Basic Transistor question

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colin mac

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Hi.

I'm looking at an example in a book which gives this circuit. The beta is given as
20. It says to calculate Ib, it's 10V - vbe = 9.3/33K = 282uA.

It then says to calculate Ic, it's Ib * beta = 282uA * 20 = 5.6mA. Now you can calculate Vrc which is 5.6mA * 1k = 5.6V.

Is the book right? Isn't that saying that regardless of Rc, the current will be Ib * beta. What happens if Rc is a lot bigger? Thanks in advance.
 

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The higher is the value of Rc then the higher is the voltage drop across it. When the voltage drop approaches the power supply voltage then the transistor is saturated and its collector voltage will be about 0.1V.
 

In the active region, Ic = Ib * Beta as you said.

But if Rc (or Ib) is large enough, the transistor will saturate as Audio said.
 
Beta is affected by collector-emitter voltage, even when the transistor is not saturated. This is called the Early effect. Below is a graph from Wikipedia that illustrates it. The Early effect is generally ignored for biasing considerations, because it is overwhelmed by the variability of beta from device to device, but it does affect the dynamic collector impedance for circuits like current sources/mirrors.
 

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I think either the Early Effect is exaggerated, it shows a power transistor at high currents or Wikipedia shows an old germanium transistor in their curves.

I looked in my 1969 Philips Transistors Manual.
The germanium AC132 and the silicon power transistor BD139 curves look like Wikipedia's curves and the BC107 low power silicon transistor curves look like my middle example.
My right curves have Mr. Early out for lunch.
 

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Then there's the Late effect...
 

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