SMPS Inductor saturates?

Do you have enough inductance to store the energy you need?

Reading your words carefully, I see that you say "the circuit will draw a couple amps" but you don't really say anything about he base drive current on your transistor switch. It still seems possible that you are running out of beta on it. At 400mA collector current and operating as a switch, you need a substantial base drive. What is the measured value, here?

Also, ignoring inductor DC resistance for now, the voltage across the inductor is V = dI/dt * L. If your dI/dt goes flat-line (doesn't rise anymore) then your V across the L is simply going to drop and the transistor collector _must_ pick up the difference and rise. So this isn't news, at all.

As I understand it, when an inductor saturates you should see current peaking assuming the transistor can supply the current. The apparent L diminishes and for a fixed voltage across it the dI/dt suddenly grows (as L drops, V/L climbs.) You seem to imagine that your switch can deliver "amps" for collector current, so if your belief is correct and if the inductor was saturating, I'd imagine you'd see a current spike until the DC resistance dominates. But you don't mention this, so I'm back to betting on your base drive.

Of course, I'm a hobbyist and not particularly knowldgeable on switchers, so perhaps I'm an easy mark for bets, too.

A smaller L would mean a faster rate of change of current for a fixed voltage across the L. A faster rate of change _could_ mean more current, if you held it on for a fixed time. But you also here say "a higher frequency" so that means a "shorter time." So depending on the exact quantities here, it may be more current, or less, or the same. Assuming everything else can deliver the current and voltage compliances, of course.

Jon

(snip) Are the currents you mention referring to the instantaneous inductor current, or the output load current?

Is the regulator operating in continuous or discontinuous mode?

What type transistor and what base current?

Another thought crossed my mind. Do you allow enough OFF time for the inductor's magnetic field to completely collapse before another ON cycle? The ON and OFF times aren't necessarily the same as the voltage across the inductor during energy release isn't usually the same as the voltage across it when you are pumping up the field. The difference between your output voltage and the supply rail, less a small drop for the diode, would be the OFF voltage, roughly, I think.

Jon

Well, I've tried different base currents of 50mA and 100mA with no change in output conditions, so I don't think that's the problem. All I need is a couple watts at 50 volts with a 4.5 to 3 volt input.

I overkilled the thing and rewound the inductor on a large core (1.25 inch diameter by 1/2 inch high) About 40 turns of heavy wire (resistance less than 200 milliohms). I can see several cycles of ringing when the inductor discharges into the load, so it appears to be very efficient. But I still can't get more than 500mA of peak inductor current.

The scope waveform on the transistor collector looks like this: Time 0, the transistor switches on and collector drops to near zero for 250 microseconds. At time 1, the transistor switches off and collector jumps to 20 volts for about 75 microseconds. At time 3, collector drops to the supply voltage (4.5 vols) and then rings about 8 cycles at

200KHz, and then flatlines at the supply voltage for 150 microseconds. The waveform then repeats and the collector drops to zero.

So, I get about 20 volts across a 1150 ohm load resistor in parallel with the capacitor. Average input current is 100mA at 4.5 volts or 450 milliwatts. Output power at 20 volts and 1150 ohms is 350 milliwatts so the efficiency is about 77%.

It works well under these conditions, but I cannot lower the drive frequency to increase the inductor current and obtain more power out. As I lower the frequency, the waveform near the point where the transistor switches off (close to time 1) starts to climb exponentionally indicating the drop on the transistor is increasing and generating lots of heat.

-Bill

(snip)

Let me think about this a moment. You are stepping 3 volts up to 50 volts with a single inductor boost converter and a 2/50=40 mA output current.

If you used the absolute maximum duty cycle (for miniumu peak current) to maintain discontinuous mode (inductor fully discharges each cycle), Assuming the inductor is lossless, the average voltage across it must be zero. So, neglecting diode drop and switching losses, if x is the fraction of each cycle 3 volts is applied, the formuls afor x is:

3 * x = 50 * (1-x)

so x = 50/53 = .943. That means the output current has to be produced by a triangular current pulse that lasts onlt 1-x = .0566 of each cycle. So the 40 mA output current must be supplied by .04 * 1/(1-x) =

707mA average current during that pulse, each cycle. But the pulse is triangular, dso it has to have a peak current of twice that value, or 1.413 A. Does this make sense, so far?

This means that the switch, diode and inductor have to function well up that current, just to get .04 A out of the converter (if I haven't made a math or logic error).

By the way, this is the best case for discontinuous mode, assuming that the inductor is always either charging or discharging. If some of each cycle is wasted with near zero current inductor ringing, the peak current requirement goes up.

If you can use continuous current mode, whenre the charrging switch comes on after only a fraction of the stored energy is dumped, the turn on switching losses are higher, but the peak current requirement goes down, because the output pulse is trapezoidal, instead of triangular, so its peak is less than twice its average during the pulse. This mode is often simplest if the switch is operated with an inductor current sensor that turns the switch on and off at two different current levels. The levels are set on the fly, based on output voltage. This concept inherently soft starts.

I had been writing up my own response last night but didn't send it, yet. I have had other things to do between now and then, but this is about where I'd gotten last night -- though my approach was different and I'd included more detailed reasoning.

But doing 2 watts at 50V from 3V takes a little thought, little of which has been in exposed by the OP so far.

I'll finish up my thoughts a little later and post them.

Jon

where are you trying to get to, and how are you trying to get there.

I don't see how that follows, (I see no mention of collector voltage when it's drawing "a couple of amps") But the only other option is that your inductor is short-circuited or has too little inductance,

are you measuring current with the scope, the current will surge if the core saturates.

if it's a core saturation problem it should. provided the rest of the circuit can handle it.

Bye. Jasen

the last boost unit i made i incorporated a current sense for the charge and voltage sense (comparator) for the drop off. this way selecting a frequency rate was much simpler to do.. this circuit simply toggles. i only needed a pulse on power initiate to start it and if for some reason the unit got over loaded it would just stop boosting kind of a safeguard.

i would say your problem after reading your other many post is that you are not saturating the coil!< also the slew rate when the transistor releases the energy is important. if this is slow you can lose energy in the process. you may want to test this with out the inductor in the circuit.

that's just a suggestion.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

I don't see the significance of collector voltage. The point is, I can stop the generator on a high level so it turns on the transistor and the current is a couple amps. The base current is about 100mA, so the gain is 20 or more. The question is why doesn't the ramp current increase beyond 500mA before the transistor starts dropping lots of voltage? I only need a gain of 5, and the gain is over 20 at DC.

Yes, the current surges, and that's the problem. The question is why? I know the gain is over 20 and I am driving the base with 100mA and I can't get more than 500mA peak before the current surges. I changed the inductor to a large core (the type used in 200 Watt supplies) and I still can't get 1% of that (2 watts).

-Bill

I think so. As I understand it, stored energy in an inductor is 0.5* L*I^2. This would indicate that most any inductor can be used if the current is adjusted for the required energy. But I like to use large inductors to reduce the current, and operating frequency, and losses in the transistor switching times.

-Bill

I think so. As I understand it, stored energy in an inductor is 0.5* L*I^2. This would indicate that most any inductor can be used if the current is adjusted for the required energy. But I like to use large inductors to reduce the current, and operating frequency, and losses in the transistor switching times.

-Bill