Commom-Collector saturation

[patroclus]

Hello,

I was wondering how could a common collector bipolar transistor saturate in DC, if it uses the same power supply for colector and base.

In saturation, Vbc = 0.5V, but if common collector, Vc = Vcc, and base cannont raise above that, so... what happends if so much current enters the base (enough to saturate the transistor)??

 

[Roff]

Common collector can't saturate unless the collector-base junction is forward-biased. If collector=vcc, then the only way it can saturate is if the base is greater than vcc (by about 0.5V, as you said).

 

[patroclus]

So, what happends if you raise base current so much?? Soon or later
Ic = Hfe*Ib
won't be able to be reach and that would mean saturation??

What would happend then??

 

[Roff]

So, what happends if you raise base current so much?? Soon or later
Ic = Hfe*Ib
won't be able to be reach and that would mean saturation??

What would happend then??

Base current and base voltage do not exist independently. Remember that the B-E and B-C junctions are both diodes. If you drive the base with a current source instead of a voltage source, you will find that the base voltage will be about 0.5V above the collector when the transistor begins to saturate.

 

[patroclus]

I know.
But imagine you drive base with voltage source, and you make the base resistor very low. then, current trough base will be greater as you lower the resistor...

 

[Roff]

I know.
But imagine you drive base with voltage source, and you make the base resistor very low. then, current trough base will be greater as you lower the resistor...

It won't saturate unless you forward bias the B-C junction by at least 0.5V. The resistor could be zero ohms, and it is still true.
Here is a quote from Wikipedia:

The BJT enters "saturation" when the base current is increased to a point where the external circuitry prevents the collector current from growing any larger. At this point, the C-B junction also becomes forward biased. A residual voltage drop of approximately 100 mV to 300 mV (depending on the amount of base current) then remains between collector and emitter.

If you short base and collector to vcc, the transistor still has Vce~0.7V, and it still has substantial beta. Therefore, the base current will be substantially less than the collector current. The only way you can force more base current is to forward-bias the C-B junction, which means the base voltage must be above vcc, as I said before.

 

[Roff]

Here is a simple simulation, with the DC operating point calculated. I put GND on the collector for simplicity.

Here is the operating point:

       --- Operating Point ---

V(b):	 -3.11236e-012	 voltage
V(e):	 -0.6535	 voltage
V(vee):	 -10	 voltage
Ic(Q1):	 0.000931538	 device_current
Ib(Q1):	 3.11236e-006	 device_current
Ie(Q1):	 -0.00093465	 device_current
I(R2):	 3.11236e-006	 device_current
I(R1):	 0.00093465	 device_current
I(V1):	 -0.00093465	 device_current

 

[patroclus]

Great! thank you very much

 

[Nigel Goodwin]

This is why you don't generally use a transistor in that way, for a saturated switch you would use common emitter, not common collector.

It's also why push-pull audio amplifiers commonly use boot-strapping, in order to get sufficient base current for the top output transistor (usually the top one in most designs).