transistor for small boost converter

[Thomas Anderson]

Hi!

I want to charge a capacitor to 300 V from something like 5-15 V.
Because I can't find 400 V MOSFETs for 10-50 mA which would have a
suitably small gate charge I tought a BJT would be the solution for
simple and fast turn off:

MPSA44
NPN Transistor
400V
300 mA
http://www.datasheetcatalog.com/datasheets_pdf/M/P/S/A/MPSA44.shtml

The specs talk about frequencies up to 20 MHz (VCE=10V).

Philips even says 100 MHz:

fT transition frequency
(IC = 10 mA; VCE = 10 V; f = 100 MHz)
20 MHz (max)

Since there is a turn on and a turn off in one period the turn off
time should be something like 1/40e6 = 25 ns or less since there may
be some duty cycle between switches. Well, things may be different
at VCE=300V and Motorola only mentions 10 MHz.

And now I am confused by the turn on/off timing diagrams in the
Motorola specs (page 4) (VCC=150V; IC/IB=10 (saturated?)):

There is ts (storage? time = BE junction discharge time?) and tf
(fall time = IC fall time?) given for IC = 1-50 mA. Do I have to add
these times? Why is the turn off time higher for smaller IC? Ohh, is
it because turn off is when IC drops below some % of initial IC and
not below some fixed value? So when I have IC = 30 mA, is my turn
off time then 3.3 us (ts) + 0.5 us (tf) = 3800 ns? That's a lot more
than 25 ns.

I wanted to use a 1 mH inductor which would loose 10 mA in 33 ns
when charging a cap at 300V:

I=U*t/L
t=I*L/U
t = 30 mA * 1 mH / 300 V = 100 ns

So I need a bigger inductor anyways (also because I have a 75 ns
diode), but if this transistor really has 3800 ns turn off time I
need a different approach. Any recommendations for a transistor? I'd
prefer the inductor current to remain below 50 mA.

Thanks!

-=-
This message was sent via two or more anonymous remailing services.

 

[Fritz Schlunder]

Hi!

I want to charge a capacitor to 300 V from something like 5-15 V.
Because I can't find 400 V MOSFETs for 10-50 mA which would have a
suitably small gate charge I tought a BJT would be the solution for
simple and fast turn off:

MPSA44
NPN Transistor
400V
300 mA
http://www.datasheetcatalog.com/datasheets_pdf/M/P/S/A/MPSA44.shtml

The specs talk about frequencies up to 20 MHz (VCE=10V).

Philips even says 100 MHz:

fT transition frequency
(IC = 10 mA; VCE = 10 V; f = 100 MHz)
20 MHz (max)

Since there is a turn on and a turn off in one period the turn off
time should be something like 1/40e6 = 25 ns or less since there may
be some duty cycle between switches. Well, things may be different
at VCE=300V and Motorola only mentions 10 MHz.

As I understand it the published transition frequency is allot like an
op-amp's gain bandwidth product. If you used the transistor in it's linear
mode as an AC amplifier, you should be able to amplify a sinusoid with a
current gain of one at the transistion frequency. That is, if fT is 20MHz,
then you should be able to put say 1mA into the base and get 1mA of
collector current. For the most part a transistor is totally useless at or
above the transistion frequency, however, even at the transistion frequency
it is still possible to get voltage gain, without current gain. Anyway,
these seemingly impressive high numbers are not especially useful for
determining how fast the transistor will operate as a switch, as in your
case (although generally speaking a higher fT would imply higher switching
speeds as well).

And now I am confused by the turn on/off timing diagrams in the
Motorola specs (page 4) (VCC=150V; IC/IB=10 (saturated?)):

There is ts (storage? time = BE junction discharge time?) and tf
(fall time = IC fall time?) given for IC = 1-50 mA. Do I have to add
these times? Why is the turn off time higher for smaller IC? Ohh, is
it because turn off is when IC drops below some % of initial IC and
not below some fixed value? So when I have IC = 30 mA, is my turn
off time then 3.3 us (ts) + 0.5 us (tf) = 3800 ns? That's a lot more
than 25 ns.

During the ts (storage time) period, you have stopped applying current to
the base, but the transistor continues to conduct basically full current and
the voltage between collector and emitter remains small. Once the storage
time runs out, only then does the collector current begin to fall and the
collector emitter voltage increase. Essentially the transistor actually
switches off during the fall time, but the storage time simply adds a delay
to the turn off event. The storage time results in pulse width distortion,
but it doesn't directly increase power dissipation to any sigificant degree.

The storage time is not a very fixed number, and depending upon how you
operate the transistor it can vary by quite a bit. In the case of operating
the transistor in the linear mode of operation, the storage time is
essentially zero. On the other hand, when operated as a fully saturated
switch, the storage time can become quite large especially for very low
forced betas (that is, say you force 25mA into the base with only a 50mA
collector current). To an extent one can improve storage time and switching
performance by applying negative base drive (with decent current capability)
to the transistor when you want it to turn off. A typical method of
implementing this might be to place a small capacitor (say a few tens to a
few hundred picofarad) in parallel with the base drive resistor, and use a
CMOS base driver (so it can both sink and source current).

Little can be done to improve the fall time.

I wanted to use a 1 mH inductor which would loose 10 mA in 33 ns
when charging a cap at 300V:

I=U*t/L
t=I*L/U
t = 30 mA * 1 mH / 300 V = 100 ns

So I need a bigger inductor anyways (also because I have a 75 ns
diode), but if this transistor really has 3800 ns turn off time I
need a different approach. Any recommendations for a transistor? I'd
prefer the inductor current to remain below 50 mA.

It would appear the MPSA44 is rather too slow to perform properly to meet
your requirements. The VN4012L from Vishay probably has adequately small
gate charge for you (like 1.5 nanocoulombs), but unfortunately it would
appear to be an "obsolete" part with no replacements.

http://www.vishay.com/docs/70207/70207.pdf

I guess tiny high voltage MOSFETs aren't in very big demand.

 

[Paul Mathews]

Look at Supertex.
Paul Mathews

As I understand it the published transition frequency is allot like an
op-amp's gain bandwidth product. If you used the transistor in it's linear
mode as an AC amplifier, you should be able to amplify a sinusoid with a
current gain of one at the transistion frequency. That is, if fT is 20MHz,
then you should be able to put say 1mA into the base and get 1mA of
collector current. For the most part a transistor is totally useless at or
above the transistion frequency, however, even at the transistion frequency
it is still possible to get voltage gain, without current gain. Anyway,
these seemingly impressive high numbers are not especially useful for
determining how fast the transistor will operate as a switch, as in your
case (although generally speaking a higher fT would imply higher switching
speeds as well).





During the ts (storage time) period, you have stopped applying current to
the base, but the transistor continues to conduct basically full current and
the voltage between collector and emitter remains small. Once the storage
time runs out, only then does the collector current begin to fall and the
collector emitter voltage increase. Essentially the transistor actually
switches off during the fall time, but the storage time simply adds a delay
to the turn off event. The storage time results in pulse width distortion,
but it doesn't directly increase power dissipation to any sigificant degree.

The storage time is not a very fixed number, and depending upon how you
operate the transistor it can vary by quite a bit. In the case of operating
the transistor in the linear mode of operation, the storage time is
essentially zero. On the other hand, when operated as a fully saturated
switch, the storage time can become quite large especially for very low
forced betas (that is, say you force 25mA into the base with only a 50mA
collector current). To an extent one can improve storage time and switching
performance by applying negative base drive (with decent current capability)
to the transistor when you want it to turn off. A typical method of
implementing this might be to place a small capacitor (say a few tens to a
few hundred picofarad) in parallel with the base drive resistor, and use a
CMOS base driver (so it can both sink and source current).

Little can be done to improve the fall time.




It would appear the MPSA44 is rather too slow to perform properly to meet
your requirements. The VN4012L from Vishay probably has adequately small
gate charge for you (like 1.5 nanocoulombs), but unfortunately it would
appear to be an "obsolete" part with no replacements.

http://www.vishay.com/docs/70207/70207.pdf

I guess tiny high voltage MOSFETs aren't in very big demand.

 

[[email protected]]

I want to charge a capacitor to 300 V from something like 5-15 V.
Because I can't find 400 V MOSFETs for 10-50 mA which would have a
suitably small gate charge I thought a BJT would be the solution for
simple and fast turn off:

MPSA44
NPN Transistor
400V
300 mA
http://www.datasheetcatalog.com/datasheets_pdf/M/P/S/A/MPSA44.shtml

[snipped: discussion of MPSA44 switching speed]

I wanted to use a 1 mH inductor which would loose 10 mA in 33 ns
when charging a cap at 300V:

I=U*t/L
t=I*L/U
t = 30 mA * 1 mH / 300 V = 100 ns

So I need a bigger inductor anyways (also because I have a 75 ns
diode), but if this transistor really has 3800 ns turn off time I
need a different approach. Any recommendations for a transistor? I'd
prefer the inductor current to remain below 50 mA.

The conventional approach is to use a transformer and a
low-saturation switching transistor.

For inspiration, have a look at camera flash circuits and the like in
Sam Goldwasser's "Various Schematics and Diagrams":

http://www.eio.com/repairfaq/sam/samschem.htm

Search for "Kodak MAX", "tiny inverter", and enjoy.

James Arthur

 

[Joerg]

Hello Fritz,

It would appear the MPSA44 is rather too slow ...

Nice British style ;-)

It reminds me of a guy in England standing behind his Volkswagen bus
looking at the engine compartment that had burned itself into a black
hole: "It would appear that the engine must have had a slight problem."

I guess tiny high voltage MOSFETs aren't in very big demand.

Nah, there's lots of them. The only problem is for situations where you
need true logic level drive which hardly any of them do. Those that do
are often from vendors where you can't get them unless you are a large
automotive company or your family name is Rockefeller.

Example: IRFR310, made by IRF and Fairchild. Can be had for about $0.25
in qty. But it really needs 10V gate drive and you'd have to slosh
around a few hundred pF of Cgs in time. If you do then it can typically
switch within the desired 25nsec.

Regards, Joerg

 

[Terry Given]

Hi!

I want to charge a capacitor to 300 V from something like 5-15 V.
Because I can't find 400 V MOSFETs for 10-50 mA which would have a
suitably small gate charge I tought a BJT would be the solution for
simple and fast turn off:

MPSA44
NPN Transistor
400V
300 mA
http://www.datasheetcatalog.com/datasheets_pdf/M/P/S/A/MPSA44.shtml

The specs talk about frequencies up to 20 MHz (VCE=10V).

Philips even says 100 MHz:

fT transition frequency
(IC = 10 mA; VCE = 10 V; f = 100 MHz)
20 MHz (max)

Since there is a turn on and a turn off in one period the turn off
time should be something like 1/40e6 = 25 ns or less since there may
be some duty cycle between switches. Well, things may be different
at VCE=300V and Motorola only mentions 10 MHz.

And now I am confused by the turn on/off timing diagrams in the
Motorola specs (page 4) (VCC=150V; IC/IB=10 (saturated?)):

There is ts (storage? time = BE junction discharge time?) and tf
(fall time = IC fall time?) given for IC = 1-50 mA. Do I have to add
these times? Why is the turn off time higher for smaller IC? Ohh, is
it because turn off is when IC drops below some % of initial IC and
not below some fixed value? So when I have IC = 30 mA, is my turn
off time then 3.3 us (ts) + 0.5 us (tf) = 3800 ns? That's a lot more
than 25 ns.

I wanted to use a 1 mH inductor which would loose 10 mA in 33 ns
when charging a cap at 300V:

I=U*t/L
t=I*L/U
t = 30 mA * 1 mH / 300 V = 100 ns

So I need a bigger inductor anyways (also because I have a 75 ns
diode), but if this transistor really has 3800 ns turn off time I
need a different approach. Any recommendations for a transistor? I'd
prefer the inductor current to remain below 50 mA.

Thanks!

ISTR that base "speedup" caps are not a good idea with HV transistors,
but I'm way too lazy to dig up the reference

why not use the npn as a cascode switch? Rb to +10V, in parallel with
diode (anode to B). bung a small FET (eg 2N7002) in the emitter leg, and
switch the FET. with FET on, base current flows thru Rb, D
reverse-biased. When FET turns off, coillector current flows thru B-C
junction, via diode to +10V, sweeping out stored charge.

this trick is older than I am, but it hasnt stopped me from using it
successfully. Very convenient if your smps has to run from 1400Vdc.
BU508 switching happily at 150kHz anyone?

Cheers
Terry

 

[Winfield Hill]

Terry Given wrote...

... 1400Vdc. BU508 switching happily at 150kHz anyone?

10kV, 600kHz, 200pF, anyone?

 

[Joerg]

Hello Win,

10kV, 600kHz, 200pF, anyone?

5kV, 20MHz, several hundred pF, 500mA. Could have gone higher in
frequency but didn't need to. Could have gone to an amp but the mains
circuit breaker prevented that. This was with two tubes QB5/1750, each
the size of a dill pickle jar.

Regards, Joerg

 

[Winfield Hill]

Joerg wrote...

Hello Win,


5kV, 20MHz, several hundred pF, 500mA. Could have gone higher in
frequency but didn't need to. Could have gone to an amp but the
mains circuit breaker prevented that. This was with two tubes
QB5/1750, each the size of a dill pickle jar.

Xc of 200pF at 20MHz = j40 ohms. Current for 5kV = 125 amps.
Hmm, 0.5A, off by 250x. Joerg, do you have an explanation?

 

[Terry Given]

Hello Win,

thats your series FET converter, right?

5kV, 20MHz, several hundred pF, 500mA. Could have gone higher in
frequency but didn't need to. Could have gone to an amp but the mains
circuit breaker prevented that. This was with two tubes QB5/1750, each
the size of a dill pickle jar.

Regards, Joerg

That'll teach me for messing with the big boys

looks like I'd better build a flyback MRC with a 9kA 6kV GTO....

how many watts Pout? my BU508 didnt even need a heatsink for 25W Pout @
750Vdc in, and cost $0.50.

I played with a few 1700V BJTs, but I never used the 2kV parts I
scrounged. Modern HOPTs could easily switch at 200-300kHz in cascode.
kinda makes ya think about single-ended off-line MRCs, using only stray
capacitance - who cares if the peak switch voltage is 4 x Vdcin

Cheers
Terry

 

[[email protected]]

ISTR that base "speedup" caps are not a good idea with HV transistors,
but I'm way too lazy to dig up the reference

why not use the npn as a cascode switch? Rb to +10V, in parallel with
diode (anode to B). bung a small FET (eg 2N7002) in the emitter leg, and
switch the FET. with FET on, base current flows thru Rb, D
reverse-biased. When FET turns off, coillector current flows thru B-C
junction, via diode to +10V, sweeping out stored charge.

this trick is older than I am, but it hasnt stopped me from using it
successfully. Very convenient if your smps has to run from 1400Vdc.
BU508 switching happily at 150kHz anyone?

Cheers
Terry

But isn't the OP's real problem his approach? He wants to use a 1mH
inductor, a small inductor current, and a large inpututput voltage
ratio, apparently in a straight boost converter.

If we take Vcc=+10V, a 1mH inductor will charge from zero to 50mA in
5uS, and discharge @ 300V in 5uS/30 = ~170nS.

Possible with specialized parts, and taking care to rip charge out of
the switching transistor to turn it off quickly, I suppose, but,
replacing the inductor with a step-up transformer would make transistor
& rectifier choices very easy.

Another choice might be to up the inductor current a lot, increasing
the discharge time enough to allow for slower diodes, and easing
switching-speed demands on the transistor. A larger output cap could
gobble up the extra energy, filtering the output.

Regards,
James Arthur

 

[Joerg]

Hello Win,

Xc of 200pF at 20MHz = j40 ohms. Current for 5kV = 125 amps.
Hmm, 0.5A, off by 250x. Joerg, do you have an explanation?

It was driving a resonant circuit ;-)

Regards, Joerg

 

[Joerg]

Hello Terry,

That'll teach me for messing with the big boys

looks like I'd better build a flyback MRC with a 9kA 6kV GTO....

how many watts Pout? my BU508 didnt even need a heatsink for 25W Pout @
750Vdc in, and cost $0.50.

It could do about 1500W but that was only limited by the fact that the
mains circuit breaker would trip if we went over that.

Didn't have a heat sink either. But holding a hand over the unit felt
like holding it over a wood stove.

Regards, Joerg

 

[Terry Given]

Hello Terry,


It could do about 1500W but that was only limited by the fact that the
mains circuit breaker would trip if we went over that.

Didn't have a heat sink either. But holding a hand over the unit felt
like holding it over a wood stove.

Regards, Joerg

Hi Joerg,

there was a fascinating paper published in Industry Apps (IIRC) a few
years back that discussed why thermionic devices rule at high power
levels. Pretty much cos they can run stinking, filthy hot. SiC has been
promising for years, but buying actual parts is a whole 'nother thing. a
bit like SiThs (the thyristor, not the evil Jedi) and MCTs

I read a paper a few weeks back about a hydraulic control system that
operated from 10F to 1000F. cool trick.

Cheers
Terry

 

[Joerg]

Hello Terry,

there was a fascinating paper published in Industry Apps (IIRC) a few
years back that discussed why thermionic devices rule at high power
levels. Pretty much cos they can run stinking, filthy hot. SiC has been
promising for years, but buying actual parts is a whole 'nother thing. a
bit like SiThs (the thyristor, not the evil Jedi) and MCTs

That is also the beauty of the old tubes. As long as the plate doesn't
exceed red-hot most of them are ok. Those QB5/1750 can sink more than 4A
each for brief periods while dropping only a few hundred volts. And they
are fast. With some tricks they are good up to 100MHz.

Regular switch gear is often quite staunch as well. Once I had to design
a test set to demonstrate defibrillation proofness of one of my
circuits. The TUEV inspector wanted me to show it. He retracted towards
a corner of the room when I told him I'd do it right now. Basically a
32uF cap is charged to 5kV and then discharged via a bunch of big coils,
resulting in a current of 50-100A IIRC. Ka-boom. After that the PBX
system at my client needed a power cycle and the SW guys were throwing
paper balls at me because the LAN went on the fritz.

I read a paper a few weeks back about a hydraulic control system that
operated from 10F to 1000F. cool trick.

Nice, but it does go below 10F at times around here ;-)

Regards, Joerg

 

[Thomas Anderson]

That sounds cool, I'm about to try it.

Possible with specialized parts, and taking care to rip charge out of
the switching transistor to turn it off quickly, I suppose, but,
replacing the inductor with a step-up transformer would make transistor
& rectifier choices very easy.

Making an inductor is just much simpler. I have some ferrite
transformers from scrounged TVs and SMPSs but I don't know how
suitable they are and I don't feel confident breaking them up,
rewinding them and fitting the ferrite back together.

But isn't the OP's real problem his approach? He wants to use a 1mH
inductor, a small inductor current, and a large inpututput voltage
ratio, apparently in a straight boost converter.

If we take Vcc=+10V, a 1mH inductor will charge from zero to 50mA in
5uS, and discharge @ 300V in 5uS/30 = ~170nS.

I now wound a 15 mH inductor and, well, I guess I could also allow
100 mA. That would make 5000 ns @ 300V. Do you think that would work?

Thanks

 

[[email protected]]

That sounds cool, I'm about to try it.


Making an inductor is just much simpler. I have some ferrite
transformers from scrounged TVs and SMPSs but I don't know how
suitable they are and I don't feel confident breaking them up,
rewinding them and fitting the ferrite back together.


I now wound a 15 mH inductor and, well, I guess I could also allow
100 mA. That would make 5000 ns @ 300V. Do you think that would work?

Yes. Your inductor will now charge from 10v in 150uS, and discharge
at a leisurely pace that easily-found diodes can handle. Terry's
cascode trick will be handy, and you're in business.

Regarding winding transformers being difficult: By winding the 15mH
you've already done most of the work! If you'd wound that <secondary>
over the original 1mH <primary> winding, you'd have a 1:4 transformer &
the switch at the primary would only see (Vcc + 75v).

Regards,
James Arthur