Kevineamon Posted April 3, 2017 Share Posted April 3, 2017 Apologies for the size of this image. I did try to resize it on Photobucket. Seems to have had little effect. So I'm having a problem with this. I've been studying the past several hours, learning how to calculate these things. The more simple type. eg. Find ib, ie, ic, Vcb, Vce... I had taken notes in class, for the answers to this new type of question above; where we have this VCE(sat) thingee... Now I've written:- IC = (VCC - VCE) / RC IC = 5.02 / 6000 = 0.8mA Question is where the heck am I getting this 5.02? At first I thought I'd gotten the 0.2v from the VCE(sat) thing wrong. That it should have been 0.2 instead of 0.02 and that I should have added instead of taking away... what I mean is (VCC + VCE). Problem is I've got another example just like this. So where am I getting this 5.02 from? Link to comment Share on other sites More sharing options...
studiot Posted April 4, 2017 Share Posted April 4, 2017 The diagram size was fine for its purpose. You are correct in saying that the collector current is almost completely controlled by the (collector) load resistor in saturation. Your question gives you the collector emitter voltage as 0.2 which makes the voltage across the load as (5 - 0.2) or 4.8 volts, by your own (correct) equation. So you have simply done this sum incorrectly. You know the base voltage and you know the supply voltage, this is dropped across the base resistor and you know the gain since the transistor is just at the onset of saturation, ie just leaving the linear amplifier condition. So you can calculate the base current by dividing the collector current by the gain. Ohm's Law will then give you base the resistor value. 1 Link to comment Share on other sites More sharing options...
Kevineamon Posted April 4, 2017 Author Share Posted April 4, 2017 Thanks Studiot. I was thinking that... Unfortunately I have worse problems now. RL circuit analysis. Yeh... Noooo.... wt??? I'm so dead in this test. Link to comment Share on other sites More sharing options...
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