Need to create a circuit equation

Thread Starter

nickb34

Joined Jun 10, 2007
4
I have a circuit that is connected to a transmission line. I need to create an equation for the circuit, but I can't seem to get it right. I think it should be:

i = ((Vcc2-v)/Rlh)+(-v/Rll)

Any ideas?
 

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hgmjr

Joined Jan 28, 2005
9,029
You will need to have some information on the circuitry connected to the junction of Rll and Rlh before you can proceed with the calculation for the value of the current i.

Do you know what the circuit looks like that is connected to this junction?

hgmjr
 

Thread Starter

nickb34

Joined Jun 10, 2007
4
The current and voltage changes based on the time (reflecting waves). I just need an equation for that specific circuit in terms of i or v. I think it should be able to be figured out with a KVL/KCL....
 

Thread Starter

nickb34

Joined Jun 10, 2007
4
I know the Thevenin's equivalent works, but it is a bit more difficult to program... but I guess I may try that if I can't get any help. Thanks
 

Papabravo

Joined Feb 24, 2006
13,697
As far as the AC behavior of the circuit is concerned, the value of the battery has no effect on the impedance of the network. Your network is essentially two resistors in parallel. Transmission lines at DC are trivial and uninteresting. As you sweep the frequency higher, the dimensions of the transmission line and the components will become significant, and again the fixed DC voltage is irrelevant.
 

Thread Starter

nickb34

Joined Jun 10, 2007
4
The DC voltage will affect the steady state value of the voltage and current of the entire circuit. For a better idea of what I am working on, I'll attach the whole circuit... Basically what happens is that the circuit is at a steady state high (Vcc1 is attached to the transmission line) the circuit is switched and will go to a steady state low. While switching, voltages and currents are reflected at discontinuities on the circuit. I am building a model of this in mathcad, trying to show the reflections in a bergeron diagram.
 

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Papabravo

Joined Feb 24, 2006
13,697
That was helpful, since your original drawing was incorrect. In it you had the negative battery terminal connected to the high resistor, and the positive battery terminal connected to GROUND. The steady state voltages and currents are straightforward. There is a voltage divider with R1(high) in series with the parallel combination of Rs(low) and R1(low), if we are neglecting the DC resistance of the transmission line. Transient solutions however complicated will asymptotically approach that value. As I said in my previous post the transmission line sees R1(high) in parallel with R1(low) as the termination impedance. The reflection coefficient will determine the magnitude of the reflections.
 
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