transistor for small boost converter

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
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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!

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Reply to
Thomas Anderson
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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.

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

Reply to
Fritz Schlunder

Reply to
Paul Mathews

[snipped: discussion of MPSA44 switching speed]

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":

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

James Arthur

Reply to
dagmargoodboat

10kV, 600kHz, 200pF, anyone?
--
 Thanks,
    - Win
Reply to
Winfield Hill

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

--
 Thanks,
    - Win
Reply to
Winfield Hill

Hello Fritz,

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."

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

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Reply to
Joerg

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.

Regards, Joerg

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Reply to
Joerg

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

Reply to
Terry Given

But isn't the OP's real problem his approach? He wants to use a 1mH inductor, a small inductor current, and a large input:output 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

Reply to
dagmargoodboat

thats your series FET converter, right?

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

Reply to
Terry Given

Hello Win,

It was driving a resonant circuit ;-)

Regards, Joerg

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Reply to
Joerg

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

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Reply to
Joerg

Hello Terry,

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.

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

Regards, Joerg

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Reply to
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

Reply to
Terry Given

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?

Thanks

Reply to
Thomas Anderson

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 over the original 1mH winding, you'd have a 1:4 transformer & the switch at the primary would only see (Vcc + 75v).

Regards, James Arthur

Reply to
dagmargoodboat

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