Help designing a step-down converter using linear ltc3630a

[Futterama]

Hi forum,

I need to regulate a 42V LiPo battery voltage down to 5V for MCU logic and 16V for a MOSFET driver.
I have found the LTC3630A to be perfect for my needs: http://cds.linear.com/docs/en/datasheet/3630Afa.pdf

So I started reading the datasheet and doing the calculations on inductor and capacitor values. When I had done the calculations in my spreadsheet, I went on to verify my calculations using the values in the design example on page 17.

The first thing I notice in the design example is that the VIN(MAX) of 80V is never used in the calculations. So I checked with the calculation discussion earlier in the datasheet and notices that the design example does not use the VIN(MAX) value when calculating the minimum inductor value, but according the the calculations on page 12, it should.

The next thing I notice is that the calculation for Cout is completely different from the calculations earlier in the datasheet. They also give different results. I get confused because I'm not used to do these kind of calculations, so I cannot see which calculation is the most correct to use and Linear doesn't seem keen to answer my web inquiry.

Please advise me so I can buy the correct value components - the ceramic capacitors are pretty expensive in these voltage/capacitance ratings ;-)

Best regards,
Martin

 

[WTP Pepper]

Resorting to old tech, try ringing Linear Tech's applications engineers. In the past I have always found them pretty spot on and can walk you through a specific design.
Much the same for those I have spoken to at TI.

 

[MrAl]

Hi,

One question that comes up is, do you really need to use the minimum value inductance or can you use a somewhat higher value anyway?

In figure 3 they seem to show a graph which indicates they did take it up to about 75 volts. They show L=22uH operating at about 130kHz, and for voltages from 0 to about 75 volts. So i would think if you went up to about 150kHz you should be ok, and 200kHz even better, as the graph would indicate something like 15uH for 200kHz so 22uH is even safer.

But you also have to consider the peak current. With the maximum 'on' time and maximum input output differential you can calculate the peak current. Then oversize by say 50 percent and you should be ok.

The inductor equation is:
v=L*di/dt

or solving for L:
L=v*dt/di

or solving for di:
di=v*dt/L

Now here di should be less than the rating for the inductor. So if the inductor is rated for 1 amp and di is 0.5 amps that should be ok.

Next, if your IC has slow turn on built in you dont have to worry about turn on surge. Turn on surge could cause a higher than typical peak current during start up because of the reasonance between L and C. I think this IC has slow turn on which they also call "soft start". If so, you should use it. Sometimes it requires a capacitor selection where that capacitor regulates the ramp up speed of the output. If the ramp is controlled there is no or little surge, but if there is no ramp control then there is most likely a surge. The surge lasts for a time depending on a number of factors but usually isnt too very long. But using the slow start, slow turnon, or soft start mechanism gets rid of this anyway. Otherwise you'd have to be sure the inductor goes into soft saturation to prevent overloading the drive circuit or causing an overvoltage at the output when the circuit is switched on.

 

[Futterama]

Hi,

I don't need the lowest inductor value possible, I would just like to do the math and make sure I don't choose a too small inductor value since this could damage the chip.

The peak inductor current is limited by the chip to 1.2A typical. As you said, a good idea is to use an inductor that can handle more than that.

I don't understand your equations. I know L is inductance, I guess dt is delta time, is this the on-time of the high-side switch (5µs at 200kHz)? I don't know the rest, di and v.

The chip also has soft start. If no capacitor is added: "The minimum soft-start time is limited to the internal softstart timer of 0.8ms."

I still can't figure out which equation is the correct one for calculating the output capacitor.

 

[alec_t]

I think the cause of confusion is that VIN(MAX) as defined on page 12 is not the same as the Absolute Maximum VIN rating (80V) as listed on page 2. It's my understanding that in your case VIN(MAX) is 42V.

 

[MrAl]

Hi,

I don't need the lowest inductor value possible, I would just like to do the math and make sure I don't choose a too small inductor value since this could damage the chip.

The peak inductor current is limited by the chip to 1.2A typical. As you said, a good idea is to use an inductor that can handle more than that.

I don't understand your equations. I know L is inductance, I guess dt is delta time, is this the on-time of the high-side switch (5µs at 200kHz)? I don't know the rest, di and v.

The chip also has soft start. If no capacitor is added: "The minimum soft-start time is limited to the internal softstart timer of 0.8ms."

I still can't figure out which equation is the correct one for calculating the output capacitor.

Hello again,

Apologies for not making that clearer. I think i meant to but then got side tracked with something else here.

The basic operation of the switch is to turn on when we need more output, and when the switch turns on it puts a voltage across the inductor. So at that point in time the inductor is connected right to the input source voltage, and the other side of the inductor is connected to the top of the output load, and that is usually positive.
So when the switch turns on, the inductor sees the input voltage minus the output voltage:
vL=Vin-Vout

where vL is simply the voltage measured across the inductor at some instantaneous time t.

So with simple math, if the input is 100 volts and the output is 10 volts, then the inductor sees 90 volts across it:
vL=100-10, so:
vL=90 volts.

During startup however there is no output voltage so the inductor sees the whole input voltage:
vL=100-0=100 volts across the inductor.

Now the inductor responds by allowing current to flow through it, but it only allows it to build up somewhat slowly because that's what inductors do. So we end up with a ramping current that starts at zero and keeps getting higher and higher as time dt progresses:
iL=vL*dt/L

So if we had an inductor that was 1uH (one microhenry) and we allowed the voltage vL=100 to exist across the inductor for 1us (one microsecond) then we would have a current at the end of the time dt=1us of:
iL=100*(1e-6)/(1e-6)
or:
iL=100 amps.

So at the end of 1us the current would be 100 amps. But if we increase the time dt to 2us, we'd end up with:
iL=100*(2e-6)/(1e-6)
iL=200 amps.

So you can see that the current got higher because we allowed the voltage to stay across the inductor for a longer time period.

If we increase the inductor value however, we get the opposite effect. If we use a 2uH inductor instead, we see:
iL=100*(2e-6)/(2e-6)
so:
iL=100 amps again.

So increasing the time period increases the peak current, and increasing the inductor value decreases the peak current.

Now the question is what to use for these values in a real circuit. In the real circuit, we have an operating frequency F instead of a time period dt. But knowing how a buck converter works, we know that the 'on' time of the switch varies from 0 to constantly on, but in most practical converters the 'on' time is limited to just under the time period of the time of one cycle of the operating frequency, or:
T=1/F

For duty cycles less than this, we'll see a different time period like T=1/(2*F) for example, which means the current does not peak as high as above. But for a converter with a large input and low output like 60 volts input and 5 volts output, we'll see a much smaller duty cycle. That would imply that we could use a much smaller inductor because if we look at shorter time periods the current does not get up that high.

So for this circuit if you were to design for a max time period of 1/F there is almost no way you can go wrong. If we replace the dt with 1/F we get:
iL=vL*(1/F)/L
or:
iL=vL/(F*L)

If we solve for L we get:
L=vL/(iL*F)

Now if vL is 50 volts and F is 200kHz and iL is 1 amp, that gives us:
L=50/(1*200000)=250uH

which might be higher than expected. If the operating frequency was increased to 500kHz, the inductor would come out to 100uH.

Does this make sense to you now?

Also note that if we can bank on a smaller duty cycle then we can make the inductor smaller because with 1/5 duty cycle the inductor value would come down to 20uH. We'd have to investigate short circuit operation to be more sure of this however.
The theoretical ideal buck converter duty cycle is D=Vout/Vin., but that is for normal operation where the converter is up and running and stabilized with a fixed constant load.

 

[Futterama]

alec_t, yes my VIN(MAX) is 42V, but in the design example on page 17 they clearly state VIN(MAX) = 80V and yet they use VIN = 24V as the VIN(MAX) value in the calculation for L(MIN). That is what confuses me. There must be some typo in there somewhere. It is also strange to say that for an application VIN = 24V and VIN(MAX) = 80V. That doesn't really make any sense to me. Input voltage is 24V but max input voltage is 80V? And they don't account for the 80V input voltage anywhere in their calculations? Strange.

Anyway, for my application, the maximum input voltage will be 42V so I will use that to calculate the inductor.

MrAl, thanks, is does make sense to me. But the calculations in the datasheet uses the output voltage to calculate the inductor value, not the voltage difference between input and output. And they multiply with some value (1-Vout/Vin).

That was the inductor. Now the capacitors. They use the same equation for Cin and Cout in the design example, but they give different equations in the section "Cin and Cout selection" on page 13. So basically I have 3 different equations for calculation of Cout but I have no clue which to use since they all give different results. I would have used the two equations on page 13 and chosen the larger value of the 2, since one sizes Cout from output ripple and the other from inductor energy and the capacitor should be able to handle both. But then the design example uses a third equation for calculating Cout.

Confused? I am.

 

[alec_t]

in the design example on page 17 they clearly state VIN(MAX) = 80V and yet they use VIN = 24V as the VIN(MAX) value in the calculation for L(MIN). That is what confuses me.

I think they've confused themselves, too. Although the 80V figure is mentioned it isn't actually used in that design example. Of course, if you were designing a circuit for general use with an unknown input voltage you would design for the worst case scenario and use 80V in that situation.

I would have used the two equations on page 13 and chosen the larger value of the 2, since one sizes Cout from output ripple and the other from inductor energy and the capacitor should be able to handle both.

I would have done the same.

 

[MrAl]

alec_t, yes my VIN(MAX) is 42V, but in the design example on page 17 they clearly state VIN(MAX) = 80V and yet they use VIN = 24V as the VIN(MAX) value in the calculation for L(MIN). That is what confuses me. There must be some typo in there somewhere. It is also strange to say that for an application VIN = 24V and VIN(MAX) = 80V. That doesn't really make any sense to me. Input voltage is 24V but max input voltage is 80V? And they don't account for the 80V input voltage anywhere in their calculations? Strange.

Anyway, for my application, the maximum input voltage will be 42V so I will use that to calculate the inductor.

MrAl, thanks, is does make sense to me. But the calculations in the datasheet uses the output voltage to calculate the inductor value, not the voltage difference between input and output. And they multiply with some value (1-Vout/Vin).

That was the inductor. Now the capacitors. They use the same equation for Cin and Cout in the design example, but they give different equations in the section "Cin and Cout selection" on page 13. So basically I have 3 different equations for calculation of Cout but I have no clue which to use since they all give different results. I would have used the two equations on page 13 and chosen the larger value of the 2, since one sizes Cout from output ripple and the other from inductor energy and the capacitor should be able to handle both. But then the design example uses a third equation for calculating Cout.

Confused? I am.

Hi again,

Dont worry too much about the slightly different equations. They included Vin in one of them and that's what counts. If you want to state the entire equation here that you are looking at, we can compare and see what is different or the same about them.

The simplest way to calculate the output capacitor is to use the simple equation:
dv=i*dt/C

In words this would read:
"The change in voltage 'dv' is equal to the current 'i' times the time period 'dt' divided by the capacitance".

For the ripple output we just have to consider the change in 'i' over one cycle, so if the peak current is Ipk and the nominal output current is Inom then the change is:
Ichng=(Ipk-Inom)*2

The equation then looks like this:
dv=Ichng*dt/C

and if we again take dt to be the 1/frequency we have:
dv=Ichng/(F*C)

and solving for C we get:
C=Ichng/(F*dv)

So for example if we have a peak current of 1 amp and nominal output current of 0.5 amps, we first have:
Ichng=(1-0.5)*2=1 amp

then we have:
C=1/(F*dv)

and for an operating frequency of 100kHz we have:
C=1/(100000*dv)

Now it's just a matter of deciding how much ripple we want to allow. Let's say we will allow 0.1 volts ripple, that gives us:
C=1/(100000*0.1)=1/10000=0.0001 Farads which is the same as 100uF.

So the output cap in this case would be 100uF. To get the ripple down 10 times lower than that (like around 10mv) we would use a 1000uF cap (ten times higher value).

These concepts are very simple for the buck circuit so once you do a couple of these circuits you'll be able to do a million of them in almost no time at all

Before i forget again, i should mention that any circuit like this should be fully tested before buying thousands of parts for say a production run, and also before even putting one such circuit into actual use. A full test would entail running the input up and down from min expected to max expected, and testing with small load, no load, and full loads, as well as what is sometimes called a "life" test.

The most important extremes are low input and full load which makes sure the correct output can actually still be obtained with little input, and high input and small load which makes sure the output can still regulate with max input and nothing much consuming any energy.

The life test is also important. Sometimes called an "infant mortality" test, it tests the total design to make sure it can run continuously without melting down or going unstable in some way. This test means running the unit at the very least over night for say 12 hours or more at full load.

 

[Futterama]

MrAl, thanks for the explanation. I have actually been contacted by an engineer from Linear. He is very helpful and has provided an Excel sheet for calculations and a spice schematic for their LTspice application. He also made some good points about EMI shielding.

I will remember to do the tests you mentioned. Unfortunately my lab supply can only deliver 30V max but I can use the 42V battery and a LM317 to run the voltage up and down.

I guess it would be a good idea to monitor the input and output voltage during the infant mortality test, luckily I have recently purchased a 2 channel PC scope with longterm logging capability (just a few samples per second though) which would be useful for this purpose.

 

[Futterama]

I have attached the Excel sheet and the LTspice schematic that the engineer from Linear sent me for others to use.