# Current Collector

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#### ren_zokuken01

##### New Member

It's just one of those things that you've just assumed to be true whenever I'm playing around with BJTs, but never actually got explicit confirmation from anybody. But, doesn't the formula I_c = hFE x I_b only determines the maximum current that can flow on the collector end? Say in the figure above, the collector current is
(9V - 0.2V)/1k ohm = 8.8mA
and not
(5V - V_be)/1k ohm x 140 = 602mA (assuming hFE = 140)
??

Well is it??

I need this very simple assumption or else I've just plugged-in garbage in my calculation.

#### Ian Rogers

##### User Extraordinaire
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(5V - V_be)/1k ohm x 140 = 602mA
I would say... This is what it can drive... doesn't mean that's what it is driving..

The limitations are in the 9v circuit....

#### ronsimpson

##### Well-Known Member
If the transistor was 100% on it, in this circuit, can only have a collector current of 9V/1k. So R2 and your 9V battery are the limiting factor not the transistor. The transistor will have some small voltage across C-E so the voltage will be about 9-0.2V.

If R2=0 ohms then the transistor will be the limiting factor.

#### Ratchit

##### Well-Known Member

It's just one of those things that you've just assumed to be true whenever I'm playing around with BJTs, but never actually got explicit confirmation from anybody. But, doesn't the formula I_c = hFE x I_b only determines the maximum current that can flow on the collector end? Say in the figure above, the collector current is
(9V - 0.2V)/1k ohm = 8.8mA
and not
(5V - V_be)/1k ohm x 140 = 602mA (assuming hFE = 140)
??

Well is it??

I need this very simple assumption or else I've just plugged-in garbage in my calculation.
Why drag "beta" into your reasoning and calculations? The transistor is operating in the saturated region instead of the active region, so the collector current is determined by the collector voltage, resistor, and voltage across the CE terminals.

Ratch

#### AnalogKid

##### Well-Known Member
I need this very simple assumption or else I've just plugged-in garbage in my calculation.
You are correct. When the possible collector current is limited by external factors such that it is less than the base current times the beta, this is called the saturation region of transistor operation. In this region, the effective collector-emitter equivalent resistance is the lowest it can be. So is the transistor power dissipation.

ak

#### crutschow

##### Well-Known Member
When hFE x Ib becomes less then 9V / R2, then the transistor will come out of saturation and the collector current will become limited to that calculated using hFE.
(Note that we are talking hFE for the actual transistor you are using which has a large variation from unit to unit as shown in the data sheet).

#### MikeMl

##### Well-Known Member
Usually, the problem is worked the other way. Say Vcc = 9V and Vbb=5V, as shown. The collector load R2 resistance is 1KΩ.

What is the maximum value of R1 that will keep the transistor saturated?

Make Vce= ~5V?

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