In this lecture we'll discuss what is known as the canonical circuit model. All PWM, continuous conduction mode DCDC converters perform the same basic functions, and as a result, their equivalent circuit models have the same general form. And the canonical circuit model is a way to represent that general form. So basically you can use the canonical circuit model. For any converter, and then we just plug in the specific element values for which ever converter you're using. one of the uses of the canonical model is that we can solve larger system problems in a generic sense for, for any converter. And then plug in the detail parameter values for whichever converter we're using to apply it to our particular situation. The canonical circuit model also is actually a way to solve the circuit for the transfer functions through circuit manipulations rather then doing algebra on paper, and so I'm going to illustrate how to do that also. So these basic functions are this. first of all the, the DC-DC converter transforms the voltage in current levels, ideally with 100% efficiency. And therefore we expect the model to contain DC transformers that represent that. We have low-pass filtering of the waveforms from the inductors and capacitors that are really there to filter the ripple, but they also filter signals at other frequencies as well. So the, the v hat and d hat variations get filtered by the L's and C's in the converter, also. we can control the, the circuit by variation of the duty cycle. And so the canonical model has sources that depend on the duty cycle variations, D-hat. So we have these independent sources in the model of any converter. And therefore all of these elements should be present in any of our converter equivalent circuits. So here is a, a development of the canonical model, based on simple physical principles. So again, we said this first order property that we're trying to obtain is the conversion ratio limb of d, with this effective dc transformer function. And so we will start our converter model by building it up beginning with this DC transformer. So the effective turns ratio is the converter conversion ratio M of D this conversion ratio is a function of the quiesce and duty cycle capital D, so that it c-, will vary if the steady state duty cycle varies. Here I've also illustrated the DC input voltage, capital v g, and the DC output voltage, capital v. So, solution of this model at this point says that capital V equals M of capital D times capital VG. The next thing we'll do is, is introduce variations in VG. So our power input source now, we will write as VG plus VG hat. So we can have variations in VG. And to first order what happens is that we would expect those variations to pass through the same conversion ratio and cause a vhat variations in the output. At this point what I'm doing is drawing the ideal transformer symbol containing both a DC line so that it passes DC. And an AC sig-, line in it so that we are now modeling how it passes the AC variations as well. So effectively, what we've done here is combine both the DC and AC quantities into a single equivalent circuit model, which really combines the, the DC models we talked about in previous weeks. With this week's AC models. The third step is introdu-, introduction of an effective low pass filter, so the, the inductors and capacitors that filter the ripple [COUGH] also filter the variations that come from VG HAT. In the canonical model what we do is we push all the inductors and capacitors to the output side of the equivalent circuit. And so we have some effective low pass filter there. It may turn out that the, the element values in a low-pass filter are not the same as the actual values in a converter. We're going to see in the example in a couple of minutes that we push an inductor through the transformer turns ratio and the inductor then becomes a function of D in this effective low-pass filter. So it is an effective low-pass filter that does depend on the values of L and C but it's not. the values may not be exactly L and C. The next step is to introduce variations in the du-, duty cycle, or the control input. So now we will represent the duty cycle as capital D plus D hat. Capital D effects the conversion ratio or transformer turns ratio, whereas the d hat causes introduction of control sources. So we have some independent sources of d hat sources, in the canonical model there are two sources one is a voltage source. A value E of SD hat and the other is a current source of value J of SD hat. In the canonical model we push all of the independent sources to the left side of the model. So we have both the VG sources and the duty cycle sources on the left. Okay we can talk about transfer functions that are predicted by the canonical model. If we look at this circuit as its drawn right now we have an output voltage, the AC output V hat. And it is the function of the two independent inputs, the VG hat input and the D hat input. So in general, we can express the output as a superposition of terms that come from the two different inputs. So there are some coefficients when we apply superposition. The coefficient of VG hat we call the transfer function gvg of s. And the coefficient of D hat we're going to call the transfer function gvd of s. And to find one of these transfer functions you set the opposite source to zero. So for example to find gvg what we do is this is the transfer function from vg hat to v hat. And it's found under the conditions that the other input or d hat is set to zero. Okay? So in order to calculate that we can just look at the circuit and see what the canonical model predicts. If we set d hat to zero that will make this d hat voltage source into a short, setting d hat to zero makes the current source into an open. And then we have the vg hat source connected up to the primary of the transformer. I would also note that to find AC transfer functions, we set DC quantities to zero, or DC biases. So the capital VG will be zero. Capital V will therefore be zero also. This doesn't mean that cat-, the DC component of EG is zero in the actual converter. It simply means that in order to solve for this transfer function, we set the DC quantities in our model to zero. Okay so with vg hat applied across the primary of the transformer, the voltage at the secondary side would be vg hat multiplied by the turns ratio m. That voltage gets multiplied by the transfer function of our low-pass filter, so that the voltage at the output will be n times that transfer function times vg hat. So then this transfer function from vg hat to v hat is n times he of s, like this. Likewise, to find the control to output transfer function, gbd of s, this is found as the transfer function from d hat to v hat, under the conditions that we set the other source, vg hat, to zero. So, let's set vg hat to zero here, and capital VG to zero. Vg hat is set to zero, so we get a short circuit, at the input of our converter. That shorts out the current source. And therefore, the voltage across the input of the transformer is e of s times d hat. The voltage coming out of the transformer then will be the turns ratio m times ed hat. And the voltage at the output of our converter then would be this quantity multiplied by the effect of low-pass filter transfer function. So I get m, or he of s, times m of d, times e of s, times d hat. [COUGH]. So this is the controlled throughout the transfer function of gvd, as shown here. We can solve other quantities here if we want as well. For example, we could find the input impedance or the output impedance and we'll be talking about those later. OK, here is the buck boost small signal model that we previously derived. Let's try to manipulate this into canonical form. So in order to do that, we need to push all the d hat sources or independent sources to the left, to the input side of, of our model. And we need to push the L's and C's to the right, and then finally in the middle we need to combine the transformers. So lets do that. I'm going to do it one step at a time. And what we'll do is we'll take this current source and push it through the transformer. When we do that since the dots are reversed on this transformer, it reverses the polarity, so it will point down. And then, we also have to divide by d prime, this turns ratio, so that on the primary side of this transformer, we get i d hat over d prime. I'm also going to push this voltage source through to here. Okay? The voltage source is in series with the inductor. We can write them in any order so, we can just push this straight through the inductor. And then when we push it through the turns ratio of this transformer we'll have to what, divide by d. So this voltage source would be vg minus v over d times d hat. OK, that's shown on this slide, here's the current source, and here's the voltage source. The next thing to do is we need to push this current source past this inductor. Now that's maybe not so obvious how to do it, but here's the trick. What we do is break this, this path and connect the current source there. Now you can't just do that. what, in order to compensate for connecting the current there instead of here, we need to put another current to compensate. And basically we're taking this current and putting into this node where it doesn't belong. So we need to take the correct back out of the node and put it where it does belong. And this current source then is the same value as that one, it's capital ID hat over d prime. So I've shown that right here. My claim then is that we're allowed to do this because the loop and node equations of this circuit are unchanged. Since we put the current into the node and then take it back out again we effectively don't change the node equation there. Now what do we do with this? We can take this second current source and push it through the transformer to here. And what? In the process of doing that we would multiply by d, the turns ratio. So this becomes d over d prime times capital I times d hat. Okay, so that takes care of that one. What do we do with this current source? Well, there's a trick for that also. What we have to do is make a thevenin equivalent of this inductor in parallel with this current, so we have basically this between these terminals here. And, let's see. We have a impedance of SL, and we have a current going that way of id hat over D prime. So this is a, we can view this as a Norton equivalent type circuit. We need to turn it into a Thevenin equivalent circuit. So it would be an impedance in series with a voltage source. And the voltage source would be the open circuit voltage between these terminals. So it would be what? Plus on the left, minus on the right, and it will have a value equal to the current times the impedance. So sL times capital I over D prime times d hat. So this is a thevenin equivalent model for this network, and we can place, put that in, in place of this. So that's what we have here. Here's the Thevenin equivalent voltage and inductance. And looks like I haven't pushed the current source through the transformer yet. Next step, we'll push this current source through the transformer, as I described previously. And once we get it here, we will further simplify by trying to push it past this voltage source. So that's on the next slide. Here is the current source written there. And to push it past the voltage source, we do the same trick that we did with the inductor. We're going to break this connection. Put the current source into this node. And then take it right back out again and put the current into the node where it belongs. Okay, so this further simplifies the circuit. We now have two current sources in parallel that we can add. So we make a composite source that would be, id hat times what, one plus d over d prime. And then as for these, we have a voltage source in parallel with a current source between these terminals. Okay, what, what is the IV characteristic between these terminals of the voltage source in parallel to the current source. Well in this case the voltage source wins and the current source makes no difference. The independent voltage source holds the voltage at, between these terminals to be the given value, regardless of the current. So in fact, you can just ignore the current source. Okay. Last step, we will push the inductor through the transformer. We'll push this voltage source to this transformer. When the inductor gets pushed through the transformer, we have to divide by the terms ratio squared or D prime squared. So we get this. Here's the effective low-pass filter, has the inductance over D prime squared, plus the capacitor. The two transformers are now combined. And the sources are combined as well. So we had, we got two voltage sources that I then combined and we have a current source. Okay, an interesting thing happened when we combined the two voltage sources. One of the voltage sources depended on the inductor impedance and it picks up an s times L factor inside it. So the coefficient of d hat is now frequency dependent, and it has terms that depend on s. In general, the e of s and j of s coefficients of the d hat sources can be functions of s. And in this case, we picked up the factor of s by pushing the current source past the inductor. Here is that e of s coefficient for the e of s d hat voltage source. We can actually use the steady state relationships for the, this converter. For example, v is, the output voltage. as a function of the input voltage, vg, and duty cycle, and our expression for the inductor current, capital I, can be simplified from the steady state model. And when we do that, we can then express e of s in this form. Okay? This frequency dependence is actually causes control problems and we're going to talk about those in the next chapter. So here then is the canonical model for the basic buck, boost and buck-boost converters. The effective low-pass filter contains an inductor and capacitor. Although the value of inductance can be a function of duty cycle, and we can get the d hat sources and the m of d transformer. And here's a handy table that gives the results, so the element values in the canonical model are listed here. And if you want to apply or find the AC model for one of these converters you can then, the work is done for you you can just simply plug into this table to get your model.