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'H-Bridge - Warning!'
1996\12\03@105618 by Martin Nilsson

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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_OUTmnTakeThisOuTspamsics.se
Box 1263, S-164 28 Kista                 Fax: +46-8-751-7230
Sweden                                   Tel: +46-8-752-1574

1996\12\03@111246 by Martin Nilsson

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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

1996\12\03@124230 by Jonathan King

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> 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                        |   .....kingKILLspamspam@spam@uicc.com
Unitrode Corp.                       |   http://www.unitrode.com
7 Continental Blvd                   |   (603) 429-8715
Merrimack, NH   03054                |   (603) 424-3460

1996\12\03@163453 by Martin Nilsson

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> I think it was right the first time?

Yes, you are right. Sorry for wasting bandwidth. (But it is good this issue
gets attention.)

-- Martin Nilsson

1996\12\04@084707 by Larry G. Nelson Sr.
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At 04:56 PM 12/3/96 +0100, Martin Nilsson wrote:
{Quote hidden}

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.NelsonKILLspamspam.....ieee.org
http://www.ultranet.com/~nr

1996\12\04@110718 by Ray Gardiner

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>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_OUTspamTakeThisOuTdsp-systems.com
Technical Director DSP Systems               http://www.dsp-systems.com
Private e-mail to:-  rayspamspam_OUTnetspace.net.au

1996\12\04@160009 by Gregg Kricorissian

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Martin Nilsson <@spam@mnKILLspamspamSICS.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 <KILLspamkingKILLspamspamUICC.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

1996\12\04@194647 by Steve Hardy

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> From: Gregg Kricorissian <RemoveMEgrkricorTakeThisOuTspamccs.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

1996\12\07@011905 by Ben L Wirz

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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.

1996\12\07@030256 by Tim Jackson

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>> 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 \        
  (== ^ ==)
    ) - (
   ( ) ( )
 ( ( ) ( ) )
 (_(_)_(_)_)

1996\12\07@130627 by nigelg

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In message  <spamBeGonePine.SOL.3.91.961206235629.28717F-100000spamBeGonespamclarion.cec.wustl.edu>> TakeThisOuTPICLISTEraseMEspamspam_OUTMITVMA.MIT.EDU writes:


{Quote hidden}

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.

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