Hello,
If the requirements are 5v input (constant, no sag at any current) and the output 12v at 300ma, the design looks very marginal no matter what you do.
We have to remember that the peak current is related to the output power and input and output voltages. The ideal duty cycle is related to the input and output voltage, and the peak current is related to the output power and the duty cycle, along with the efficiency. The efficiency plays a role too as it may only be 85 percent or even less.
The transistor drop could be as high as 1v maybe even higher. The drop across Rsc could be roughly 0.4 volts. We end up down to maybe 3.5v effective input voltage if that. If the duty cycle was 0.58 in the ideal case, it will have to be even much less which drives the peak current up even higher. It seems like the limit of 1.5 amps just may not be enough.
We haven't even looked at the minimum duty cycle given the limitations of a boost circuit with resistive losses, which may be greater than that required to get the right output, if that was even possible.
The 22uH did not sound high enough for this device right off the bat. If anything, maybe 250uH or greater.
We should go over this design more carefully. With all these limitations I'd be surprised if this ever works with the given input/output requirements. I guess there is a chance that things are not as bad as they read from the data sheet, but one build may work while the next one fails.
A good, simple simulation may help decide also.
So ok, 22uH is definitely out. Probably 100uH min, but the startup surge is going to be too high and may cause a problem. 200uH would be much better. 20uf output cap, 1000uf low ESR input cap. This might just work.
A drawback with this chip is not only the use of internal bipolar transistors, but also no slow start that came in with more modern chips. No slow start means higher input startup surge current and higher output surge voltage at startup.
But with a 200uH inductor, I'll get only 560mA max current, right? (considering 35kHz frequency)
Hi,
What do you mean a max of 560ma?
Did you change the output spec from 12v at 300ma, to 12v at something even greater than 500ma?
Hello!
I meant that if I plug in the value for the 200uH inductor of into the Iout max formula (attached), I get 560mA.
Hello again,
Ok, that sounds like it may be the maximum current deviation not the absolute maximum output current. If not, then the formula can not be correct, or they mean something else by it. BTW, where did you get that formula from?
What changes when you increase the inductor is the ripple current, not the average current, unless the controller maxes out at it's max current handling ability.
For example, with 12v out and 0.325 amps out, the average current with a 200uH inductor is about 1 amp, and the deviations are about 185ma above that and 185ma below that, approximately.
With a 100uH inductor, the average current is still 1 amp, but the deviations above and below that are roughly twice that as with the 200uH inductor. If we go too much lower than that though, the average current plus the positive deviation will exceed 1.5 amps, so we have to watch out for that.
Another consideration with all of these switcher circuits is the ripple current. We want to keep it low enough so the output filter capacitor does not get banged too hard with the changing current. We have to choose a cap that can take it, and the lower the ripple is the better as the cap would last longer. Choosing the inductor controls the ripple current, so we have some control over that.
The other consideration with the inductor (and the capacitor) is the initial surge current when the power is first applied. With no slow start, the current will exceed 1.5 amps with a 100uH inductor and 20uf cap, but would probably be ok with 200uH. You can test for that.
The calculation of the average current goes with the power out and the power in. If you had say 4 watts out and 4 watts in (for a theoretically ideal circuit) then if the output voltage is Vout and higher than Vin, then the average current input will be higher than the average output current:
Vout*Iout=Vin*Iin
so:
Iin=(Iout*Vout)/Vin
With Iout=0.325 and Vout=12 and Vin=3.9 volts, we get:
Iin=1.0 amps.
That is the average current which means the average DC current into the input.
Note I used 3.9 volts for the input voltage, not 5.0 volts. That is to compensate for some of the losses in the circuit that makes the input look lower for a simpler calculation.
I also assumed 35kHz.
Thank you for taking the time to explain how the inductor is involved in all this and the formulas ! I got mine from ChatGPT and it wouldn't surprise me if it was incorrect although it seemed like it understood perfectly what I was asking 
So if I wanted to power a 12V stepper motor that draws 1.2A from 5Vin, it would draw 2.8A from the input.
How did the switching frequency fit into your calculation above?
Hello again,
[1] On ChatGPT...
OMG, ChatGPT is totally bogus on technical issues for the most part. Maybe for some simpler stuff it might work, but after a lengthy 'conversation' with it some time ago I determined that it is not to be used for technical issues especially advanced ones. I had asked it about a special numerical calculation for the sine of an angle, and it would give me an answer. After I told it that was wrong, it would apologize, then give another wrong answer, then after I told it that was wrong too, it gave another wrong answer, and this went on for a while. One of the answers it gave me even mentioned a mathematician that gave that result, and when I checked it, it was totally bogus too. I went back the next day and gave it the same answer and it said that it didn't work, and that there was no such algorithm.
After that I realized you can't use if for stuff like this.
On that same note, see what it gave for the duty cycle D of the boost converter:
D=Vout/(Vout+Vin)
which is totally bogus. The correct expression is:
D=(Vout-Vin)/Vout=1-Vin/Vout
To prove the first one is wrong, just enter Vin=5 and Vout=5.01 and see that it comes up with a duty cycle close to 1/2, which is by far and wide the worst estimate in the history of the universe, or at least in the history of boost converters. The real duty cycle would be close to zero, which means the switch is always off, and thus the input goes right to the output (not including the diode drop which is considered zero in theory with ideal elements).
To modify the duty cycle to include efficiency, which can be low, we can multiply Vin times the efficiency Eff:
D=1-Vin*Eff/Vout
With Vin=5 and Eff=0.78 and Vout=12, D comes out to 0.675 which is very reasonable.
Note Eff=0.78 is the fraction for 78 percent efficiency.
[2] On the stepper motor current drive...
If you have 5v input and 12v output at 1.2 amps, then the input current in the ideal case is 2.88 amps.
The expression was:
Vout*Iout=Vin*Iin
and solving for Iin:
Iin=Vout*Iout/Vin
and we can include the efficiency estimate in that formula as:
Iin=Vout*Iout/(Vin*Eff)
and again with Vin=5 and Vout=12 and Iout=1.2 amps, Iin comes out to 3.69 amps.
I used 78 percent (0.78) because that's a reasonable efficiency for a bipolar with all those other resistive losses, but it could be 70 to 85 percent. I doubt we would see 85 percent with this chip though, maybe not even 80 percent.
[3] On the switching frequency...
The switching frequency fits in with the duty cycle to estimate the excursion above and below the average DC current in the inductor for one. If the duty cycle is 0.675 and the frequency is 100kHz, then the period is 10us and the switch 'on' time is 6.75us. Using the familiar:
di=v*dt/L
and with v=5 and dt=6.75us and L=200uH we get:
di=169ma
and half of that is:
di/2=84.5ma
so the excursions above and below the average DC current would be about 85ma.
If the DC current was 1 amp, then the upper current peak would be 1.085 amps.
What we assume in the above is the current waveshape is a very close approximation to a true sawtooth wave, even though it curves very, very slightly, but this is usually a good approximation.
[4] The input current surge...
Lastly, the input surge is a little more complicated. With feedback it's even more complicated. I think the best bet here is to do a simulation, but we could look into the math behind it too as there are calculations that could estimate this.
[5] Summary
Just to recap a little here...
1. It's not a good idea to use ChatGPT for technical issues, especially advanced issues, even mildly advanced sometimes.
2. You understood the input current calculation very well.
3. The frequency and duty cycle work together to allow the calculation of the absolute peak current [see also #6 below].
4. The input surge is another matter which requires a little more thought when there is no slow start mechanism.
[6] Unusual Chip Activity, another note on the frequency and duty cycle...
This chip happens to be an unusual switching converter because there are times when it might skip an on or off cycle. This means the ripple on the output goes higher. The other thing is it is somewhat random when this happens so the output could look very different than a more typical switching converter. The peaks and valleys can go higher (and lower) sometimes than other times. This is one of the things I did not like about this chip. There are of course advantages too though for using this chip.
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