Power Electronics PLAYLIST: https://tinyurl.com/Power-Electronics-Playlist Unit 1 - Introduction to Power Electronics Part 3 - Steady-State Equivalent Circuit ...

this brief video is a review of circuitmanipulations using ideal transformers the basic rules for ideal transformers in which we push an element a circuit element through a transformer to roughly what we call reflecting to the other side goes as follows if we have a voltage and we push it through a transformer that has a turns ratio of n1 to n2 then as seen between the secondary terminals of the transformer effectively we have a voltage source whose value is multiplied by the turns ratio so we get into over in 1 times V as the effective voltage source is seen by the second from the secondary terminals likewise with a current source we can push it through a transformer also but we have to divide by the turns ratio so if we have an n1 to n2 turns ratio here then as seen between the secondary terminals we effectively have a current source of value I that is divided by n2 over n1 or we can write in 1 over in 2 times I and the third manipulation is if we have some impedance and call it Z and we push that through a transformer then we effectively see an impedance between the secondary terminals that is the turns ratio squared times C so we

get into over in one squared Z as the effective impedance seen between those secondary terminals one little case to be careful with here with Z is if Z is a resistor so if we have a resistor then effectively between the secondary terminals the value will be the resistance times the turns ratio squared likewise if we have an inductor whose impedance is SL then we would multiply that impedance by the same turns ratio squared which is as if the inductance was multiplied by n2 over n1 squared so sometimes we'll just simplify this to say there's an effective inductor a value in two over n1 squared L if we have a capacitor recall that the capacitor the impedance of a capacitor is 1 over SC so the impedance here would be in two over in 1 squared times 1 over SC but as far as the effective value of the capacitance you can see that since impedance varies inversely with capacitance this is as if we have a capacitor that gets divided by the turns ratio squared instead of multiplied so it's as if we have a capacitor of value C that is in 1 squared over in 2 squared okay those are the rules for pushing simple elements through transformers

let's do an example now suppose we have say a voltage source and let's take some larger circuit maybe with a resistor or one some transformer with a turns ratio let's say 1/2 in a maybe we have another resistor here the value R 2 and another transformer that has the dots reverse so the primary dot is on top and the secondary dot is on the bottom of the winding and let's give this a turns ratio of in p21 and what we would like to do is find the effective network with respect to the output terminals and in fact maybe we'll will even solve for the output voltage V out so - one good way to solve this circuit is to use these circuit manipulations and push the elements through the transformers so what I'm going to do here is push them through the Transformers one at a time so let's start by pushing the voltage source and R 1 through the first transformer so what we will get with respect to the secondary terminals of this first transformer will have the voltage source pushed through and in the process it gets multiplied by n a so we have n sub a V is the voltage effective voltage and it's in series with an effective resistor that would be the

resistance r1 multiplied by the turns ratio squared so in a squared that's what we see effectively between the secondary terminals of this first transformer okay so then I'll copy the rest of the circuit after that okay next let's push the elements through the second transformer okay so here's our still our output terminals so we'll push the elements to see what how they appear effectively between the secondary terminals of this second transformer so when we push our to through the transformer we'll get a resistor on the secondary side and it gets multiplied by the turns ratio squared but here the turns ratio is one over in B so effectively we would get our 2 over in B squared it's the effective resistance the fact that the dots are reversed doesn't affect impedances because we the dot is like two having a minus sign on the turns ratio but the impedance changes by the turns ratio squared and the minus sign cancels out okay next let's push the R is our one resistor through so we'll get an effective resistor here that is the in a squared R one and then we have to divide by the N B squared for

this impedance also okay and then finally we push the voltage source through the transformer so we have a voltage of value in a V and then we'll divide by this in B and since the dots are reversed we have to reverse the polarity of the voltage source so I'm going to draw it with minus on top and plus on the bottom okay so this is the effective network that is seen between the secondary terminals of this last transformer with this circuit now we can solve so what is the output voltage we could just use the divider voltage divider formula so we would take the voltage source in a over in B times V and because the polarity is reversed we have a minus sign and then that voltage gets multiplied by the divider ratio of these two resistors so it would be times the divider ratio r2 over in B squared all over what the the same r2 over in B squared plus the first resistor in a squared over in B squared times r1 so that is the solution then for the output voltage so we've taken a fairly complex circuit and by manipulating it reduced it to something where we can just write the answer in the end one other thing I

would point out is that you do have to be a little bit careful in manipulating here if we had wanted to solve for some other signal that's not at the output like suppose we wanted to find this current right here and I call that I a ia it does not explicitly appear in this final circuit and in fact when we push the elements through the transformers we have to be careful on the second circuit I is this current and what happens to i8 on the third circuit well when we push these elements through the transformer we end up pushing ia through the transformer also so here's here's a current related to ia but I ain't got push through this turns ratio and the current on this side would actually be in B times ia okay so we could solve this circuit for that current but we would be solving for ndia and then we have to divide by NB to get back what the actual ia is