Created by Created April 02, 2012 Last modified March 07, 2013 Tags

### Summary

A simple ideal behavioural transformer with any turns ratio.
Perfectly linear, no parasitics. No primary inductance so it works down to DC.

### Description

Edited 120629. Corrected error in describing the ratio for CCCS1: was incorrectly given as 1/N. Corrected to N. Ratio changed from 1:2 to 1:3 to clarify waveform plots. Added 1k source resistor and 1k*N^2 load resistor to illustrate impedance transformation action more clearly.

 do you know how to model this: http://electronics.stackexchange.com/questions/59539/pspice-homework-question with your schematic? by eggie5 March 03, 2013 Open a new schematic in the CL Editor. Cut and paste everything inside the dotted box in this schematic (Simple behavioural transformer 01) into your schematic. Read the instructions in Simple behavioural transformer 01 about how to set the turns ratio. Set the turns ratio to 2 instead of 3. Draw the rest of the circuit around it as shown in: http://electronics.stackexchange.com/questions/59539/pspice-homework-question Note that you have to find the answers analytically. You can use the model to check your answers or to give you a clue as to what sort of numerical answer you should get but the simulation alone will not give you the answer the question is looking for. Hint: 1) Find the Thevenin equivalent of the voltage source, the 80R and the 20R; 2) Find: Zsec = Zpri*N^2 where: Zpri = the source impedance seen looking out of the primary; Zsec = the load impedance seen looking out of the secondary; N = the turns ratio pri:sec; 3) Calculate the effective impedance to ground that the right hand end of RL "sees" through: 360R//(40R+Zsec) where // means in parallel with; 4) add result of (1) and result of (3); 5) Use this (4) as total source resistance, Rtot; 6) Connect Rtot to voltage source in series with RL to ground; 7) Calculate value of RL to give max power in RL; 8) Calculate power in RL. Sanity check against simulation. by signality March 03, 2013 @signality, thanks for your prompt reply! I've adapted your transformer to fit my problem: https://www.circuitlab.com/circuit/bchtyd/quiz-4/ I'm not sure if i'm supposed to ignore Z_L on these calculations. Assuming I am. 1) The Thevenin eq. of the voltage source is just a 500V source in series w/ a 100 ohm resistor right? 2) Zpri= Z_L/n^2 = (360+40)/2^2 = 100, where n=2/1=2 and Z_L=360+40=400 is this correct? Zsec = 100 * 4 = 400 where Zpri=100 and n^2=4 is is this correct? 3) 360 // (40 + 400) = 208; correct? 4) R_tot= 208 + 100 = 308; where 100 comes from #1 correct? 6) So we'll have a 500V source in seris with R_tot in series with R_L where R_L=400ohm is this correct? by eggie5 March 03, 2013 Please accept my apologies: I've mislead you. My method is wrong. I'm trying to think of the correct way to analyse this. Sorry! The stuff about the modelling is right, it's the maths I've got wrong. I'll post back if I get my head round it. by signality March 04, 2013 OK, cracked the first part of the problem (a). I haven't thought about (b). 1) Assume RL = open circuit. 2) Reflect 20R resistor the secondary side. Remember that Zsec = Zpri*N^2 where: Zpri = the source impedance seen looking out of the primary; Zsec = the load impedance seen looking out of the secondary; N = the turns ratio pri:sec; and Vsec = N*Vpri. 3) Reconnect LH end of RL to LH end of 40R resistor. 4) Reflect values of 80R and voltage source to secondary side but leave connected to primary and set transformer to 1:1, i.e. N=1. 5) Short pri and sec sides of transformer together then delete transformer; 6) Note that 40R and RL are in parallel and swap positions of the parallel pair and the 360R so 40R and RL are now in ground side of 360R. 7) Find thevenin equivalent of reflected voltage source and reflected series resistor (was 80R) and leftmost resistor to ground (was 20R); 8) Absorb 360R into resulting series resistance to thevenine equivalent voltage source; 9) Simplify expressions; 10) Find thevenin equivalent of first thevenin equivalent voltage source and first thevenin equivalent series resistor and 40R; 11) Simplify expressions; 12) You should now have a voltage source with a single series resistor, Rx, connected to the top end of RL with the bottom end to ground. You should also have an expression for the value of Rx. Set RL = Rx for maximum power transfer. Hint: once you get to this point set the voltage source to a DC value equal to the final thevenin equivalent RMS AC source voltage. 13) Find power in RL (sanity check: should be same as in Rx). 14) Run transient sim of simplified circuit with AC source and with DC source set to RMS value of AC source. Run sim over exactly 1 cycle of AC source. Plot power in load of DC sim and of AC sim over 1 cycle of AC source. Compare average of instantaneous AC power in load over 1 cycle of AC sim with DC power in load of DC sim. 15) Plug value of RL back into original circuit. Copy original circuit and set source to DC of same alue as RMS value of original AC source. Run transient sim of both circuits over exactly 1 cycle of AC input. Plot power in load of DC sim and of AC sim. Compare average of instantaneous AC power in load over 1 cycle of AC sim with DC power in load of DC sim. 16) Compare analytical and simulation results. Note that simulation results may have small rounding errors. Phew. by signality March 04, 2013 At least I think the methodology is right but when I plug the value for RL back into the original circuit I get a different power in RL from when I put it into the reflected and simplified circuit. Need to check: I may have made a mistake in my sums. This is a much harder problem than I thought it was to begin with. :) by signality March 04, 2013 Finally, after 3 false starts and making myself look pretty dim .. I think I have a solution. Firstly I made a mistake in how I drew the circuit. I had put the 20R to ground instead of in series. So I set off solving the wrong problem. I also had a couple of mistakes in my reflection and Thevenin equivalent transformations. The hard bit was working out how the value of RL in the original circuit has to be related to the reflected circuit when the 1:2 transformer ratio is reduced to 1:1 so that the transformer can be replaced by a bit of wire. I'm still not convinced I truly understand the solution: could I find the solution for a generalised network without a clue like the one for this example that: Pwr = Vth^2/(4*RL)? Not without a huge amount more head scratching and guess work. Even that clue is misleading. It refers to the value of RL that gives max power in the reflected equivalent circuit. It is not the value of RL that gives max power in the original circuit: https://www.circuitlab.com/circuit/d966dh/dc-sweep-find-max-power-in-load-01/ BTW, this wiki page is quite handy: http://en.wikipedia.org/wiki/Equivalent_impedance_transforms by signality March 06, 2013