Post by thread:H-Bridge

Ben L Wirz wrote: > You have used the L293D or the L293 to 1 Amp? The L293 is rated > at .6 A and the L293D is rated at 1 Amp, or they can be bridged for > double capicity. Can I also ask where you found the L293's if thats what > you used, I haven't been able to find them in anything less then 200 min > order. I'm afraid you should not try to increase the current capacity by paralleling outputs of bipolar devices, such as the 293. If you do, the maximum allowable current will be not be the sum, but the minimum of the individual maxima. The reason is as follows: Bipolar transistors have negative temperature coefficient, i.e. resistance goes down as they heat up. For two paralleled outputs, the one with lower on-resistance will take more current, and will heat up more, resistance becomes even lower, etc. The positive feedback quickly causes current imbalance. MOSFETs are different, and have positive temperature coefficient, so from this point of view, their outputs can be paralleled. However, there can be other problems with paralleling MOSFETs. I have heard quite a number of reports of people blowing up their 293s by paralleling outputs, many of them from builders of the 6.270 board, where the assembly manual at one point suggested piggy-backing of two 293s. Martin Nilsson http://www.sics.se/~mn/ Swedish Institute of Computer Science E-mail: spam_OUTmnTakeThisOuTsics.se Box 1263, S-164 28 Kista Fax: +46-8-751-7230 Sweden Tel: +46-8-752-1574

In my previous message, I wrote: > Bipolar transistors have negative temperature coefficient, > i.e. resistance goes down as they heat up. For two paralleled outputs, > the one with lower on-resistance will take more current, and will heat > up more, resistance becomes even lower, etc. The positive feedback > quickly causes current imbalance. > > MOSFETs are different, and have positive temperature coefficient, so > from this point of view, their outputs can be paralleled. However, > there can be other problems with paralleling MOSFETs. I just realized I switched the words "positive" and "negative" - sorry if it caused any confusion. -- Martin Nilsson

> In my previous message, I wrote: > > > Bipolar transistors have negative temperature coefficient, > > i.e. resistance goes down as they heat up. For two paralleled outputs, > > the one with lower on-resistance will take more current, and will heat > > up more, resistance becomes even lower, etc. The positive feedback > > quickly causes current imbalance. > > > > MOSFETs are different, and have positive temperature coefficient, so > > from this point of view, their outputs can be paralleled. However, > > there can be other problems with paralleling MOSFETs. > > I just realized I switched the words "positive" and "negative" - sorry > if it caused any confusion. I think it was right the first time? MOSFETs have a positive TC -> their resistance does goes up with temperature. when paralleled, the FET getting hot increases its resistance, decreasing the current it carries, tending to cool it down, ... So even FETs without matched parameters (Vth, gm) will share reasonably well. BJT's have a Vbe going down as temp increases and beta increasing. Between BJTs, the hotter transistor increases in beta, which decreases Vce, causing it to carry more current. The effect is not linear, and the power dissipation goes up, causing it to get still hotter ... The explaination above is generally true. There are other factors which can cause either type of transistor to behave in the opposite manner. Regards Jonathan King --------------------------------------------------------------------- Jonathan King | .....kingKILLspam@spam@uicc.com Unitrode Corp. | http://www.unitrode.com 7 Continental Blvd | (603) 429-8715 Merrimack, NH 03054 | (603) 424-3460

At 04:56 PM 12/3/96 +0100, Martin Nilsson wrote: {Quote hidden}>Ben L Wirz wrote: > >> You have used the L293D or the L293 to 1 Amp? The L293 is rated >> at .6 A and the L293D is rated at 1 Amp, or they can be bridged for >> double capicity. Can I also ask where you found the L293's if thats what >> you used, I haven't been able to find them in anything less then 200 min >> order. > >I'm afraid you should not try to increase the current capacity by >paralleling outputs of bipolar devices, such as the 293. If you do, >the maximum allowable current will be not be the sum, but the minimum >of the individual maxima. > >The reason is as follows: > >Bipolar transistors have negative temperature coefficient, >i.e. resistance goes down as they heat up. For two paralleled outputs, >the one with lower on-resistance will take more current, and will heat >up more, resistance becomes even lower, etc. The positive feedback >quickly causes current imbalance. > >MOSFETs are different, and have positive temperature coefficient, so >from this point of view, their outputs can be paralleled. However, >there can be other problems with paralleling MOSFETs. > >I have heard quite a number of reports of people blowing up their 293s >by paralleling outputs, many of them from builders of the 6.270 board, >where the assembly manual at one point suggested piggy-backing of two >293s. > >Martin Nilsson http://www.sics.se/~mn/ >Swedish Institute of Computer Science E-mail: mnKILLspamsics.se >Box 1263, S-164 28 Kista Fax: +46-8-751-7230 >Sweden Tel: +46-8-752-1574 > > You can increase the current capability this way but you need to put a small resistor in series with each bipolar device to level the current sharing. You are correct that simple parallel connections will cause the current to not share equally and devices will blow up. Also the total capacity will be less than the sum of the capacity. There is also a possible problem due to switching times if one device turns on before the others but the technique will work to a point. Larry G. Nelson Sr. .....L.NelsonKILLspam.....ieee.org http://www.ultranet.com/~nr

>At 04:56 PM 12/3/96 +0100, Martin Nilsson wrote: >>Ben L Wirz wrote: >> >>> You have used the L293D or the L293 to 1 Amp? The L293 is rated >>> at .6 A and the L293D is rated at 1 Amp, or they can be bridged for >>> double capicity. Can I also ask where you found the L293's if thats what >>> you used, I haven't been able to find them in anything less then 200 min >>> order. >> >>I'm afraid you should not try to increase the current capacity by >>paralleling outputs of bipolar devices, such as the 293. If you do, >>the maximum allowable current will be not be the sum, but the minimum >>of the individual maxima. >> <snip sensible comments about blowing things up> I have used a lot of different H-Bridges over the years and would definately NOT run any in parallel. It is far easier and cheaper in the long run to get a higher spec device there are lots around... the best of those that I have used are:- 1. MPM3002 motorola keeps dropping them and then re-instating. otherwise unbeatable for high current 8A 100V 2. IR8200B IR part that nat semi now makes an exact equivalent is also excellent. Pretty well indestructable if you watch out for voltage spikes 50v 3A, but i have run at 5-6A for short periods. 3. L6203 sgs no thermal overload or I sense outputs but if you can add these externally good for 48V 5A. 4. discrete fets with hi-side drivers are generally required if you want to go much above 50v or so. IR2110's and any suitably rated N-channel fets x4. Most failures of these devices IMHO are caused by voltage spikes, snubbers transorbs mov's whatever you can do to kill those fast transients will prolong the device life. Ray Gardiner, Shepparton, Victoria, Australia EraseMErayspam_OUTTakeThisOuTdsp-systems.com Technical Director DSP Systems http://www.dsp-systems.com Private e-mail to:- rayspam_OUTnetspace.net.au

Martin Nilsson <@spam@mnKILLspamSICS.SE> stated quite correctly, that: >MOSFETs are different, and have positive temperature coefficient, so >from this point of view, their outputs can be paralleled. However, >there can be other problems with paralleling MOSFETs. Quite true. The sales blurb would have you believe that paralleling MOSFETs is an easy thing to do. However, two problems result from multiple devices: 1. they require increased drive due higher input capacitance (Cgs), and 2. the pos resistance tempco of the device (0.5 to 0.6 % per degree C) can work against you with the thermal time constant of the heat sink... The whole idea is that the MOSFET device which is hogging the most current will "automatically" throttle itself back as its die heats up internally. A number of years back I fabricated an array of parallel MOSFETs to handle higher currents, and regardless of care taken in matching them, I invariably had trouble with popping individual devices in the array, inspite of the fact that they were well within their SOA, and quiescent current through the array kept decreasing as the heatsink got hotter (another annoying trait in linear applications). After Martin second-guessed himself, Jonathan King <KILLspamkingKILLspamUICC.COM> went on to advise: >I think it was right the first time? >MOSFETs have a positive TC -> their resistance does goes up with temperature. >when paralleled, the FET getting hot increases its resistance, Correct. the tempco of MOSFET devices is often a subject of confusion, because writers often omit the measure (eg: resistance) or context (eg: channel). The device's "Channel resistance" (Rds) tempco is positive without doubt, but confusion often results because a MOSFET's Vgs threshold has a negative tempco. To prevent unwittingly perpetuating the confusion, it's best follow Jonathan's example above: spell out what you mean in words. Hope this helps, .... Gregg

> From: Gregg Kricorissian <RemoveMEgrkricorTakeThisOuTccs.carleton.ca> > [lots of good stuff snipped] I would recommend the book 'Power Electronics' by B.W.Williams (McMillan) as an excellent reference on this type of stuff. Gregg stated a Rds(on) coefficient of +0.6%/K hovever said reference indicated only +0.1%/K. Never mind, but I would be interested in knowing the tempco of the threshold voltage since this is not mentioned, and could be an important consideration in paralleling mosfets especially if the gate driver has trouble pushing all the capacitance. (Currently I'm designing a 2KW 20KV HF AC supply for running my laser tube - PIC controlled, of course, so hopefully Jory won't zap me for digressing!). A good trick for current sharing two devices is ---+---- | - | | load | | - | | ---+--- | | 3 | | 3 o 3 | | 3 o 3 | | 3 | | | | |- |- || || |- |- | | | | --+-----+--- The funny thing with '3's is supposed to be a transformer with windings as shown, tightly coupled as per bifilar wound toroid. Any tendency for one device to draw higher current causes a higher voltage to be impressed on the other device, forcing the latter device to take its fair share. This works with switching designs only. The transformer core must not saturate from prolonged imbalance. This design is much more efficient than the usual resistors. Regards, SJH Canberra, Australia

Hello Everyone, On Tue, 3 Dec 1996, Martin Nilsson wrote: > The reason is as follows: > > Bipolar transistors have negative temperature coefficient, > i.e. resistance goes down as they heat up. For two paralleled outputs, > the one with lower on-resistance will take more current, and will heat > up more, resistance becomes even lower, etc. The positive feedback > quickly causes current imbalance. I've been thinking about this for a week now and I'm not sure I understand. I thought resistance always increased with increasing temperature. The basic idea being that as you heat something up the mocules move around faster making it that much harder for the electrons to move through the conductor which is in effect increasing the resitance. Take super conductivity for example, you decrease the temperature to the point where the mocules are moving very slowly which makes the resistance very low. Now I can believe that increasing the temperature affects the transitors Beta, but I find it hard to believe that it decreases the resitance. The total emitter current is given by: Ie=Ieo(e^(Vbe/nVt)-1) Vt=kT/q As T increase, the (Vbe/nVt) term will decrease. Raising the exponent to a smaller power results in a smaller Ie. So it looks to me like increasing the temperature results in smaller current flow through the eimitter which will cause a smaller current in the collector. The one thing I may be overlooking is that if the Beta does change with temperature in a real part. Does anyone have some insight for me? Thanks, Ben P.S. Sorry for the Non-PIC related post.

>> Bipolar transistors have negative temperature coefficient, >> i.e. resistance goes down as they heat up. > I've been thinking about this for a week now and I'm not sure I >understand. I thought resistance always increased with increasing >temperature. The problem is that the forward-biased voltage drop decreases with increasing temperature. |----|>|----| ---+ +--- |----|>|----| Consider the simplified version (above) of two bipolar transistors in parallel. Essentially you're dealing a pair of diodes in parallel. Nominally, a conventional silicon diode has a forward voltage drop of 600mV. If the diodes are absolutely identical, no problem. If, for argument's sake, the upper diode in the diagram has a voltage drop of 600mV and the lower one a voltage drop of 590mV, then, theoretically, the bottom diode will conduct all of the current and the upper diode will conduct nothing. As the bottom diode heats up, its voltage drop decreases, ensuring that the voltage drop difference becomes even greater and so on. In reality, both components would conduct some current, with the bottom one conducting more than the top one and with this difference increasing as the bottom one heated up faster due to its conducting a higher current. Although the voltage drop in a bipolar transistor is lower than a simple diode the same principle applies. Tim. /\_/\ / o o \ (== ^ ==) ) - ( ( ) ( ) ( ( ) ( ) ) (_(_)_(_)_)

In message <spamBeGonePine.SOL.3.91.961206235629.28717F-100000spamBeGoneclarion.cec.wustl.edu>> TakeThisOuTPICLISTEraseMEspam_OUTMITVMA.MIT.EDU writes: {Quote hidden}> I've been thinking about this for a week now and I'm not sure I > understand. I thought resistance always increased with increasing > temperature. The basic idea being that as you heat something up the > mocules move around faster making it that much harder for the electrons > to move through the conductor which is in effect increasing the resitance. > Take super conductivity for example, you decrease the temperature to the > point where the mocules are moving very slowly which makes the resistance > very low. > > Now I can believe that increasing the temperature affects the > transitors Beta, but I find it hard to believe that it decreases the > resitance. > > The total emitter current is given by: > > Ie=Ieo(e^(Vbe/nVt)-1) > > Vt=kT/q > > As T increase, the (Vbe/nVt) term will decrease. Raising the > exponent to a smaller power results in a smaller Ie. So it looks to me > like increasing the temperature results in smaller current flow through > the eimitter which will cause a smaller current in the collector. > > The one thing I may be overlooking is that if the Beta does change > with temperature in a real part. Does anyone have some insight for me? From a purely practical point of view I can assure you this happens, amongst other things I sometimes repair audio amplifiers. These are almost always class AB amplifiers, and these have a tendency to 'thermal runaway'. In class AB amplifiers the output transistors are biased on slightly, to avoid (reduce!) crossover distortion, this makes them prone to 'thermal runaway'. To avoid this the bias setting arrangement usually includes some kind of temperature compensation, in very old amplifiers thermistors were often used, but almost all modern amplifiers use a transistor connected as a VBE multiplier to do this. A VBE multiplier is simply a transistor with a preset connected to all three terminals, the slider feeds the base, and the outer connections connect to collector and emmitter. This basically gives a simple variable zener diode, and this is then connected between the bases of the two output transistors, and adjusted to give the required bias current. This transistor is then fastened to the same heatsink as the output transistors so it monitors the temperature. As the output transistors heat up, they pass more current, then they get hotter, and pass more current still, this is called 'thermal runaway' and will lead to distruction of the amp. However as the heatsink gets warmer, so does the VBE multiplier, and this also passes more current. By doing this it reduces the bias current for the output transistors and stops the runaway. Nigel. /--------------------------------------------------------------\ | Nigel Goodwin | Internet : RemoveMEnigelgTakeThisOuTlpilsley.demon.co.uk | | Lower Pilsley | Web Page : http://www.lpilsley.demon.co.uk | | Chesterfield | | | England | | \--------------------------------------------------------------/