Why is Vce(sat) Constant ?

vcesat

I have an uncleared but very basic doubt for a very long time ... since i started studying semiconductor....

Why is Vce(sat) (Collector Emitter voltage of a transistor at saturation) constant (around 0.2V) for all transistors whatever be the biasing voltage, power handling, power rating, beta, size etc. Once transistor is switched on, there won't be any depletion layer and the semiconductor will act just as a conductor (with some finite resistance), right ? Then why it is so ?

Thanks in advance

transistor vce voltage drop

It is because both the junctions in the transistor are forward biased at saturation. Under this condition, for npn transistor the emitter to base voltage is ~+0.7V and than between the base to collector ~0.5V (base p, collector n). The difference between these saturation diode voltages is because of the difference in doping profile of the emitter and collector. If you remember the contact potential of a diode is an increasing function of ration of the equilibrium majority carrier to the minority carrier. In transistor the emitter is much densely doped w.r.t. the collector which leads to higher junction contact potential at the base-emitter junction than the other junction.

vce drop

Thanks for the immediate reply.

Then one more doubt : If diode cut-in voltage is a function of doping concentration, why are almost all silicon diodes having forward drop around 0.7V and germanium around (0.3V) almost irrespective of part number. Usually power diodes will have high doping concentration, right ? for them also the drop is 0.7V!!! why ?

I am having the idea that all Si diodes have 0.7V and Ge diodes have 0.3V drop and the forward voltage drop mostly depends only on the type of semiconductor material (Si or Ge).

Please clarify this also...

- SUjO

Because if you set the electron density in the n side to a fixed value in case of both Si and Ge p-n jn diode the hole density in the n side will be different. If you remember in usual notation:

n * p = n(i)^2 = p(p) * n(p)

where n, p are electron and hole density in the n side and n(i) is intrinsic concentration for the particular material. n(i) is a function of Eg the band gap and therefore for different material n/p are different. Contact potential, which is the maximum forward voltage across the diode is

Vc = kT/q * ln[n/p] = 2kT/q * ln[n / n(i)]

n(i) is decreasing function of band gap [~exp(-Eg/kT)], so for Si n(i) is small compared to the case of Ge. Therefore n/p ratio for Si is higher than Ge and hense the contact ptential is too.

I hope this is clear.

sujojk said:

Thanks for the immediate reply.

Then one more doubt : If diode cut-in voltage is a function of doping concentration, why are almost all silicon diodes having forward drop around 0.7V and germanium around (0.3V) almost irrespective of part number. Usually power diodes will have high doping concentration, right ? for them also the drop is 0.7V!!! why ?

I am having the idea that all Si diodes have 0.7V and Ge diodes have 0.3V drop and the forward voltage drop mostly depends only on the type of semiconductor material (Si or Ge).

Please clarify this also...

- SUjO

Click to expand...

you should not mix this concept of diode modelling with actual voltage drop across diodes. whatever you have quoted is merely a diode model (i.e. 0.7 V drop for Si diodes) called constant voltage drop model. it is used for the sake of ease in analyzing diode circuits using linear approximations.

Voltage that appears across a diode is determined by the external circuit i.e. supply voltage and the biasing resistor. for instance a Si-diode being forward biased with a voltage source of 1.2V and a biasing resistor of 21kΩ has a voltage drop of approximately 0.45V (which offcourse is not equal to 0.7V)

If there is still a confusion, do ask

Regards,
Saad

To my understanding the 0.7V, as was mentioned was meant to be the maximum voltage. In case of the situation of 1.2V supply, 21kΩ the diode is not saturted. The diode is subjected to electrical circuit and has therefore to satisfy the kirchoff's law. This limits the voltage across the diode.

You must look at datasheets to see that the base-emitter voltage of a 2N3904 is typically 0.6V at 100uA collector current and is 0.8V at 100mA collector current. Its max and when the transistor is saturated is higher.

The base-emitter voltage for a 2N3055 is typically 1.9V at 9A collector current.

saad said:

Voltage that appears across a diode is determined by the external circuit i.e. supply voltage and the biasing resistor. for instance a Si-diode being forward biased with a voltage source of 1.2V and a biasing resistor of 21kΩ has a voltage drop of approximately 0.45V (which offcourse is not equal to 0.7V)

If there is still a confusion, do ask

Regards,
Saad

Click to expand...

Yes, I agree. But we were discussing about transistors in saturation region which means both junctions are fully saturated. 1.2V with 21k resistor will not saturate the junction. It will be working on the lower flat area of the VI curve.

However, I got the thing !!! thanks to all... especially to subharpe