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SMPS Inductor saturates?

B

Bill Bowden

Jan 1, 1970
0
I'm building a small Boost regulator using an inductor, transistor
switch, diode and filter cap, but I cannot get enough current into the
inductor before the transistor starts dropping a large portion of the
supply voltage. I don't think the problem is related to the transistor
gain or base drive since the circuit will draw a couple amps with just
a DC input on the transistor base resistor, and all I need from the
inductor is about 500mA peak.

I have the thing connected to a square wave generator and scope and it
works well at about 73% efficiency on small loads. But as I reduce the
frequency to increase the ramp time and peak current, and power output,
the voltage across the transistor (while it is on) starts to increase
dramatically around 400mA of current, and I need more than that.

It looked like the inductor might be saturating, so I reduced the
number of turns and ran it at a higher frequency, but that didn't solve
the problem. It still runs out of gas around 400mA.

I'm going to try a larger core to see if that helps, but I thought
using fewer turns at a higher frequency would allow more current?

-Bill
 
H

Homer J Simpson

Jan 1, 1970
0
It looked like the inductor might be saturating, so I reduced the
number of turns and ran it at a higher frequency, but that didn't solve
the problem. It still runs out of gas around 400mA.

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

Jonathan Kirwan

Jan 1, 1970
0
I'm building a small Boost regulator using an inductor, transistor
switch, diode and filter cap, but I cannot get enough current into the
inductor before the transistor starts dropping a large portion of the
supply voltage. I don't think the problem is related to the transistor
gain or base drive since the circuit will draw a couple amps with just
a DC input on the transistor base resistor, and all I need from the
inductor is about 500mA peak.

I have the thing connected to a square wave generator and scope and it
works well at about 73% efficiency on small loads. But as I reduce the
frequency to increase the ramp time and peak current, and power output,
the voltage across the transistor (while it is on) starts to increase
dramatically around 400mA of current, and I need more than that.

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.
It looked like the inductor might be saturating, so I reduced the
number of turns and ran it at a higher frequency, but that didn't solve
the problem. It still runs out of gas around 400mA.

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.
I'm going to try a larger core to see if that helps, but I thought
using fewer turns at a higher frequency would allow more current?

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
 
J

John Popelish

Jan 1, 1970
0
Bill said:
I'm building a small Boost regulator using an inductor, transistor
switch, diode and filter cap, but I cannot get enough current into the
inductor before the transistor starts dropping a large portion of the
supply voltage. I don't think the problem is related to the transistor
gain or base drive since the circuit will draw a couple amps with just
a DC input on the transistor base resistor, and all I need from the
inductor is about 500mA peak.

I have the thing connected to a square wave generator and scope and it
works well at about 73% efficiency on small loads. But as I reduce the
frequency to increase the ramp time and peak current, and power output,
the voltage across the transistor (while it is on) starts to increase
dramatically around 400mA of current, and I need more than that.
(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?
 
J

Jonathan Kirwan

Jan 1, 1970
0
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

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
 
B

Bill Bowden

Jan 1, 1970
0
Jonathan said:
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?

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.
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.

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.
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


-Bill
 
Bill said:
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.
(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).
 
...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.
 
J

Jonathan Kirwan

Jan 1, 1970
0
(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).

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
 
J

jasen

Jan 1, 1970
0
I'm building a small Boost regulator using an inductor, transistor
switch, diode and filter cap, but I cannot get enough current into the
inductor before the transistor starts dropping a large portion of the
supply voltage.

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

I don't think the problem is related to the transistor
gain or base drive since the circuit will draw a couple amps with just
a DC input on the transistor base resistor, and all I need from the
inductor is about 500mA peak.

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,
I have the thing connected to a square wave generator and scope and it
works well at about 73% efficiency on small loads. But as I reduce the
frequency to increase the ramp time and peak current, and power output,
the voltage across the transistor (while it is on) starts to increase
dramatically around 400mA of current, and I need more than that.

It looked like the inductor might be saturating, so I reduced the
number of turns and ran it at a higher frequency, but that didn't solve
the problem. It still runs out of gas around 400mA.

are you measuring current with the scope, the current will surge if the
core saturates.
I'm going to try a larger core to see if that helps, but I thought
using fewer turns at a higher frequency would allow more current?

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

Bye.
Jasen
 
J

Jamie

Jan 1, 1970
0
Bill said:
I'm building a small Boost regulator using an inductor, transistor
switch, diode and filter cap, but I cannot get enough current into the
inductor before the transistor starts dropping a large portion of the
supply voltage. I don't think the problem is related to the transistor
gain or base drive since the circuit will draw a couple amps with just
a DC input on the transistor base resistor, and all I need from the
inductor is about 500mA peak.

I have the thing connected to a square wave generator and scope and it
works well at about 73% efficiency on small loads. But as I reduce the
frequency to increase the ramp time and peak current, and power output,
the voltage across the transistor (while it is on) starts to increase
dramatically around 400mA of current, and I need more than that.

It looked like the inductor might be saturating, so I reduced the
number of turns and ran it at a higher frequency, but that didn't solve
the problem. It still runs out of gas around 400mA.

I'm going to try a larger core to see if that helps, but I thought
using fewer turns at a higher frequency would allow more current?

-Bill
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.
 
B

Bill Bowden

Jan 1, 1970
0
jasen said:
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,

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

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
 
B

Bill Bowden

Jan 1, 1970
0
Homer said:
Do you have enough inductance to store the energy you need?

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
 
B

Bill Bowden

Jan 1, 1970
0
Homer said:
Do you have enough inductance to store the energy you need?

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
 
J

jasen

Jan 1, 1970
0
(hmm... how come I only see this today?)
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.

???

above you imply that it plateaus at 500mA, maybe I'm misreading that.

By surge a sudden spike at the end of the ramp.
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).

Are you using the same number of ampere-turns and the same frequency as the
200W supply?

Bye.
Jasen
 
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