0:12

We will use a boost converter as an example.

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

and the converter is operated to produce an output

voltage of 400 volts with 2 kilowatts of output power.

The inductance is 250 microhenries.

The output filter capacitance is 33 microfarads.

We operate a converter at 100 kilohertz,

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

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

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

so that the cross-over frequency for

the current control loop is equal to 10 kilohertz.

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

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

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

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

1:29

Assuming our current control loop works perfectly,

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

given the specified input voltage and output voltage,

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

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

is, neglecting losses, equal to 11.8 amps.

2:03

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

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

the equivalent current sensing resistance Rf,

and the DC value of the current 11.8 amps,

which gives us the control input of around 3 volts.

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

2:34

If the sensing resistance of Rf,

we use to sense the inductor current of 11.8 amps,

the power loss on that sense resistor would

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

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

3:24

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

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

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

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

And then the current iL is observed as the voltage drop

across the current sense resistance Rs.

3:58

The voltage drop is then amplified by a simple inverting

op-amp configuration with a gain equal to the ratio

of the two resistors, Rs2 over Rs1.

So as an example, if we choose 10 milliohms for

the current sensing resistance, the power dissipation

on that current sense resistance would be around 1.4 watts,

which can be considered acceptable and practical.

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

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

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

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

0.25 ohms for the equivalent current sense resistance.

This is just one possible way of sensing current.

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

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

current sense resistance represents the total value of the gain between

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

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

the current flowing in the power converter.

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

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

Remember that the design is based on our expression for

the current control loop gain.

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

a gain of the pulse width modulator and

the control to output transfer function Gid.

5:55

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

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

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

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

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

which is really the ratio of iL hat over d hat

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

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

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

We have analytical expressions for the low frequency gain, and

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

And similarly, we can find analytical expressions for

the corner frequencies f0 and the zero frequency fzi,

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

7:06

The zero frequency comes out to be around 120 Hertz,

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

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

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

we are in position to sketch the

magnitude and phase responses of the uncompensated current loop gain.

Let's do that.

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

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

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

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

a low frequency gain in front,

which now includes the factors associated with the current sensing

resistance Rf

and the gain of the pulse width modulator.

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

be 10.8 dB.

8:16

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

the uncompensated current-loop gain.

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

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

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

8:46

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

the magnitude response is well below zero dB,

which really means that our current-loop compensator

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

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

The amount of gain needed to do so

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

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

we have a simple asymptotic behavior of the uncompensated

current-loop gain that can be found as follows.

You see in the expression for uncompensated current control loop,

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

In the range of frequencies around the crossover frequency,

we are well pass the zero frequency, so

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

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

So in the denominator, we can approximate the transfer

function with just s square over omega_0 square.

So that in this range of frequencies,

Ti(s) is going to behave as Tiu0

times s over omega_zi over s squared

over omega_0 squared and

you see that the result is equal

to Tiu0 omega_0 squared over omega_zi

over s or a constant over s.

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

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

And we have indeed,

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

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

This is summarized right here.

So, asymptotic behavior of the uncompensated loop gain around

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

When we plugin analytical expressions for these three values that

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

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

the equivalent current sensing resistance Rf,

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

11:52

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

It is in fact possible to come up with this asymptotic

behavior of the uncompensated current control loop gain

simply by observing the behavior of the equivalent circuit model of

the boost converter.

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

frequency compared to the corner frequencies of the converter,

12:36

The output voltage perturbation can be assumed to be

essentially 0 at such high frequencies.

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

response on the input side.

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

independent inputs other than d_hat are set to 0.

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

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

And so the equivalent circuit model behaves simply,

as a controlled voltage source V d_hat

in series with inductor L and

the current of interest is iL_hat.

So you see here immediately that iL_hat

over d hat is going to be equal to V,

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

13:49

So Vout over sL, the impedance of the inductor in

series with that voltage source.

In the control loop, we have additional gains.

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

of the uncompensated loop gain around the cross-over frequency

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

14:42

Around the cross-over frequency, the behavior

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

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

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

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

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

In fact, we could go even for

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

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

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

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

In addition to this inverted zero, we also typically place

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

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

the current-loop compensator and filters out switching ripple or

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

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

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

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

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

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

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

16:57

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

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

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

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

And that's exactly how we choose Gcm.

18:15

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

or larger amount of phase margin

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

on the high frequency end and we would have a narrower

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

So as a compromise we choose this factor of 2.5.

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

where to place the zero frequency below the crossover frequency and

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

With this particular choice,

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

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

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

It is quite simple.

The reference value for the control input is brought to

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

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

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

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

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

20:19

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

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

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

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

we get the zero, pole and

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