0:12

We will use a boost converter as an example.

Â In this boost converter the input DC voltage is set to 170 volts,

Â and the converter is operated to produce an output

Â voltage of 400 volts with 2 kilowatts of output power.

Â The inductance is 250 microhenries.

Â The output filter capacitance is 33 microfarads.

Â We operate a converter at 100 kilohertz,

Â with a pulse width modulator, that is assumed to have VM of four volts.

Â The equivalent current sensing resistance is assumed to be 0.25 ohms.

Â The objective is to resign the current control compensator, Gci(s),

Â so that the cross-over frequency for

Â the current control loop is equal to 10 kilohertz.

Â And so that we have at least 45 degrees of phase margin.

Â Notice the switching frequency is 100 kilohertz and now the goal is to design

Â the cross-over frequency to be 1/10 of the switching frequency.

Â First, let's see the steady-state solution.

Â 1:29

Assuming our current control loop works perfectly,

Â meaning the input current is exactly equal to the reference value,

Â given the specified input voltage and output voltage,

Â we find that the duty cycle of the converter should be equal to 0.575.

Â The DC input current, which is equal to the DC value of the inductor current,

Â is, neglecting losses, equal to 11.8 amps.

Â 2:03

And assuming the control input equals the sensed value of the current, in DC sense,

Â the control input would need to be equal to 0.25 ohms, the assumed value for

Â the equivalent current sensing resistance Rf,

Â and the DC value of the current 11.8 amps,

Â which gives us the control input of around 3 volts.

Â Now at this point here you may be concerned about one particular detail.

Â 2:34

If the sensing resistance of Rf,

Â we use to sense the inductor current of 11.8 amps,

Â the power loss on that sense resistor would

Â be Rf IL squared, in the order of 35 watts.

Â Well, that's clearly a lot, and would be considered unacceptable, not practical.

Â 3:24

One practical approach is illustrated here, where a relatively small value of

Â the sense resistance is placed in series with the current we're interested in.

Â In this case, the inductor current which flows also in this branch right here.

Â The circuit ground is assumed to be placed at the source of the power transistor.

Â And then the current iL is observed as the voltage drop

Â across the current sense resistance Rs.

Â 3:58

The voltage drop is then amplified by a simple inverting

Â op-amp configuration with a gain equal to the ratio

Â of the two resistors, Rs2 over Rs1.

Â So as an example, if we choose 10 milliohms for

Â the current sensing resistance, the power dissipation

Â on that current sense resistance would be around 1.4 watts,

Â which can be considered acceptable and practical.

Â It is a very small fraction of the output power of 2 kilowatts in this converter.

Â But to reach the desired value of the equivalent current sense resistance,

Â we could employ a gain in the sensing path of Rs2 over Rs1 = 25.

Â So 25 times 10 milliohms then gives us this assumed value of

Â 0.25 ohms for the equivalent current sense resistance.

Â This is just one possible way of sensing current.

Â There are a number of other ways that are used in practice.

Â But in any case, it is certainly important to understand that the equivalent

Â current sense resistance represents the total value of the gain between

Â the sensed current and the voltage value that shows up in the control loop.

Â And it doesn't mean that that resistance is inserted in series with

Â the current flowing in the power converter.

Â Now that we have current sensing and steady-state solution out of the way,

Â let's proceed with the design of the current control loop.

Â Remember that the design is based on our expression for

Â the current control loop gain.

Â The product of Rf, Gci, which is really what we want to determine,

Â a gain of the pulse width modulator and

Â the control to output transfer function Gid.

Â 5:55

The first step is really to determine that control to output transfer function Gid.

Â We would typically do that by circuit analysis of an equivalent small

Â signal circuit model that we already know we have for the converter.

Â For the boost example, here is the boost small signal equivalent circuit model.

Â And by circuit analysis you can find out the Gid(s)

Â which is really the ratio of iL hat over d hat

Â when all other independent inputs are set to 0, is equal to

Â a low frequency gain times a transfer function that has a 0

Â in the numerator and a pair of poles in the denominator.

Â We have analytical expressions for the low frequency gain, and

Â we can plug in numerical values to find what that low frequency gain is.

Â And similarly, we can find analytical expressions for

Â the corner frequencies f0 and the zero frequency fzi,

Â and evaluate those values given the parameter values for the example.

Â 7:06

The zero frequency comes out to be around 120 Hertz,

Â and the center frequency of the pair of poles is 745 Hertz.

Â The Q factor of the pair of poles is relatively large, 12.4.

Â So now that we have the duty cycle to current transfer function,

Â we are in position to sketch the

Â magnitude and phase responses of the uncompensated current loop gain.

Â Let's do that.

Â So here is the expression for the uncompensated current loop gain.

Â You will notice immediately that this is simply a scale factor, Rf times 1 over Vm

Â times the expression we had found for the duty cycle to current transfer function.

Â So we have zero in the numerator, a pair of poles in the denominator, and

Â a low frequency gain in front,

Â which now includes the factors associated with the current sensing

Â resistance Rf

Â and the gain of the pulse width modulator.

Â The low frequency gain of the un- compensated current-loop gain comes out to

Â be 10.8 dB.

Â 8:16

So, here's a sketch of the magnitude response of

Â the uncompensated current-loop gain.

Â We have a low frequency gain, zero at 120 hertz, followed

Â by a pair of polls at 745 Hz with relatively high Q factor,

Â then magnitude response rolling off at -20 dB per decade.

Â 8:46

And we see that at the frequency that is the desired cross-over frequency,

Â the magnitude response is well below zero dB,

Â which really means that our current-loop compensator

Â will have to provide an additional gain in the loop, so

Â that we can bring cross-over frequency to ten kilohertz.

Â The amount of gain needed to do so

Â can be found very simply by noting that around the cross-over frequency,

Â which is well above the zero frequency and well above the frequency of the poles,

Â we have a simple asymptotic behavior of the uncompensated

Â current-loop gain that can be found as follows.

Â You see in the expression for uncompensated current control loop,

Â we have a low frequency gain, zero and a pair of poles.

Â In the range of frequencies around the crossover frequency,

Â we are well pass the zero frequency, so

Â that we can approximate the numerator with just s over w zi.

Â We are well pass the frequency of the pair of poles.

Â So in the denominator, we can approximate the transfer

Â function with just s square over omega_0 square.

Â So that in this range of frequencies,

Â Ti(s) is going to behave as Tiu0

Â times s over omega_zi over s squared

Â over omega_0 squared and

Â you see that the result is equal

Â to Tiu0 omega_0 squared over omega_zi

Â over s or a constant over s.

Â And indeed, we see that in the part of the magnitude response,

Â we have a rolloff at minus 20 dB per decade in that range of frequencies.

Â And we have indeed,

Â a simple expression, a constant over s, that represents the behavior of

Â the uncompensated current-loop gain around the cross-over frequency.

Â This is summarized right here.

Â So, asymptotic behavior of the uncompensated loop gain around

Â the cross-over frequency can be represented as a constant over s.

Â When we plugin analytical expressions for these three values that

Â we have in the numerator, we obtain a very simple expression for

Â that constant, which depends only on the gain and the pulse-width modulator,

Â the equivalent current sensing resistance Rf,

Â the dc value of the output voltage, and the value of the inductance L.

Â 11:52

Let's look at this result here from an intuitive point of view.

Â It is in fact possible to come up with this asymptotic

Â behavior of the uncompensated current control loop gain

Â simply by observing the behavior of the equivalent circuit model of

Â the boost converter.

Â At frequencies around the cross-over frequency, which is a relatively high

Â frequency compared to the corner frequencies of the converter,

Â 12:36

The output voltage perturbation can be assumed to be

Â essentially 0 at such high frequencies.

Â What that really means is that when you look at the iL_hat

Â response on the input side.

Â And of course, in the process here, we assume that vg_hat =0. All

Â independent inputs other than d_hat are set to 0.

Â You can solve the circuit here by noting that this voltage here is approximately 0,

Â which means that the primary side of this transformer is approximately 0.

Â And so the equivalent circuit model behaves simply,

Â as a controlled voltage source V d_hat

Â in series with inductor L and

Â the current of interest is iL_hat.

Â So you see here immediately that iL_hat

Â over d hat is going to be equal to V,

Â which is equal to the output voltage capital V=Vout.

Â 13:49

So Vout over sL, the impedance of the inductor in

Â series with that voltage source.

Â In the control loop, we have additional gains.

Â Rf and 1 over Vm, which is why we can say that the asymptotic behavior

Â of the uncompensated loop gain around the cross-over frequency

Â can be approximated as a simple expression as shown right here.

Â 14:42

Around the cross-over frequency, the behavior

Â is going to depend on the type of the current-loop compensator we select.

Â Remember, we have found that the uncompensated loop gain has a roll off of

Â minus 20 dB per decade around the cross-over frequency.

Â This is what makes it very easy to design a compensator around that response,

Â which is a simple PI or lag compensator shown right here.

Â In fact, we could go even for

Â a simple just proportional compensator if we wanted to do so.

Â But typically, we do employ a PI compensator with an inverted zero

Â in the numerator, because we want to make sure that the DC error in

Â the control loop is going to go to zero when the loop is well designed.

Â In addition to this inverted zero, we also typically place

Â a high-frequency pole in the response of the compensator.

Â And that pole is what brings in the low-pass nature to

Â the current-loop compensator and filters out switching ripple or

Â any noise associated with sensing the current in the switching power converter.

Â So, a most typical compensator in the current control loop is of this type.

Â Again, Gcm, this gain is going to be selected, so

Â that we position the cross-over frequency where we want.

Â And then the zero and the pole, inverted zero and the pole, those are placed

Â around the cross-over frequency to obtain an adequate amount of phase margin and

Â to make sure that our loop is stable and well behaved.

Â 16:57

And so we can conclude that around the cross-over frequency, importantly, the loop gain

Â is a simple constant over s, and that constant can be moved up and

Â down by the choice of G sub cm in order to position the crossover frequency or

Â the frequency where the magnitude of Ti is equal to 0db wherever we like.

Â And that's exactly how we choose Gcm.

Â 18:15

Positioning fz lower and fp higher will result in a higher,

Â or larger amount of phase margin

Â but the filtering of the high frequency noise would then be compromised

Â on the high frequency end and we would have a narrower

Â range of frequencies where our loop gain is very large on the lower frequency end.

Â So as a compromise we choose this factor of 2.5.

Â Typically you can choose that factor to be anywhere between, let's say, 2 and 5 for

Â where to place the zero frequency below the crossover frequency and

Â where to place the pole frequency above the desired crossover frequency.

Â With this particular choice,

Â we get a phase margin of about 46 degrees which meets the objective.

Â The last step is to implement the PI (lag) compensator and

Â the typical op-amp based implementation is shown right here.

Â It is quite simple.

Â The reference value for the control input is brought to

Â the plus input of the op-amp, around the op-amp a circuit is constructed

Â by having a resistance of R1 in front from the sensed value

Â of the current of interest to the negative input of the op-amp and

Â in the feedback circuit, we have an RC network that gives us the ability

Â to position the inverted zero and the additional high frequency pole.

Â 20:19

The zero frequency is determined by the values of R2 and C2,

Â it's right here and the pole frequency is determined by the values

Â of C3 and R2 and that is the expression that we have right here.

Â And you can verify that with these numerical values that are shown right here

Â we get the zero, pole and

Â the gain values that are very close to the values that we have computed.

Â