Transistor Hard Saturation - pg224 - Malvino

'A designer who wants a transistor to operate in the saturation region under all conditions often selects a base resistance that produces a current gain of 10. '

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I don't understand what he's doing with the math (it's not complicated math but he's trying to illustrate an idea that I don't get).

  1. betaDC=Ic/Ib but we only have control over Ib so.. how has he computed Ic=2mA?
  2. Then he does Ib = 2mA/50 okay but then again the normal way would be to work out Ib first and multiply with the given BetaDC to work out Ic..

If he wants IcSat directly then that's at Vce=0 so you do Vcc/Rc = IcSat What's he doing in Link-2 (Link-1 is his reference ckt)

Reply to
veek
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He isn't deriving anything, he's just calculating some values. It is not a mathematical proof.

He isn't calculating it, he picked it as an example by setting the Vcc and Rc. 20 V / 10 kohms = 2 mA.

Again, he has a max Ic already. Now he wants to pick an Ib that will saturate the transistor. Since the beta is 50, use a gain of 10 which will use a lot more base current than needed and put the transistor into saturation.

I don't see any links.

He picks a value for Ic and sets Rc and Vcc to get that. Now to set the value of Rb he knows the beta is 50, so using a value of 10 is assured to provide much more base current than needed and puts the transistor into saturation. So Ib = 2 mA / 10 = 0.2 mA which with Vbb = 10 V makes Rb = 100 kohms. Er, um, wait, that's not right! Ib of 0.2 mA with Vbb of 10 volts would be 50 kohms (ignoring the 0.7 volt Vbe drop).

Don't know why he is showing 100 kohms on the schematic, but do you see the logic here?

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Rick C
Reply to
rickman

Thanks yeah I got it - I did it the hard way as well - calculated Ib for beta=10 and beta=50.

(He could have just said it out - instead of showing me the math - he's done these calculations before so the math is totally unimportant in that context)

Reply to
veek

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