Lost Secrets of the H-Bridge, Part II: Ripple Current in the DC Link Capacitor

Okay! Now let's just make sure we can run this, to show ripple current in the load inductance. Remember, we're normalizing ripple current to \( I_{R0} = VT/L \).

Now we have to figure out DC link capacitor current, given the inductor current. As I mentioned earlier,

$$I_{CDC} = (s_A - s_B) I_L - I_S$$

where s_A and s_B are the switching functions of nodes A and B: 1 if the high-side switch is on, 0 if the low-side switch is on.

Let's write some Python to simulate this, given what we know so far:

That's it! Really! Now let's add this to our PWM graph:

That wasn't so hard! Now we can simulate any situation where we know D_a, D_b, and I_Ldc. The symbolic analysis is a bit tricky (and grungy), so let's divide and conquer.

Superposition

Superposition is a very powerful tool. It's a way of understanding linear systems by separating out pieces, analyzing the results, and then adding them back together. Linear systems allow you to do this: the response of a system to the sum of two inputs is the same as the sum of the response of the system to each input separately.

In this case, how should we divide the problem?

One way is to split up the capacitor current by separating out the contribution due to the DC current I_S:

This particular split is useful for determining zero-to-peak and peak-to-peak currents. We can see pretty clearly from the middle graph that the nonzero parts of this funky waveform are equal to the current in the inductor, and we know its maximum current from the information we had last time on the inductor ripple current values. To be clear on the terminology again, we normalized to \( I_{R0} = VT/L \), and the times for nonzero capacitor ripple current are from t₁ to t₂, and from t₄ to t₅.

$$\begin{eqnarray} I_L(t_0) & = & I_L(t_3) = I_L(t_6) = 0 \cr I_L(t_1) & = & -I_L(t_5) = D \times (|D| - 2 D_0)/4 \cr I_L(t_2) & = & -I_L(t_4) = D \times (2-|D| - 2 D_0)/4 \cr D & = & D_a - D_b \cr D_0 & = & (D_a + D_b) / 2 \end{eqnarray}$$

We later figured out that

$$\max(|I_L(t_1)|,|I_L(t_2)|) = \underbrace{\frac{|D|(1-|D|)}{4}} _ {I_R/4} + \underbrace{\frac{|D| |D_0 - \tfrac{1}{2}|}{2}} _ {I_{R2}/4} $$

Also, last time, we analyzed current ripple with I_Ldc = 0, so we'll have to add that in:

$$\mathrm{peak}(I_L) = \frac{I_R + I_{R2}}{4} + |I_{Ldc}| $$

This is also the peak-to-peak value of I_cdc. Because its waveform is asymmetric, there are two different zero-to-peak values, one positive and one negative:

$$\begin{eqnarray} \mathrm{peak_+}(I_{cdc}) & = & \frac{I_R + I_{R2}}{4} + |I_{Ldc}| - I_S = \frac{I_R + I_{R2}}{4} + |I_{Ldc}|(1-|D|) \cr \mathrm{peak_-}(I_{cdc}) & = & -I_S = -DI_{Ldc} \end{eqnarray}$$

These equations cover the case where there is positive power flow out of the H-bridge -- that is, \( I_S = DI_{Ldc} > 0 \). For regeneration, where \( I_S = DI_{Ldc} < 0 \) and power flows back into the H-bridge, the peak-to-peak current is the same, but the zero-to-peak current signs are reversed:

$$\begin{eqnarray} \mathrm{peak_+}(I_{cdc}) & = & -I_S = -DI_{Ldc} \cr \mathrm{peak_-}(I_{cdc}) & = & -\frac{I_R + I_{R2}}{4} - |I_{Ldc}| - I_S = -\frac{I_R + I_{R2}}{4} - |I_{Ldc}|(1-|D|) \end{eqnarray}$$

Superposition choice 2: RMS

To analyze RMS current we'll choose a different separation of capacitor currents, which I'll call pulse-and-ramp:

The ramp capacitor current \( I _ {CDC\_ramp} \) is the capacitor current that would result only from the inductor ripple current, and not from its DC value, so here we assume \( I _ {Ldc} \) = zero.

The pulse capacitor current \( I _ {CDC\_pulse} \) is the capacitor current that would flow with the average load current \( I _ {Ldc} \) at its real value, but with large enough inductance that the inductor ripple current were zero.

Got that? Here it is again, more concisely:

Ramp capacitor current results from inductor ripple current. It is independent of the DC inductor current.

Pulse capacitor current results from DC inductor current. It is independent of the inductor ripple current.

Why did we choose this? First, it's a little easier conceptually, since these are two different causes. More importantly, it's because the mean value of both currents is zero, and the two currents are orthogonal signals:

\( \int I _ {CDC\_ramp} I _ {CDC\_pulse} \enspace dt = 0 \)

I'm not going to prove this, but if you're clever, you can see it by noting these things:

• the ramp current is nonzero at the same times when the pulse current is nonzero
• during these two intervals, the pulse current has a constant value of \( I_{Ldc} \mathrm{sign}(D) - I_S \)
• the mean value of the ramp current is zero

OK, so they're orthogonal signals — so what? It turns out that the RMS value of the sum of orthogonal signals is the root-sum-square of each signal's RMS value. We can calculate the RMS value for each of them independently, and merge the results together into one final calculation. (This isn't true for non-orthogonal signals.)

Let's start with the ramp capacitor current.

Ramp capacitor current

Let's plot the ramp capacitor current by itself, and remember what the inductor currents are (for \( I_{Ldc} = 0 \)), to keep things fresh in our mind:

$$\begin{eqnarray} \text{with} \hspace{0.2em} I_{Ldc} & = & 0: \cr I_L(t_0) & = & I_L(t_3) = I_L(t_6) = 0 \cr I_L(t_1) & = & -I_L(t_5) = D \times (|D| - 2 D_0)/4 \cr I_L(t_2) & = & -I_L(t_4) = D \times (2-|D| - 2 D_0)/4 \cr D & = & D_a - D_b \cr D_0 & = & (D_a + D_b) / 2 \end{eqnarray}$$

The RMS of the ramp capacitor current should be less than the RMS inductor ripple current, since the waveforms are the same except that capacitor ripple current is zero outside the time intervals \( [t_1, t_2] \) and \( [t_4, t_5] \). The length of each of these intervals is just \( |D_a - D_b|/2 = |D|/2 \). We're also going to define \( D_{0dev} = D_0 - 0.5 \), which is the deviation of center-aligned PWM from the optimal common-mode duty cycle \( D_0 = 0.5 \).

Again, we have to help sympy a little bit with square roots, but this isn't too hard to manually simplify. While we're at it let's add back our normalized ripple current I_R0 = VT/L:

$$ RMS(I _ {CDC\_ramp}) = I _ {R0} \frac{|D|^{3/2}}{4\sqrt{3}}\sqrt{12(D_0-\tfrac{1}{2})^2 + (1-|D|)^2} $$

In the "nice" case where D₀ = 0.5, this simplifies to

$$ RMS(I _ {CDC\_ramp}) = I _ {R0} \frac{|D|(1-|D|)}{4\sqrt{3}}\sqrt{|D|} $$

and in both cases, \( RMS(I _ {CDC\_ramp}) = \sqrt{|D|} \times RMS(I _ {L\_ripple}) \).

Let's graph the RMS capacitor current, as a function of duty cycle. If you go through some more algebra, you'll find out it has its maximum value at D = ±0.6. While we're at it, let's plot the RMS capacitor ripple for common-mode duty cycles different from 0.5: (the results are the same whether we go above 0.5 or below 0.5)

Let's just spot-check this for a few values to make sure we haven't made any obvious mistakes:

Pulse capacitor current

Pulse capacitor current is much simpler. It is equal to \( I_{Ldc} (1-|D|) \mathrm{sign}(D) \) in the intervals \( [t_1,t_2] \) and \( [t_4,t_5] \) (when voltage is applied across the load), a total time of \( |D|T \), and is equal to \( -DI_{Ldc} \) during the rest of the time when the load is short circuited, a total time of \( (1-|D|)T \).

The RMS value of this waveform is easy to calculate, since it's piecewise constant: we just take the square root of the sums of squares of each constant, weighted by the total fraction of the time period it is active:

$$\begin{eqnarray} RMS(I _ {CDC\_pulse}) & = & \sqrt{|D| \times { I _ {CDC}}^2 \big| _ \text{Vdc across load} + (1-|D|) \times {I _ {CDC}}^2 \big| _ \text{shorted load} }\cr & = & I _ {Ldc}\sqrt{|D|(1-|D|)^2 + (1-|D|)D^2} \cr & = & I _ {Ldc}\sqrt{|D|(1 -|D|)}\sqrt{(1 -|D|) + |D|} \cr & = & I _ {Ldc}\sqrt{|D|(1 -|D|)} \end{eqnarray}$$

Total RMS capacitor current

We figured out that \( RMS(I _ {CDC\_ramp}) = \sqrt{|D|} \times RMS(I _ {L\_ripple}) \)

Now if we combine RMS values for ramp and pulse currents together, we get:

$$\begin{eqnarray} RMS(I _ {CDC}) & = & \sqrt{RMS(I _ {CDC\_ramp})^2 + RMS(I _ {CDC\_pulse})^2} \cr & = & \sqrt{ |D| \times RMS(I _ {L\_ripple})^2 + I _ {Ldc}^2 |D|(1-|D|)} \cr & = & \sqrt{ |D| } \sqrt{RMS(I _ {L\_ripple})^2 + (1-|D|)I _ {Ldc}^2} \end{eqnarray}$$

With \( D_0 = 0.5 \), this simplifies to

$$\begin{eqnarray} RMS(I _ {CDC}) & = & \sqrt{ |D| \times \frac{{I_R}^2}{48} + I _ {Ldc}^2 |D|(1-|D|)} \cr & = & \sqrt{ |D| (1-|D|) } \sqrt {\tfrac{27}{4}D^2(1-|D|)\left(\frac{I _ {R0}}{18}\right)^2 + I _ {Ldc}^2 } \end{eqnarray}$$

Blah, blah, blah, algebra, magic numbers, blah blah. So what? Here's what the function \( \frac{27}{4}D^2(1-|D|) \) looks like:

It has its maximum value of 1 at \( D=\pm \frac{2}{3} \).

This means that in order for ramp current and pulse current to have equal contributions to RMS capacitor ripple, even at \( D=\pm \frac{2}{3} \), \( I_{Ldc} \) would have to be 18 times smaller than \( I_{R0} \).

Here's what this case looks like:

This is so near to \( I_{Ldc} = 0 \), that in normal cases, when \( I_{Ldc} \) is much larger, the net impact of inductor ripple current to the RMS capacitor current is essentially negligible. We'll talk more about this and other practical issues a bit later.

Edge-aligned PWM

So far, so good. But let's just cover the edge-aligned PWM case as well, just to be thorough.

Again, we can do the same partitioning for analysis of peak and RMS currents. Fortunately the edge-aligned PWM case is easier to analyze.

Peak capacitor current, edge-aligned PWM

The peak-to-peak capacitor current is the maximum load inductor current, which is \( \mathrm{peak}(I_{L}) = |I_{Ldc}| + \tfrac{1}{2}I_R = |I_{Ldc}| + \tfrac{1}{2}|D|(1-|D|)I_{R0} \).

The zero-to-peak values of capacitor current for positive power flow \( DI_{Ldc} > 0 \) are

$$\begin{eqnarray} \mathrm{peak_+} (I_ {CDC}) & = & |I_{Ldc}|(1-|D|) + \tfrac{1}{2}I_R \cr \mathrm{peak_-} (I_ {CDC}) & = & -DI_{Ldc} \end{eqnarray}$$

and for regenerative (negative) power flow \( DI_{Ldc} < 0 \) are

$$\begin{eqnarray} \mathrm{peak_+} (I_ {CDC}) & = & -DI_{Ldc} \cr \mathrm{peak_-} (I_ {CDC}) & = & -|I_{Ldc}|(1-|D|) - \tfrac{1}{2}I_R \end{eqnarray}$$

RMS capacitor current, edge-aligned PWM

Again, let's split up capacitor current the other way:

Here's the contribution of ramp current, which is nonzero only for a time DT where it ramps from \( -I_R/2 \) to \( I_R/2 \):

Take the square root of this and we just get

$$RMS(I _ {CDC\_ramp}) = \frac{I_R}{2\sqrt{3}}\sqrt{|D|} = \sqrt{|D|} \times RMS(I _ {L\_ripple})$$

and again, if we've already calculated the load inductor RMS ripple current, all we have to do to get the RMS of capacitor ramp current is multiply by this factor of \( \sqrt{|D|} \).

For pulse current, the RMS value is the same as in the center-aligned PWM case (same amplitude, half the frequency of center-aligned PWM):

$$RMS(I _ {CDC\_pulse}) = I _ {Ldc}\sqrt{|D|(1 -|D|)} $$

The total RMS current, just as in the center-aligned PWM case, is

$$RMS(I _ {CDC}) = \sqrt{|D|}\sqrt{I _ {L\_ripple}^2 + (1-|D|)I _ {Ldc}^2}$$

When we simplify it in terms of \( I_{R0} \) we get a slightly different answer, since the edge-aligned case produces twice the ripple current as center-aligned PWM:

$$\begin{eqnarray} RMS(I _ {CDC}) & = & \sqrt{ |D| \times \frac{{I_R}^2}{12} + I _ {Ldc}^2 |D|(1-|D|)} \cr & = & \sqrt{ |D| (1-|D|) } \sqrt {\tfrac{27}{4}D^2(1-|D|)\left(\frac{I _ {R0}}{9}\right)^2 + I _ {Ldc}^2 } \end{eqnarray}$$

but the end conclusion is still about the same: maximum contribution of ramp current to RMS, compared to pulse current, is at \( D = \pm \frac{2}{3} \), and even at that point the DC current has to be pretty small for the ramp current to contribute equally.

Harmonic amplitudes

Capacitor ramp current

For both the edge-aligned PWM case, and the center-aligned PWM case with common-mode duty cycle \( D_0 = \frac{1}{2} \), the capacitor ramp current waveform is zero except for a linear segment from \( -I_{pk} \) to \( I_{pk} \), with slope = \( (1-|D|)I _ {R0} \). For the edge aligned case, this occurs once per period for a time DT. For the center-aligned case, it occurs twice per PWM period for a time |D|T/2. The wave shape is the same for both, but at twice the frequency for center-aligned PWM, and the peak current \( I_{pk} \) is half as much.

(Here I'm going to be a little bit sloppy, and shift the waveform in time so that the ramp in the waveform is centered at t=0.)

We'll evaluate the Fourier coefficients for the edge-aligned case; the center-aligned case includes an additional factor of \( \frac{1}{2} \). Since there is only one linear segment of the waveform, we have only one integral to evaluate:

$$A_k = \frac{2}{T}\int _ {-T/2}^{T/2}{e^{2\pi jkt/T}}I _ {CDC\_ramp}(t)dt = \frac{2}{T}\int _ {-|D|T/2}^{|D|T/2}{e^{j2\pi kt/T}}(1-|D|)I _ {R0}\frac{t}{T} dt$$

Let's use sympy to help get those Fourier coefficients.

It appears as though the general term is $$A_k = \frac{j(1-|D|)I _ {R0}(\sin k\pi |D| - k\pi |D| \cos k\pi |D|)}{k^2\pi ^2}$$ but let's double-check this just to be sure:

Not bad, although because of the step discontinuities there are Gibbs phenomena, so let's look at 100 Fourier coefficients:

The harmonic content of the center-aligned case with \( D_0 \neq \frac{1}{2} \) is possible to calculate as well, but I'll leave that as an exercise for the interested reader.

Harmonic content of capacitor pulse current

Here the waveform is pretty simple just a two-level piecewise constant waveform.

Except for its DC value, the harmonic amplitude of this is the same as any other waveform shifted up by a constant, so let's consider a square wave shifted so that its level near t=0 is at 0 and its level near t=± 0.5 is at \( sI_{Ldc} \) where \( s=\mathrm{sign}(D) \).

$$A_k = \frac{2}{T}\int_{-T/2}^{T/2}{e^{-2\pi jkt/T}}sI_{Ldc}dt $$

This looks like it's going to be pretty simple (smells like a sine function to me), and we could do it by hand, but let's give sympy a try:

Excellent! It's just $$A_k = \frac{2sI_{Ldc} \sin k\pi|D|}{k\pi} = \frac{2I_{Ldc} \sin k\pi D}{k\pi} $$

Let's verify:

There's those Gibbs ears again!

So the total harmonic amplitude, including both ramp and pulse components of capacitor current, is

$$A _ k = \underbrace{\frac{j(1-|D|)(\sin k\pi |D| - k\pi |D| \cos k\pi |D|)}{k^2\pi ^2}} _ {\text{ramp current}} + \underbrace{\frac{2I _ {Ldc} \sin k\pi D}{k\pi}} _ {\text{pulse current}}$$

3-phase PWM

Yet another exercise for the reader. Or you'll see it sometime in an appnote at http://www.microchip.com/motorcontrol when I get a chance to write one.

Now what?

Oh, my head aches with algebra! Thank you for your patience if you've read this far. It's time to summarize these results.

Summary

Here's a summary of capacitor ripple current statistics for H-bridge PWM.

Remember our definitions:

\( D = D_a - D_b \) is the effective load duty cycle

\( D_0 = (D_a + D_b)/2 \) is the common-mode duty cycle

\( I_{CDC} \) is the current flowing out of the DC link capacitor

\( I_S = DI_{Ldc} \) is the supply current

\( I_{R0} = V_{DC}T/L \) is a reference current that simplifies calculation of these statistics.

\( I_R = |D|(1-|D|)I_{R0} \) is a differential duty-cycle dependent ripple current

\( I_{R2} = 2|D||D_0-\tfrac{1}{2}|I_{R0} \) is a common-mode duty-cycle dependent ripple current

\( I_{Lpk} \) = Mean-to-peak load current ripple

\( I_{Lrms} \) = RMS load current ripple (does not include DC)

Here are statistics about \( I_{CDC} \):

Information from Part 1:

^*1st harmonic of capacitor current in center-aligned PWM has low amplitude if D₀ is close to 0.5

These statistics about DC link capacitor current can be expressed independent of the PWM type:

* \( m=1 \) for edge-aligned PWM, \( m=2 \) for center-aligned PWM with \( D_0 = \tfrac{1}{2} \). Calculating \( A_k \) for the general case of center-aligned PWM (\( D_0 \neq \tfrac{1}{2} \)) is an exercise for the reader.

Now that we have all these gory equations, next time we'll talk about some of the practical issues related to capacitor and inductor ripple current.

Happy switching!

© 2013 Jason M. Sachs, all rights reserved.

Stay tuned for a link to this code in an IPython Notebook. I'll put it up on the web, and I do plan to release this in an open-source license, I just have to do a little more homework before I pick the right one.