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using Ohm's law, i need to find the current for $$I_C , I_B$$ in terms of: $$V_B, V_C, V_{DC1},V_{DC2}$$ for the following ciruit.

enter image description here

my solution is: $$V_{DC2}+{I_B}{R_B}={I_C}{R_C}+{V_{DC1}}+{V_E}$$ But this seems to be wrong. why? Thank you

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  • \$\begingroup\$ You're overthinking this. Notice that \$I_C\$ is the same as the current through one of the resistors. And you can write Ohm's law for this resistor in terms of two of the voltages you are given. Same thing for \$I_B\$. \$\endgroup\$ – The Photon Apr 28 '16 at 2:19
  • \$\begingroup\$ You need the DC current gain of the transistor. \$\endgroup\$ – soosai steven Apr 28 '16 at 2:20
  • \$\begingroup\$ @soosaisteven, no he doesn't, if he's allowed to take the node voltages as given. \$\endgroup\$ – The Photon Apr 28 '16 at 2:21
  • \$\begingroup\$ @The Photon Do you mean that i can simply make $$I_B=V_B/R_B,,I_C=V_C/R_C$$? Thank you for your response \$\endgroup\$ – Hasnain Ali Apr 28 '16 at 2:26
  • \$\begingroup\$ No, \$V_C\$ is the voltage between the collector node and ground, and the resistor is not connected between those two nodes. \$\endgroup\$ – The Photon Apr 28 '16 at 2:27
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This is a horrible question, because it's not related to how we'd normally calculate the operating point of this circuit. It's so annoying that rather than try to teach you anything, I'm just going to give you (part of) the answer. I'm doing this because the question isn't well-posed to teach you anything anyway.

You can get \$I_C\$ very simply as

$$I_C = \frac{V_{DC1}-V_C}{R_C} $$

That's just a simple statement of Ohm's law for RC, in terms of the voltages you were asked to find the current in terms of. So it answers the question as stated.

You can get the current for \$I_B\$ by a very similar statement of Ohm's law for \$R_B\$. I'll let you do that part yourself.

In the real world, of course, this has nothing to do with how you'd find \$I_C\$ in a real circuit, so just give your instructor the answer he wants but don't imagine you're learning much from it.

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