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Shown below is the question which is doing my head in.

I'm on part (ii) and I'm struggling. It's only 2 marks so it must be a simple answer but I cant solve what \$V_{ceq}\$ is. Below the question is what I've done. I don't know what \$R(L)\$ is so I cant get the voltage across it. I've been given \$V(A)\$ so I must have to use an equation with that. Any ideas?

Thanks

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Because RL is not given, we have to make a suitable assumption. One common rule is to make VCE equal to the remaining DC voltage drops in the path. That means:

VCE=5V and IC*(RL+RE)=5V.

From this with IC*RE=2V we can derive a suitable RC value of RC=3V/1mA=3 kOhms.

For the remaining part it is recommended to calculate the resistors R1 and R2 based on the assumption that the current through R2 is app. 10 times larger than the base current.

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  • \$\begingroup\$ What do you mean about making V(CE) equal to the remaining voltage drops? So V(CE) is just V(CC)/2? Thanks \$\endgroup\$
    – Physicsman
    Aug 26 '15 at 13:13
  • \$\begingroup\$ Yes - that`s what I mean. It is just something like a "rule of thumb", but if we have no other requirement... \$\endgroup\$
    – LvW
    Aug 26 '15 at 13:59
  • \$\begingroup\$ Ok i'll keep that in mind. I'll have a better look at this later. Is what you're saying the same as what the first answer says? \$\endgroup\$
    – Physicsman
    Aug 26 '15 at 14:25
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You are missing some information from this question I think. It cannot be solved in the current state. Are you sure they didn't tell you RL? It's usually assumed to be 50 Ohms in these kinds of questions. You either need that or IS, and you can use this equation to solve for vCE. enter image description here

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  • \$\begingroup\$ Usually 50 Ohms? Typing error? \$\endgroup\$
    – LvW
    Aug 26 '15 at 7:18
  • \$\begingroup\$ I tried doing that but I think I ended up getting a huge voltage that didn't really make sense \$\endgroup\$
    – Physicsman
    Aug 26 '15 at 13:08
  • \$\begingroup\$ I got I(s)= 2.6*10^-16 and V(CE)=353V \$\endgroup\$
    – Physicsman
    Aug 26 '15 at 13:16
  • \$\begingroup\$ Well typically an amplifier matches to 50 ohms load. Im not sure about his specific situation though. \$\endgroup\$ Aug 26 '15 at 14:18
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If the quiescent current is 1mA, the quiescent \$V_{out}\$ will be \$V_{CC} - (0.001 * R_L)\$, for example, 10 - (0.001 * 5000) = 5V for \$R_L\$ = 5kΩ.

I don't see where they ask for \$V_{CEQ}\$, and it wouldn't be my first assumption since it's not a direct performance measure. Are you sure they're not asking for the quiescent \$V_{out}\$?

\$V_{CE}\$ can of course be calculated easily once \$V_{out}\$ (\$V_C\$) is known.

Without further information, like the value of \$R_L\$, the answer for \$V_{out}\$ will be in symbolic form, not a specific number.

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  • \$\begingroup\$ Hi, I thought the quiescent voltage is the Q point voltage which is V(CEQ)? \$\endgroup\$
    – Physicsman
    Aug 26 '15 at 13:09
  • \$\begingroup\$ I expect the general term "quiescent voltage" could be applied to quite a few voltages, like \$V_B\$, \$V_E\$, \$V_C\$, and \$V_{CE}\$. So its an ambiguous term. In my travels I haven't detected a convention by which the meaning of that term is understood without elaboration. It could mean Vce if that's what you're instructor has said. I just don't see it in the question as posted. Note that Vce can be determined just as easily as Vc. It would be Vc - Ve, => Vcc - (1mA * \$R_L\$) - 2 => 8 - 0.001 \$R_L\$. As I mentioned, the result will be with \$R_L\$ in symbolic form until that is specified. \$\endgroup\$
    – scanny
    Aug 26 '15 at 15:05

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