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Could somebody please check to see if what I have done is right? I posted a similar question a few days ago and someone said V(CE) is equal to V(CC)/2. Is it ok to do this?

So I got V(CEQ) = 6V and then I(C)=1mA

I tried to solve I(B) on the left side first but got a negative value for it? So that can't be right.

I don't need any help with part b)

Thanksenter image description here

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Task 1 - find Ib.

We know that VBE is 0.7V (given).{quick check - is this possible with the given values? - without the base connected R2 would produce a drop of about 0.75V - this is a good result because we can always drop this voltage by taking current into the base}.

For VBE = 0.7 and R2 = 6.3k by Ohm's law I(R2) = 0.7/R2 = 0.11111 mA

The 11.3 volt drop (12 - 0.7) across R1 gives us a current of I(R1) = 11.3/93.7k = 0.120598mA

By Kirchoff's law the difference in current must be the base current 0.12 - 0.11 mA (2dp) = 0.01mA

Task 2 - the collector current

We know that current gain is 100 (given)

So Ic = 100Ib = 100 * 0.01mA = 1mA

Task 3 - find the voltage at the collector.

By Ohms law the voltage drop across Rc = Ic*Rc = 1(mA) * 6k = 6V

So the voltage at the collector (with respect to ground is 12 - 6 = 6V

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  • \$\begingroup\$ Hi, thanks a lot for the help. I didn't realise you can just disconnect the base to get the resistance for R2. And do you just ignore R(S) because of the capacitor? \$\endgroup\$
    – Physicsman
    Commented Aug 27, 2015 at 20:32
  • \$\begingroup\$ @Physicsman no problem - I don't usually give a (full) answer to homework questions but I thought this might be useful example for tackling similar problems for anyone starting out. Learning to break down a problem into smaller (easier) steps is far more important than just getting 'the right answer'. Good luck with your studies. \$\endgroup\$
    – JIm Dearden
    Commented Aug 27, 2015 at 20:40
  • \$\begingroup\$ Thanks. I have an exam in 12 hours and am just clearing up the things I don't understand: Could you check a question called: "Amplifier Gain. OP amp?" It hasn't really been answered yet and it seems like its a simple method \$\endgroup\$
    – Physicsman
    Commented Aug 27, 2015 at 20:42

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