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I am doing a course on Coursera "Intro to electronics" as electronics always was my weak point.

Now we discuss npn transistors and I have simplified circuit diagram (see below) where I am supposed to solve for VB. I am feeling really embarrassed to ask this but I want to understand this...

Given are Rb1=15000, Rb2=82000, Vcc=15 and IB=40 mA

My idea was to to a KVL and KCL and I came up with these equations: $$ V_{cc} - I_{b2}R_{b2}-I_{b1}R_{b1}=0 \\ I_{b2} - I_{b1}=I_{b} \\ and \\ V_{b}=V_{cc} - I_{b2}R_{b2} $$

and this becomes

$$ I_{b2}R_{b2} + I_{b1}R_{b1}=V_{cc} \\ - I_{b1}+I_{b2}=I_{b} $$ $$ \pmatrix{ R_{b2} & R_{b1} \cr - 1 & 1}\pmatrix{I_{b1} \cr I_{b2}} = \pmatrix{15 \cr 0.04} $$

When I solve this for IB1, IB2 I get 0.04 mA and 15 mA. This gives me VB = VCC-IB2*RB2 = 14.5055 V

... which incorrect!! :-(

The solution's hint is to use superposition.

My two question: 1. What's wrong with my solution way? 2. How do I go about with a superposition?

EDIT: ad 2: after some more thinking: can superposition in this case mean that I

a) have one circuit without IB.

AND

b) one circuit with both voltage set to ground. Thus I get RB1 and RB2 in parallel?

enter image description here

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    \$\begingroup\$ You might want to work on your title, its rather meaningless \$\endgroup\$ – PlasmaHH Sep 11 '18 at 8:42
  • \$\begingroup\$ Yes. That's right. Do you have a good suggestion since I don't have a clue on how title this simple problem... :-( \$\endgroup\$ – Andreas K. Sep 11 '18 at 8:49
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    \$\begingroup\$ Ok. I changed it. Hope this has sufficient meaning. Thanks. \$\endgroup\$ – Andreas K. Sep 11 '18 at 8:51
  • \$\begingroup\$ "1 Voltage, 1 Current" isn't a circuit. Can you try once more to clear up the title? Also, please start early on with the schematic in your question. People are differnt, but I look at any schematic first and then read the question. \$\endgroup\$ – winny Sep 11 '18 at 11:16
  • \$\begingroup\$ I am really sorry. I tried my best to find a good title for this problem. And thanks for the hint with the schematic. \$\endgroup\$ – Andreas K. Sep 11 '18 at 13:54
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Your KVL looks like this: $$ I_{b2}*R_{b2} + I_{b1}*R_{b1}=V_{cc} \\ - I_{b1}+I_{b2}=I_{b} $$

And if we plug the numbers into WolframAlpha

http://www.wolframalpha.com/input/?i=15+%3D+X2*82+%2B+X1*15,+X2+-+X1+%3D%3D+40

We get

\$I_{b1} = -33.66mA\$ and \$I_{B2} = 6.34mA\$ and \$Vb = 15V - 82k\Omega \cdot 6.34mA = - 504.8V \$

Now we can try the superposition

First I remove the current source, so we are left with this:

schematic

simulate this circuit – Schematic created using CircuitLab

The voltage divider output voltage is :

$$V_B' = Vcc*\frac{R_1}{R_2 + R1} = 2.319V$$

Next, we are short-out the voltage source so we are left with this:

schematic

simulate this circuit

Therefore \$V_B'' = - R_1||R_2 * I_B = -507.21V\$

And finally

$$V_B = V_B'+V_B'' = 2.319V + (-507.21V) = -504.8V$$

And we are done.

As you can see there is something wrong with your equivalent circuit.

I suspect that the \$I_B\$ (base current) is equal to \$40\mu A\$ (not \$40mA\$) and in this case \$V_b = 1.812V\$ seems reasonable.

Try to show us the original circuit.

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  • \$\begingroup\$ One question: why −R1||R2∗IB? Especially the "-"? \$\endgroup\$ – Andreas K. Sep 11 '18 at 16:19
  • \$\begingroup\$ And yes you are right. They seemed to mix up milli and micro. \$\endgroup\$ – Andreas K. Sep 11 '18 at 16:20
  • \$\begingroup\$ @AndreasK. I put a "minus sign" because of the Ib current direction through resistors. The Ib current is flowing from lower to the upper part of a resistor. So the positive voltage is at the bottom and negative at the top of a resistor (current flow from "+" to "-"). Do you get it? \$\endgroup\$ – G36 Sep 11 '18 at 16:45

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