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.