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CIRCUIT ANALYSIS: Find the Thevenin equivalent of transistor model  CCCS, IVS, 2 res 
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#1
Jan2907, 01:57 AM

P: 492

1. The problem statement, all variables and given/known data
Find the Thevenin equivalent at terminals ab for the transistor model below. 2. Relevant equations KVL, KCL, v = i R, [itex]R_{TH}\,=\,\frac{V_0}{I_0}[/itex] 3. The attempt at a solution I took out the independent voltage source and added a test voltage and current. EDIT: I also rearranged the circuit elements as suggested by mjsd. I have no idea how to solve this circuit though. The resistor on the left looks as though both ends are connected to the same node! Does that mean that the resistor on the left does not affect the [itex]R_{TH}[/itex]? Any tips on how to solve for [itex]I_x[/itex] in order to get [itex]R_{TH}[/itex]? 


#2
Jan2907, 03:24 AM

P: 144

Don't take out the voltage source with those 6V. The voltage difference between a and b is just the voltage over the 2kΩ resistor. Find that voltage and you have [itex] V_{TH}[/itex]. And what would be the current through a and b if they were connected? That's your [itex] I_x[/itex]. You can then calculate the [itex] R_{TH}[/itex].



#3
Jan2907, 03:49 AM

P: 492

So I leave in the 6V volt source? What about the 1V test voltage? Leave that as well?



#4
Jan2907, 03:49 AM

HW Helper
P: 860

CIRCUIT ANALYSIS: Find the Thevenin equivalent of transistor model  CCCS, IVS, 2 res
mmmm..... a circuit model for a Current Amplifier circuit... see input current [tex]I_0[/tex] is amplifiered by 20 times!? You have already been given the Norton equivalent circuit for ab,... so simple to convert to Thevenin eh? Still need to find [tex]I_0[/tex] of course, but that's too easy for someone like you...



#5
Jan2907, 04:13 AM

P: 492

[tex]I_0\,=\,\frac{6\,V}{3000\Omega}\,=\,0.002\,A[/tex]
[tex]20\,I_0\,=\,0.04\,A[/tex] Does this mean that [itex]21\,I_0[/itex] of current goes through the bottom wire between the CCCS and the [itex]2\Omega[/itex] resistor? Still not understanding how to get the [itex]I_x[/itex] so that I can calculate the [itex]R_{TH}[/itex]. How am I given the Norton equivalent? I don't see where that comes from. 


#6
Jan2907, 05:01 AM

HW Helper
P: 860

I have already told you that this is a model for a Current Amplifier circuit..
I guess you don't know about this thing called an amplifier model....well... which is really just a black box which somehow "amplify" the input current or voltage on one side and turn it into something bigger on the other side (ie. modelled by the controlled current source in this case)... by the way , no current flows on the bottom wire.... you can only find two loops in this diagram.... or visualise it by keep shortening this wire until the wire carrying I_0 touches the the wire with the CCCS on it. then you will see that there are 4 wires at that node .. now the rest is obvious..... Now, can you see the Norton equivalent of ab? 


#7
Jan2907, 06:27 AM

P: 492

Ok, I modified the diagram, does that seem right?



#8
Jan2907, 06:36 AM

P: 492

[tex]I_1\,=\,\frac{V_1\,\,V_2}{2000\Omega}[/tex] < Right?
Still problems! I do KVL for the upper right loop) [tex](2000\Omega)\,I_1\,+\,(1\,V)\,=\,0\,\,\longrightarrow\,\,I_1\,=\,\frac{1\,V}{2000\Omega}\,=\,0.0005\,A[/tex] KVL for bottom Loop) [tex](6\,V)\,+\,(3000)\,I_0\,=\,0\,\,\longrightarrow\,\,I_0\,=\,\frac{6\,V}{3000\Omega}\,=\,0.002\,A[/tex] KCL @ [itex]V_2[/itex]) [tex]20\,I_0\,+\,I_1\,+\,I_0\,=\,I_0\,+\,I_x\longrightarrow\,\,I_x\,=\,20\,I _0\,+\,I_1[/tex] [tex]I_x\,=\,20\,(0.002\,A)\,+\,(0.0005\,A)\,=\,0.0405\,A[/tex] Now I get [itex]R_{TH}[/itex]. [tex]R_{TH}\,=\,\frac{1\,V}{0.0405\,A}\,\approx\,24.69\Omega[/tex] Does that look right? How do I get the [itex]V_{TH}[/itex] now? 


#9
Jan2907, 07:57 PM

HW Helper
P: 860

final hint: did you realise that if you happen to choose Vx to be 2V, your subsequent Ix and R_TH will be different from your 24.69?
you have an expression in Vx and Ix, you need to isolate Vx and Ix on one side such that you can solve for Vx/Ix. Now, if you look at this circuit, you may think where can I get another indep equation in Vx and Ix such that when put together with the first one will allow me to solve for the ratio Vx/Ix. In fact, for this question, introducing Vx and Ix unnecessarily complicates the matter. hint: you may take out 6V source if that is still not clear to you... but that's not necessary 


#10
Jan3007, 06:42 PM

P: 492

Find Thevenin equivalent at terminals ab Is this how to get [itex]R_{Th}[/itex]? In the first diagram, if you ground out the bottom node (b), the resistor on the left hand side of the circuit (after "turning off" voltage source) grounds itself out. Both ends of the resistor are connected to the bottom node. [tex]R_{Th}\,=\,2000\Omega[/tex] Now, to get [itex]V_{Th}[/itex], I redrew the first diagram and added a current and a ground node. [tex]i_1\,=\,\frac{V_1}{2000}[/tex] [tex]I_0\,=\,\frac{6\,V}{3000\Omega}\,=\,\frac{1}{500}\,A\,=\,0.002\,A[/tex] [tex]20\,I_0\,=\,0.04\,A[/tex] KCL at Node b) [tex]20\,I_0\,+\,I_0\,\,I_0\,+\,I_1\,=\,0\,\,\longrightarrow\,\,I_1\,=\,20\,I_0\,=\,20\,(0.002\,A)\,=\,0.04\,A[/tex] Since [itex]V_1\,=\,V_{Th}[/itex] (Right?) then we can get [itex]V_1[/itex] from [itex]I_1[/itex]. [tex]0.040\,A\,=\,\frac{V_1}{2000\Omega}\,\,\longrightarrow\,\,V_1\,=\,80\,V[/tex] So, [itex]R_{Th}\,=\,2000\Omega[/itex] and [itex]V_{Th}\,=\,80\,V[/itex]? EDIT: Thanks for your help mjsd! 


#11
Jan3007, 07:53 PM

HW Helper
P: 860

R_TH is correct... V_TH is not. this is because you missed one branch in your KCL for node b)... but that's again an overkill.... you effectively have a Norton equivalent given to you... do u see that now? When you have properly worked out I_1 you will see that.. I hope



#12
Jan3007, 07:57 PM

HW Helper
P: 860

by the way, redrawing the diagram is also not necessary. (that's just to illustrate you only have two loops in the circuit..ie. for clarity only)



#13
Jan3007, 08:53 PM

P: 492

I have edited my post above using your suggestions, hopefully it's right now. Thanks!



#14
Jan3007, 09:52 PM

HW Helper
P: 860

after thought: you have given yourself a lot of unnecessary work to do in this question. All you have really shown (to yourself), by doing all those extra work, is really that the first loop containing the 6V and 3000 ohm is effectively disconnected from the second loop with the CCCS and 2000 ohms (except via I_0). So to find Thevenin is actually trivial.



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