) 0.865V? What is the difference between VBE(sat) and VBE(on)?

Thanks~~

 

[cabraham]

Vbeon must be specified at some value of colector current & some temp, usually 25C. If the bjt is in its active region, i.e. Vce>0.7V or so, then Vbeon is the nominal forward voltage drop across the b-e junction.

To saturate the device, a base current greater than Ic/hFE is needed. Most general purpose bjt devices spec saturation at Ic/Ib equal to 10. In this case, the base is overdriven so that Vce drops to a low value, known as Vcesat. This is a condition where the device acts like a closed switch.

Because the base current is higher than its value when the device is in the active region, Vbe is higher as well, & this is Vbesat. Since Vbesat occurs when base current is increased, then Vbesat>Vbeon. But the voltage across a forward biased p-n junction increases only as the logarithm of the current, a large increase in Ib results in a relatively small increase in Vbe. Hence Vbesat is larger than Vbeon by a small margin.

Did this help?

Claude

 

[alexhong81]

Thanks Claude.
See whether my interpretation of your explanation is correct:

BJT in active region, VCE>VCE(sat): VBE(on) = nominal base emitter voltage.
BJT in saturation region, VCE<VCE(sat), Ib>Ic/HFE
BJT in active region, Ib higher, VBE(sat) higher.

1. What happens when VBE is lower than VBE(on)?
2. What happens when VBE is higher than VBE(on)?
3. What happens when VBE is lower than VBE(sat)?
4. What happens when VBE is higher than VBE(sat)?

Thanks again Claude

 

[logicman112]

alexhong81: BJT in active region, VCE>VCE(sat)

BJT is in active region when: Nearly VCE > VBEon

1. What happens when VBE is lower than VBE(on)?
All Transistor currents are nearly zero: IB=IC=IE=0

2. What happens when VBE is higher than VBE(on)?
For simple Bias circuits VBE can not be higher than VBEon. Driving BE diode above VBEon may lead to Avalanche break down. But transistor CAN work with over voltages, its work is highly dependent to the type of transistor and the circuit and we will have to consider the precise transistor physical models like Ebers-Moll or Gummon-Poll models.

3. What happens when VBE is lower than VBE(sat)?
There is no VBEsat for first order circuit analysis of transistor bias circuits. We have VCEsat!

4. What happens when VBE is higher than VBE(sat)?
The question is wrong!

Microelectronic circuit design by Richard Jaeger is highly recommended.

 

[alexhong81]

Thanks...guess i will have to do some reading up first...