# Find equivalent resistance seen by a capacitor.

Discussion in 'Homework Help' started by Whyregister, Sep 14, 2012.

1. ### Whyregister Thread Starter New Member

Sep 12, 2012
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Where Req2 is where the capacitor is, looking into the circuit.

It took me one page of working out using KVL using a test voltage Vt, and the associated current iT, then doing Vt/iT and getting a large equation in terms of all the circuit variables to get Req2 = 16K, when the answer is 15k doing some method I do not really understand.

Would it make sense to use a test voltage and get the equivalent resistance?

Can someone explain what exactly was done?

Rc1 = 15k
Re2 = 9k
Rpi2 = 1.4k
gm2 = 107.1mS
Rc2 = 15k
RL = 10k

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But you can also find separately resistance looking into the emitter.

Now it should be obvious that

Rin = Vin/I1

I1 = Vin/Rpi - (-Vin *gm) = Vin/Rin + Vin*gm = Vin *(gm + 1/Rpi) = Vin (gm*Rpi + 1)/Rpi

So

Rin = Vin/I1 = (Vin)/(Vin (gm*Rpi + 1)/Rpi ) = Rpi/(gm*Rpi +1)

And this Rin resistance (resistance seen from emitter terminal when we looking in to emitter) is connect parallel to Re2

So we end-up with this simply diagram

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3. ### WBahn Moderator

Mar 31, 2012
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We can't guess at why you got 16kohms unless you show us your work. We are not mind readers.

The equation given is correct and the answer is correct (I get just under15.01kohms).

By inspection, you can see that the elements to the right of the dependent current source have no affect on the resistance seen at the port - change either of those components and all it will do is change the voltage on the right side of the current source in order to keep the same current flowing.

Redraw the circuit with the port on the left side. Then, from the top terminal, go out through Rc1 along the top branch and put the 'ground' node at the far top/right of the diagram. Now draw the return node (the right side of the port in the figure) as the bottom terminal in the port and just draw it going out to the right. Now, between the ground done and the return node you have RE2, rpi2, and the dependent current source (with the current going down from ground to the return node).

Now just focus on the current source and rpi2. Notice that if the voltage from the top node to the bottom node is vbe2, then the total current going through the current source and rpi2 is trivially easy to calculate. From then you can calculate the effective resistance that these two elements represent. You will find that it is a bit over 9ohms. This is then in parallel with RE2 (9kohms), so that combination is 9ohms. The 9ohms is then in series with 15kohms, which is 15kohms.

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4. ### Whyregister Thread Starter New Member

Sep 12, 2012
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Jony, I do not understand how you moved from the first diagram to the second diagram, why did you get rid of Rc1 and Re2? Also what program do you use to make that?

WBahn, I couldn't really show my work because it's a long page of writing and it would be difficult to take a picture of it in focus, and I figured such a method is not really viable, even if it were correct.

I do not understand what you mean by "Notice that if the voltage from the top node to the bottom node is vbe2, then the total current going through the current source and rpi2 is trivially easy to calculate"

From what I see ok, we have Vbe2 at that node, so therefore the current going through Rpi2 is just Vbe2/Rpi2, the total current going through the current source is just gmVpi2 isn't it?

Most importantly I don't understand how 'by inspection' we can eliminate Rc2 and RL, I had a gut feeling that they had nothing to do with it, but I do not understand why exactly.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I simply split the circuit to simplify the analysis.
Look at this diagram where I show which part of the circuit I split and find Rin of this part of a circuit.

And as you can see Ic + IL must be equal I3 and I3 is equal to gm*Vbe
And this is why we can eliminate Rc2 and RL from the equations.

Rin = Vin/I1

I1 = I2 - I3 = Vin/Ri - (-gm*Vbe)

So we end-up with this simply diagram

library for Inkscape
http://elportal.pl/ftp_05/201207_Inkscape.zip

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6. ### Whyregister Thread Starter New Member

Sep 12, 2012
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Hmm...shouldn't it be I3 = -(Ic + IL)? Since the currents through Rc2 and RL go to ground, and not into the node?

I still do not understand why we can disregard the resistors because the resistor currents add up to another current coming out of a node. Is there a formal rule/law regarding this?

I do not recall seeing or learning something like this before.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The directions of Ic and IL are arbitrary.

Because we have a current source so I3 always will be equal to gm*vbe regardless Rc and RL value. And the rule/law we call it a current Kirchoff's law.
This current source determines the value of a current not resistors in the circuit.

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8. ### Whyregister Thread Starter New Member

Sep 12, 2012
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Hmm, interesting! Quite a bit to digest there and comprehend, thanks so much for all the insight and information.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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When we looking for Rin all we need is to find I1 current and we don't need to bother about voltage across Rc and RL. So the total current going through the current source is just gm*Vbe.

Simply find Rin and Va for this simple circuit:

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10. ### Whyregister Thread Starter New Member

Sep 12, 2012
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What is that double circle thing?

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Ideal constant current sources

12. ### Whyregister Thread Starter New Member

Sep 12, 2012
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OK.

I am very confused. For the first one. I calculated Rin = Vin/In = 1V/1A = 1 ohm

and for Va: (1 - Va)/100 = 1A, which means Va = -99V

What's confusing is if there's 1A flowing in the circuit, it must mean there's V = 100*1 = 100V drop across the resistor, but that seems weird.

13. ### Whyregister Thread Starter New Member

Sep 12, 2012
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Using similar kind of reasoning for the second one, Va = 1*500 = 500V

Rin for that would be Vin/Iin, and since Vin and Iin are 1, Rin = 1Ω

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In this example I want to show you that in the circuit that contains a current source in series with the resistor the input resistance (and input current) is determined by current source. The series resistor has no influence on the input resistance. So resistor RB can be eliminate from the circuit and we still can find Rin. So we have a very similar situation that you had in you circuit (we can ignore Rc2 and RL).

As you can see the series resistor with the current source has only effect on Va voltage and has no influence on input resistance.

As for the confusing part. In real life it is impassible to buy a current source, we only can buy a voltage source. But circuit theory has no problem with that. And yes the correct answer are -99V and 500V.

In summary we can ignore series resistor with ideal current source when we want to know the current that is flow through circuit.
But this statement holds as long as we have a ideal current source.
But if we take into account a real current source with internal resistance Rs (1/hoe or 1/h22 or ro for BJT ) we cannot ignore Rb in our equations if we want to find Rin.

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15. ### Whyregister Thread Starter New Member

Sep 12, 2012
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So in this new case, if we consider two meshes, and the 1st mesh as having current i1 = Iin, and 2nd mesh i2 = 1A

1 = 100i1 + 200i1 - 100*1A
i1 = Iin = 0.67A

Therefore Rin = 1/0.67 = 1.49Ω

Iin = (1 - Va)/100, since we have Iin, Va = -66V.

16. ### Whyregister Thread Starter New Member

Sep 12, 2012
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I think I'm starting to see it now, I'm surprised I didn't have a realisation about this earlier. You've been a massive help!

Feb 17, 2009
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