Emitter switching: Damn! Who said that bipolars were slow and hard to drive? - Page 2 - diyAudio
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Old 2nd April 2006, 08:35 PM   #11
poobah is offline poobah  United States
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Eva,

Another fun thing you can try sometime is to parallel a MOSFET and a bipolar or IGBT. You then drive with separate timed circuits... always turning the FET on first and off last. You get the low switching losses of the FET and the low conduction losses of the bipolar.


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Old 2nd April 2006, 09:18 PM   #12
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The problem with emitter switching is that the base of the top transistor briefly sees the full collector current during turn off. As can be expected, some transistors don't like this treatment, and will fail after a while. Too bad, as you can get pretty spectacular results otherwise.
ST has just come out with a series of transistors expressly designed for emitter switching, meant to replace >1000V MOSFETs, which are huge and expensive. It'd be interesting to find out what they did to tailor the transistor for ths kind of operation (if anything), and whether it works as well as advertised. If I remember, I'll also ask a colleague at work who has more experience with emitter switching to go into some more detail about the conditions where he experienced transistor failures. I've personally used emitter switching with a Philips BUP23C, but never saw any failures that could be attributed to the base giving out. However, the design never went into production - maybe I was lucky...
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Old 2nd April 2006, 09:26 PM   #13
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The plot thickens - I just checked the ST website and did a search using the key words "emitter switched". The ST device in question has the MOsFET and bipolar transistor integrated together. It has a rating of 1500V/8A, with (supposedly) a square SOA.
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Old 2nd April 2006, 10:11 PM   #14
Eva is offline Eva  Spain
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poobah:

I proposed that in another thread, but I haven't still tried it. Due to the N-MOSFET-feeding-a-P-bipolar nature of IGBTs, I have doubts about how much a slow "standard-speed" IGBT would take to saturate when there is another device shunting it and preventing it from receiving enough drive current, and how much would it take to desaturate back when the C-E voltage is not allowed to rise.


Onra:

Yes, I built it in a breadboard because I wanted to investigate the turn-off behaviour, as my other bipolar base drive system has some drawbacks that I'm desperately trying to overcome. That thread covers most of my previous work with bipolars: 0-15V 0-120A adjustable PSU attempt with average current control.


wrenchone:

My circuit is not exactly like the ones shown in the application notes. See the 560pF feedback capacitor across the MOSFET and try to figure out what it does. See also the 1uF base capacitor, that in hard switching conditions could easily destroy the B-E junction, but mine is not emitter-hard-switching...

Actually, the first thing that I found out is that hard switching over the emitter was not going to produce the performance that I wanted. Paradoxically, with bipolar transistors the fastest turn-off behaviour with no current-tail is only obtained when the rate at which charge is extracted from the base is intentionally limited, and the optimum value seems to be somewhat smaller than Ic. In the shown working conditions I got no current tail at all as the emitter does not seem to get pulled up during the collector voltage rise period. I have to verify that for lower currents and lower forced gains, though.
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Old 2nd April 2006, 10:39 PM   #15
mzzj is offline mzzj  Finland
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Nice circuit exept I dont like parts count in your circuits Especially all custom-wound magnetics wich are costly solution unless you order million of them from little guys in china

IPP60R099CS costs something like 5 euros, has 0.099mohm on-resistance and switches quite easily under 50ns
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Old 2nd April 2006, 11:30 PM   #16
Eva is offline Eva  Spain
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For the moment, the only custom-wound thing in that circuit is a 20mm diameter toroid ferrite core with a 1 turn primary and a 5 turn secondary. That core may be shared for two switching cells in a half bridge. How much could it cost?
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Old 3rd April 2006, 12:18 AM   #17
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Conrolled turnoff for HV bipolars is a pretty old concept, and was covered quite extensively in Philips and other application notes in the 80's, when HV bipolar transistors were the only economical alternative for switching power supplies. If you can dig them up, the app notes make for really interesting reading. I hadn't run into the concept in the context of emitter switching, but it makes sense in retrospect.

For those unfamiliar with the concept (not you, Eva...), there is an optimum rate of turnoff for HV bipolar transistors. This is mostly to do with the high resistivity collector region needed in order to get a high breakdown voltage in a bipolar transistor. Unless one somehow keeps the transistor out of saturation (not a good idea - high VCE drop), you need to fill the collector region with charge for a low VCE drop during turnon (forward bias the base-collector junction), then pull all that charge out during turnoff. If you try to withdraw charge too rapidly by driving the base-emitter junction too hard, some charge remains trapped in the collector region, resulting in a turnoff tail. If you don't drive the base hard enough, you get slow turnoff for the usual obvious reasons. Some designs used an inductor in series with the base to control the rate of turnoff. The inductor would also reverse avalanche the base-emitter junction. Motorola published an application note about this and concluded that reverse avalanching the BE junction did not harm the device, assuming a reasonable reverse current.

Eva, I would be interested to see the upper transistor base current during turnoff. It may be that the capacitor on the gate to drain of your bottom MOSFET is smearing out the base current pulse in the upper transistor during turnoff so that it is less likely to damage the device. BTW - you sure get a lot of mileage out of 13009s.....
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Old 3rd April 2006, 12:52 AM   #18
Eva is offline Eva  Spain
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I have these Philips application notes about HV bipolars, they are worth reading.

What the capacitor across the MOSFET does is to force the drain voltage to have fixed rising and falling slopes. That in turn forces the 1uF capacitor to be charged and discharged through the base at constant rates, and thus, with constant currents.

My rates are currently 8V/us for turn-on and 6V/us for turn-off, yielding capacitor currents of 8A and 6A respectively. That's what I have empirically found to be optimum in my clumsy test fixture. For turn-off 6A may seem a small value, particularly because 2.4A from the current transformer have to be substracted from that, yielding a 3.6A real value, but the transistor likes that rate, it seems to have stopped conducting almost completely by the time the collector voltage starts to rise.

I've also found out by myself that the BE junction of bipolar transistors can whitstand reverse avalanche without causing any damage to the device as long as the pulse energy and total dissipation are kept under control. One of the undesirable features of my previous base drive system is that, when one set of transistors is turned on, it briefly drives the other set into reverse avalanche due to drive transformer leakage inductance. That's not destructibe but steals precious base current.
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Old 3rd April 2006, 02:05 AM   #19
Eva is offline Eva  Spain
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The 47 ohm resistor that dissipates the energy stored in the inductor just got too hot and desoldered from the diode when I was testing at 12A. That killed the MJE13009 instantaneously, but after replacing it I noticed that this little transistor survives unclamped turn-off at low current levels

That capture shows a nearly 1.000 Volt collector turn-off spike. Inductor current was 4.5A before turn-off, but most of that energy is transferred to the capacitances (and to the resistivity of the own iron-powder core) instead of avalanche dissipated in the bipolar.
Click the image to open in full size.

Note how the transistor becomes slower at low Ic levels.
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Old 4th April 2006, 02:27 AM   #20
Eva is offline Eva  Spain
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Updated circuit:
Click the image to open in full size.

To my surprise, connecting the 560pF capacitor that way not only solves all the MOSFET parasitistic oscillation issues that I previously had, but also produces better waveforms. It compensates for high B-E impedance during turn-on, that otherwise would reduce the amount of base boost current during the first 200ns, and it also allows for better B-E reverse biasing at turn-off.

I also rearranged the current sense transformer to avoid persistent base drive after collector current has ceased. It should have been that way from the beggining. Forced gain is now 4, as it produces smaller conduction losses without degrading turn-off performance too much.

This is how Vbe looks during a switching period at Ic=11A on a resistive (and slightly inductive) load. It was measured across transistor legs to avoid breadboard resistance. Scales are 1V/div and 1us/div. The 0V level corresponds to that small horizontal red mark on the left border of the grid. Note the increased Vbe due to 8A of Ib boost during the first microsecond, the progressive charge removal at the end of the cycle, and the convenient B-E reverse biasing during Vce rise. Some inductive artifacts are also shown.
Click the image to open in full size.


And this is how Vce looks in the same conditions and measured directly across transistor legs. Note the dramatic effects of those 8A of base drive boost applied during the first microsecond, that seem to eliminate dynamic saturation effects almost completely, and the ultra-fast desaturation process.
Click the image to open in full size.

I'm seriously considering that topology
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