This week we will continue our discussion of frequency response and bode plots and we will apply the material to power converter transfer functions. So in this lesson I want to work through construction of the transfer functions, of the buck-boost converter that we modeled in Chapter 7. So again, here is that equivalent circuit model that we found. And what I want to do now is, work out the asymptotes of the magnitude and phase of the control to output and line to output transfer functions. So, we've discussed this before, as well. If we view the output voltage, v hat, as the output of our circuit. and we have the control input, d hat, and the power input, vg hat. Then we can express the output v hat at, through super position as a linear combination of terms that arrives from each of the two inputs. And we've already talked, then, about how to find those transfer functions. The control to output transfer function Gbd is found by first setting the other input vg hat to zero like shown here, and then find the ratio of v hat over d hat and that ratio then we define as the control to alpha transfer function GBD. Likewise, to find the line to alpha transfer function Gvg of s, we set d hat to 0, and we find the ratio of v hat over v g hat. So let's work through the solution of each of those, and then construct the transfer functions. So first, define Gvg of s. So we'll set d hat to 0, what that does to the d hat voltage source here is to turn it in to a short circuit. And to the two current sources that depend on d hat, when we set d hat to 0 they become open circuits with 0 current. So then the circuit reduces down to this one. Okay. Next we can push all the elements to the right through the transformers. So if we push the inductor through this transformer we'll have to divide by d prime squared as shown here. And when we push the vg hat source through both transformers, we will let's see, multiply by D, divide by D prime, and then we have to reverse the direction because the dots are reversed. So I get vg hat times minus D over D prime for that source. So, now we have this circuit. And finally, we need, we can solve this using the voltage divider formula. So if we, view this impedance as the parallel combination. R in parallel with one over sc. And then we have this impedance as sl over d prime squared. Then using the divider, voltage divider formula, we can write the transfer function this way. So we have the coefficient minus D over D prime out in front and then multiplied by the voltage divider formula. Okay. At this point, I would say R in parallel with 1 over sc is a term that comes up an awful lot. And it's worth working it out, and writing it down, and memorizing what it is. R in parallel with 1 over sc is r times 1 over sc over R plus 1 over sc. If you multiply top and bottom by sc, you get R over 1 plus sRC. So we can plug that in. So it's shown here and here. And next we can multiply top and bottom by 1 over sRC to turn this into a rational fraction. So if you do that, you get this expression. Now, what I would say, at this point, is we still aren't quite done. Because we need to write the, poles and zeros in normalized form. Where the coefficient of s to the 0 power is 1. So we should divide the top and bottom by R to get that to happen. And if we do that, then we'll get 1 in the numerator and this coefficient of s to the 0 power in the denominator will become 1 as well. So that's shown on the next slide right here. And, finally, here's the answer that we get. Okay, we recognize this form now as having a gain out in front that we're going to call Gg0. And then it has a quadratic denominator. So we'll write the denominator in the standard quadratic form. which we discussed last week like this. Okay? By acquainting like terms in the expression that we have in the standard form, we can get equations for the salient features. So Gg0 is apparently equal to minus d over d prime for this converter. the coefficient of s squared in our normalized form is 1 over omega not squared. And that should be the same, then, as this coefficient, LC over d prime squared. So you can solve for omega naught. And we'll get what, d prime over root LC. And then finally the coefficient of s to the first power would be 1 over Q omega naught in our standard form. That should be equal to L over d prime squared R. [SOUND] So you can plug this expression for omega naught in here. And then solve for omeg-, or, for q. And get an equation for q. Then, here is the result. Here's a summary of our, our salient feature expressions. So we have Gg0, omega naught, and q, written as, as nice simple, formulas. Okay? Now let's work out the control to output transfer function. So this is v hat over d hat. Again, with vg hat. [SOUND] Equal to 0. So if we let vg hat be 0, that shorts out this vg hat source. Shorting the source shorts out the current source on the input side as well, which n shorts the primary of the transformer. So effectively in our model, the second area just looks like a short circuit. So then the circuit reduces down to this. We can continue, we can push this voltage source through the transformer, we'll have to divide by the D prime terms ratio, and reverse the polarity because the dots are reversed. So that gives us this source. We can also push the inductor through the transformer. We'll have to divide by D prime squared in that case and we get this. And so the circuit now reduces to this. Well we still have two D hat sources so we have quite a bit of work to solve for the output voltage. Here, we can view v hat as actually being a superposition of terms coming from the two sources. Even though the two sources are both d hat sources. We can still write v hat is equal to one transfer function times the first source. Plus the second transfer function times the second source. So then here is that, here is the, for the first transfer function times the voltage source. With the current source that would have been here set to zero. and here's the second case where we've set the voltage source to zero. And then we get the current source driving these impedances. So, for the first case, we can, again, use the voltage divider formula. The output voltage is this coefficient. Multiplied by the divider ratio and it's the same divider ratio as in the Gvg transfer function. In the second case, we have the current source driving a parallel combination of impedances so we can write that the transfer function here is the coefficient I multiplied by the parallel combination of all three elements. By use of superposition, then, we can add these two together to get the total transfer function, Gbd. And here it is. Now, this is still a pretty big long expression. And it will take quite a bit of algebra and work to simplify it. And write it in a, a standard form. Frankly I think that it's, it's easier actually to manipulate the model into canonical form and then solve it that way. but this is still a valid way to do it and if you are able to do the algebra this is, maybe, the most straightforward way. So what I will say, I'm going to do all of this algebra here, but if you simplify this expression, you can get it down to, to this one, in which have a DC gain. We have a right half plain zero in the numerator, and we have a quadratic denominator that is, in fact, the same denominator as in Gvg. So this transfer function then can be written in this standard form, where the coefficient Gd0, is this term here. the right half-plane zer, zero frequency, omega z, would be 1 over this quantity. And the denominator, Q and omega nought are the same as before. So then here is a summary of the equations for the salient features in this case. One other thing I would note is that, to simplify the expression for omega z. The zero from this form into this form. We use the dc relationships that come from solving the dc model. Okay. So we have a nice analytical result, where we have equations for the salient features. At this point, if we like, we can plug in numerical values, and construct a bode plot. So, I've done that here for this example using these values for the element values in the duty cycle. So if you plug those in to our expressions, you get these quantities. so the dc gain of Gbg is 3 and 1/2 db, or 1.5. The dc game of the control to output transfer function is 187 and 1/2 volts. It has dimensions of volts because the duty cycle is dimensionless, while the output is voltage. expressed as db, this corresponds to 45.5 dV volts. Then we have our quadratic poles at 400 Hz, with a queue of four, which is 12 dB. And our right half plane zero turns out to be at 2.65 kilohertz. So then here is the plot of that. So this is the control to output transfer function. We have our DC gain of 45.5 dB volts. We have two poles at 400 Hz. After 400 Hz, the function starts to decrease with a -40 dB per decade slope. And we have a q of 4, which is 12 db at f naught, so there's a resonance here and this mech, value of the function at 400 hertz would be 12 db plus 45 and a half db. That's actually db volts. At the right half plane zero frequency of 2.6 kilohertz, the zero adds plus 20 dB per decade slope. So the composite slope of the, the entire transfer function then will be minus 40 plus 20, which gives us minus 20 dB per decade slope. Beyond 2.6 kilohertz. Here are the phase asymptotes. We start at zero degrees. I actually haven't included the minus sign in the transfer function here. Although, since we have a negative output voltage. There is actually a minus sign, which would be another 180 degrees of phase. And that's not included in this plot. If we work out the phase asymptotes, the right half plane 0 will have a break frequency at 1/10 of 2.6 kHz, which is 260 Hz, and it's shown right there. And that's the first break frequency that we get to. And beyond this 260 hertz, the right half plane zero will contribute a minus 45 degree per decade slope all the way up to ten times 2.6 kilohertz, which would be 26 kilohertz. So we have minus 45 degrees per decade there over that range. And then, in addition, the complex zeroes contribute a slope here that's very large. The slope, recall the formula, is minus 180 Q degrees per decade, with Q as four, so that would be four times 180. Negative degrees per decade. So we get a very large slope over this range of frequencies, extending from 10 to the minus 1 over 2q times 400 hertz. That works out to be 300. Up to 533 hertz, which is 10 to the plus 1 over 2q times 400 hertz. So here are the composite phase asymptotes. And at high frequency, we finally end up above 26 kilohertz with a constant phase asymptote of minus 260, sorry, minus 270 degrees. Here are the asymptotes for the line to alpha transfer function of the buck loose converter. And again they don't include the minus sign from the polarity inversion. This uh,transfer function looks similar to the control to Alpha transfer function, except that it has a dif, different DC gain, three and a half dB in this case, and it doesn't have the right f plane zero. So, the high frequency asymptote is minus 180 degrees, rather than minus 270. But otherwise, it's similar. Okay? So, we constructed the, control to output and line to output transfer function of the buckbrush converter. I went through, really, the, Direct algebraic solution of the model and developed the expressions for the salient features, which we then used to construct the isotopes.
