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'[EE]: Analog Design Challenge - explain this'
2000\06\19@094546 by Russell McMahon

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part 1 2082 bytes content-type:text/plain; (decoded 7bit)

Attached is a simple circuit whose operation may not be as clear as it
appears.
if it IS clear then your explanation will be welcome :-)
I have little doubt that someone well schooled in transistor models will
have no problem with this but it doesn't seem to fully match what I should
have learned when they taught it to me or what I use very successfully day
to day - I have no problems withe employing transistors as a rule.

While I understand (now) what it does I find that it fits less than well
with my understanding of transistor operation and I haven't found a good
explanation on the web. Searches under "xth quadrant" operation of
transistors deal with similar issues but this is not a usual way to operate
transistors.

I realise I'm inviting derision by the excessively competent here but the
exercise may be useful to all.
How hard can it be ? :-)


So:


Voltage Vin varies from 0 to +10 volt DC.
The battery is an (approx) 6 volt, rechargeable.
Q1 is a standard PNP and Q2 an NPN.

The real circuit is somewhat more complex than this but this is the heart of
it.

Intended operation:
===============

When Q2 is off (base grounded) Q1 is off.
- Load is not supplied by battery.
- When Vin exceeds Vbattery plus 1 diode drop the battery will charge via
the 10r resistor.

When Q2 is on (base driven high via resistor (not shown).
Q1 is on (base pulled low via 1K).
Battery will supply Rl if Vin is less than approx Vbatt plus a diode drop.
When Vin rises to above a diode drop above Vbatt Q1 CE will be reverse
biased, load will be supplied by Vin and battery will charge via 10r.

QUESTIONS:
++++++++++

1.    When Vin is about 10 volts what will the main current flows be (eg
some or all of via 10r, battery, 10r, Q1, load as appropriate ).

a) When Q2 is off.
b) When Q2 is on.

2.    Discuss. Explain in terms of transistor action (or any appropriate
transistor model).

3.    What simple alteration/addition would you make to make operation
better defined? .

4.    Throw rocks etc as required



Russell McMahon






part 2 16248 bytes content-type:image/jpeg; (decode)


part 3 2 bytes
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2000\06\19@121615 by Dan Michaels

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Russell McMahon wrote:
........
>While I understand (now) what it does I find that it fits less than well
>with my understanding of transistor operation and I haven't found a good
>explanation on the web. Searches under "xth quadrant" operation of
>transistors deal with similar issues but this is not a usual way to operate
>transistors.
..........
>Intended operation:
>===============
>
>When Q2 is off (base grounded) Q1 is off.
>- Load is not supplied by battery.
>- When Vin exceeds Vbattery plus 1 diode drop the battery will charge via
>the 10r resistor.
>

Looks correct.
=============

>When Q2 is on (base driven high via resistor (not shown).

Yeah - don't forget that base series R.
===============

>Q1 is on (base pulled low via 1K).
>Battery will supply Rl if Vin is less than approx Vbatt plus a diode drop.
>When Vin rises to above a diode drop above Vbatt Q1 CE will be reverse
>biased, load will be supplied by Vin and battery will charge via 10r.
>

Looks correct. Q1 supplies current to RL whenever diode in main
path [upper right] is off - due to Vin < Vbatt+0.7v. When that diode
turns on, due to increasing Vin, Vin supplies RL, and Q1 will go into
operation in its reverse or inverse mode [quadrant 3]. There may be
some charging of the battery via this pathway.
=============

>QUESTIONS:
>++++++++++
>
>1.    When Vin is about 10 volts what will the main current flows be (eg
>some or all of via 10r, battery, 10r, Q1, load as appropriate ).
>
>a) When Q2 is off.
>b) When Q2 is on.
>

For high Vin, most of current flows thru main path diode [would help
if it had been labelled] into RL. If Q1 is off, a little current will
flow into the battery via the Q1 C-B diode and 1K base R inverse
operation]. Q2 on, in this situation, will have a minor effect.

Putting a diode, pointing to the right, between Q1 emitter and battery
will prevent blackflows, but I would suspect they have little effect
here, due to presence of main charging pathway to battery, 10R + diode.

- DanM

2000\06\19@164002 by steve

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I'll step up for possible humiliation.

> For high Vin, most of current flows thru main path diode [would help
> if it had been labelled] into RL. If Q1 is off, a little current will
> flow into the battery via the Q1 C-B diode and 1K base R inverse
> operation]. Q2 on, in this situation, will have a minor effect.

I don't know that it would be that minor. There would be 10V minus a
diode drop on the collector of Q1. Q1's emitter and the other side of
the 1k resistor to the base is fixed at 6V. Take off a diode drop for
the B-C junction and you have 10-0.6-0.6-6.0=2.8V across the 1k
giving rise to 3mA.
Just because one pin is labelled E and the other C, it doesn't mean
the device has to be used that way round. The doping is such that it
works best in the traditional way but it is still two lumps of P with
a slice of N in the middle.
So if we swap the E and C labels (ie the arrow) on Q1, we now have a
forward biased B-E junction and that will  turn Q1 on.
Since the battery is an infinite current sink, the only current limit
is the base resistor and the gain of the backwards transistor. Is
that enough for smoke ?

> Putting a diode, pointing to the right, between Q1 emitter and battery
> will prevent blackflows, but I would suspect they have little effect
> here, due to presence of main charging pathway to battery, 10R + diode.

I agree with the solution, but assuming my hypothesis above is
correct, the effect might be quite good.
The disadvantage is that it puts another diode drop between your
battery and the load. An alternative may be to put a schottky
diode between the collector and the base of Q1 (cathode to base). The
current path will still exist but the C-E junction won't get
high enough to turn on the backwards transistor.

Steve.

======================================================
Steve Baldwin                Electronic Product Design
TLA Microsystems Ltd         Microcontroller Specialists
PO Box 15-680, New Lynn      http://www.tla.co.nz
Auckland, New Zealand        ph  +64 9 820-2221
email: spam_OUTstevebTakeThisOuTspamtla.co.nz      fax +64 9 820-1929
======================================================

2000\06\19@181812 by Dan Michaels

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Steve wrote:
......
>> For high Vin, most of current flows thru main path diode [would help
>> if it had been labelled] into RL. If Q1 is off, a little current will
>> flow into the battery via the Q1 C-B diode and 1K base R inverse
>> operation]. Q2 on, in this situation, will have a minor effect.
>
>I don't know that it would be that minor. There would be 10V minus a
>diode drop on the collector of Q1. Q1's emitter and the other side of
>the 1k resistor to the base is fixed at 6V. Take off a diode drop for
>the B-C junction and you have 10-0.6-0.6-6.0=2.8V across the 1k
>giving rise to 3mA.
>Just because one pin is labelled E and the other C, it doesn't mean
>the device has to be used that way round. The doping is such that it
>works best in the traditional way but it is still two lumps of P with
>a slice of N in the middle.

Steve, you're right. The battery will be charged via both pathways.
First, 10R+diode, second, diode#2 + transistor. Assuming Vbatt = 6v,
then the 1st pathway current is about (10-.7-6)/10 = 330 mA. The 2nd
pathway will add some current to this.

Re forward vs reverse [inverse?] current gains in BJTs, the reverse
is generally a *lot* lower than the forward, by probably 10X or so,
FWIUnderstand. This is generally related more to the geometry of the
device rather than just the doping, AFAIK. BJTs are designed to
maximize forward HFE by fiddling these parameters, and as a result
reverse current gain is typically much less.

So, I should expect the contribution to charging would be minimal
when the BJT is operating in reverse. Probably << 330 mA.

In the ckt the way it is drawn, I think it's hard to tell exactly.
By inserting a resistor at a strategic point, it could be controlled
more exactly. However, I suspect the ckt was not originally
designed with the intent of actually using pathway #2 for charging.

regards,
- DanM

2000\06\19@195312 by Russell McMahon
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Thanks for the several replies to this so far on and off list.

So far NOBODY has fully seen what happens.
I agree with most of the analysis so far which is all largely based on
common expectations of how a transistor works. In this case something more
arcane is happening. Steve's comment on the fact that reversing C & E still
gives a lower gain transistor is an acknowledgment that things are not
always what they seem but it goes further than this. I'll wait another day
or so to see what else is said and then comment on what "really" happens. It
has certainly changed how I will design certain circuits in future.


RM

2000\06\19@204737 by Scott Dattalo

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On Tue, 20 Jun 2000, Russell McMahon wrote:

> Thanks for the several replies to this so far on and off list.
>
> So far NOBODY has fully seen what happens.
> I agree with most of the analysis so far which is all largely based on
> common expectations of how a transistor works. In this case something more
> arcane is happening. Steve's comment on the fact that reversing C & E still
> gives a lower gain transistor is an acknowledgment that things are not
> always what they seem but it goes further than this. I'll wait another day
> or so to see what else is said and then comment on what "really" happens. It
> has certainly changed how I will design certain circuits in future.

I'll take a shot in the dark.

Suppose Q2 is off and the battery is installed. Q1's CB junction is forward
biased and the PNP, which isn't exactly designed to be symmetrical will
none-the-less start conducting. Nothing new here - a previous poster already
noted this. But suppose Q2 is turned on while Q1 is in this inverted state. What
will happen? Will the CB junction retain this forward bias or will the
transistor (Q1) suddenly turn around and behave normally? I personally don't
know. However, since Russel is throwing this out as a mystery, I'll guess that
Q1 will stay in this inverted state. If this is the case, then turning on Q2
will have the effect of driving Q1 even harder in this backward mode. The
battery will get even more current from the power supply - only limited by the
(output) diode and the reversed biased CE junction of Q1

Now, suppose you remove the supply voltage while Q2 is still turned on. Will Q1
escape this reversed state? Again, I don't know, but I'll guess that it
doesn't. If this is the case then Q1 will look like an open circuit and the
battery would be unable to supply the load. But this is just a guess... Since
I'm way out on a limb that's starting to break, I'll further guess that Q1 can
get out of this state by turning off Q2.

Dmitry probably has a better answer that only takes 12 instructions :)

Scott

2000\06\19@225829 by steve

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> So far NOBODY has fully seen what happens.

This isn't the heat-it-up-till-it-glows, opto powered transistor
thing, is it ?

Steve.

======================================================
Steve Baldwin                Electronic Product Design
TLA Microsystems Ltd         Microcontroller Specialists
PO Box 15-680, New Lynn      http://www.tla.co.nz
Auckland, New Zealand        ph  +64 9 820-2221
email: .....stevebKILLspamspam@spam@tla.co.nz      fax +64 9 820-1929
======================================================

2000\06\20@074638 by John Hallam

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Here are a few thoughts:

On Tue, 20 Jun 2000, Russell McMahon wrote:

{Quote hidden}

       In this case, Q1 conducts in inverse mode as described by another
poster and about 3mA of base current flows.  The load is powered by the
10V supply and the battery charges via the 10R and via Q1 at a rate
depending on Q1's reverse hFE.  So far so good.

> b) When Q2 is on.

       Now an extra ~8.5mA of base current flows in the inverse
conducting Q1 (Q1 base is nailed at ~9V by the 1N4001 and the collector PN
junction diode drop).  The result is that the Q1 current goes up by about
4x.  The load is powered as before and the battery charging current is
about 4 times what it was in case (a).  The circuit probably fails the
nose test at this point (if it didn't before).

       It looks to me as though it should revert to intended behaviour
when Vin drops below the battery-powered Q1 collector voltage, but
applying the 10V again will resume smoke-generation mode.  Or is it a
bomb? ;-)

> 3.    What simple alteration/addition would you make to make operation
> better defined? .

       Someone mentioned adding a diode between Q1 collector and the
load/1N4001 junction.  A reasonably fat Schottky diode would do the job
while keeping the lost voltage down, depending on current needed for the
load off battery.

> 4.    Throw rocks etc as required

       Sorry, run out of rocks.  Threw them at students...

John Hallam.

2000\06\20@122648 by Russell McMahon

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Tom

>Though not mentioned in many textbooks, bipolar transistors
>will work if connected backwards (exchange collector and
>emitter roles). This can sometimes catch you by surprise.
>Also, as you increase the reverse bias across the *real*
>collector-emitter electrodes, if there is any form of
>connection between real emitter and the base (as you have with the
>1k base/emitter resistor), then the transistor can begin
>to conduct. The "gain" in this case is quite low, but the
>conduction is quite real.
>
>Is this the effect that you are seeing? You can eliminate it by
>adding a diode in series with Q1's collector, but this gives
>you an unwanted extra diode drop. Bummer.



Yes in part - but it's not the whole answer.
When you swap collector and emitter on an NPN you still have an NPN - just
lower gain. I am aware of this configuration. People, have explained this as
either doping profile or physical construction. I don't know enough about
physical construction to be sure. All that theory was looong ago :-) but I'm
fairly comfortable with day to day practice.

In this instance the original base-emitter resistor which holds the
transistor off normally effectively becomes a base-collector resistor when
the transistor is considered inverted and it is turned on rather than off by
this resistor.
However, even when this resistor is removed and a base to original-collector
(now base to effective emitter) resistor is added the transistor still
conducts. There is effectively no stopping it :-). I suspect this is a
modification of the reverse breakdown of the junction but I'm not sure.


Mayhaps someone can explain what I see in conventional terms.

You are correct that a diode fixes the problem.
This is what I decided was the tidiest solution and what I alluded to in the
original questions but it will have to be a
Schottky as the available headroom is already tight.
I considered using a FET (you replace the diode drop with Rdson x I and this
can be arbitrarily low by spending more $ on the FET) but the component
count to drive it is unpleasant  and the design is very price sensitive,
which is why I am spending excessive time trying to make a minimalist design
work rather than just adding components - in this application every resistor
counts.



RM

2000\06\20@131021 by Dan Michaels

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Russell McMahon wrote:
>Thanks for the several replies to this so far on and off list.
>
>So far NOBODY has fully seen what happens.
>I agree with most of the analysis so far which is all largely based on
>common expectations of how a transistor works. In this case something more
>arcane is happening. Steve's comment on the fact that reversing C & E still
>gives a lower gain transistor is an acknowledgment that things are not
>always what they seem but it goes further than this. I'll wait another day
>or so to see what else is said and then comment on what "really" happens. It
>has certainly changed how I will design certain circuits in future.
>

But since you are teasing us ........ [here's another story]

I used a ckt similar to this to create an RS-232 network - ie, a bunch
of PNP transistors with their collectors all tied together driving
a single Tx line - ie, "wired-OR". Problem was, if one of the source
ckts was "disconnected" from the line by simply turning it off, its
PNP operated in the inverse mode to keep the ckt powered up - or at least
partially powered up - and it loaded down the Tx buss. Solution was to
put a diode between the collector and the common Tx bus.

- DanM

2000\06\20@133305 by Dan Michaels

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Russell McMahon wrote:
..........
>When you swap collector and emitter on an NPN you still have an NPN - just
>lower gain. I am aware of this configuration. People, have explained this as
>either doping profile or physical construction. I don't know enough about
>physical construction to be sure. All that theory was looong ago :-) but I'm
>fairly comfortable with day to day practice.
>

I believe "most" bjts are built as planar type.
For NPN, take a lightly doped N-type silicon wafer [this is the collector],
mask off and embed fairly heavily with P-type impurities [this is base],
mask smaller and embed very heavily with N-type impurities [this is emitter].

 C    B       E
-----+----+--------+-----+--------
    |    |N+      |     |
    |P   +--------+     |  <-- base "height" very small
    +-------------------+
N
----------------------------------

So, it's generally both doping AND geometry. High forward gain
derives from having a very narrow [vertically, here] base region
and a very large collector which surrounds the emitter, so that
all "emitted" electrons are effectively "collected". The heavy
doping in the emitter region enhances the #electrons emitted,
given the B-E field. From this, it's pretty clear why inverse
operation is not as efficient.

- DanM

2000\06\20@134548 by Peter Schultz

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-----Original Message-----
From: Dan Michaels [oricomspamKILLspamLYNX.SNI.NET]
Sent: Tuesday, June 20, 2000 10:32 AM
To: .....PICLISTKILLspamspam.....MITVMA.MIT.EDU
Subject: Re: [EE]: Analog Design Challenge - explain this


Russell McMahon wrote:
..........
>When you swap collector and emitter on an NPN you still have an NPN - just
>lower gain. I am aware of this configuration. People, have explained this
as
>either doping profile or physical construction. I don't know enough about
>physical construction to be sure. All that theory was looong ago :-) but
I'm
>fairly comfortable with day to day practice.
>
Officially called inverz mode, used in early days for discrete D/A
converters for achieving very low and stable saturation..
Peter

2000\06\20@135349 by Dan Michaels

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Peter Schultz wrote:
........
>Officially called inverz mode, used in early days for discrete D/A
>converters for achieving very low and stable saturation..
>Peter
>

So, Vec[sat] < Vce[sat] would apparently be due to the much
heavier doping in the emitter region w.r.t. the collector region?

2000\06\20@140016 by Peter Schultz

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Yes,
That is true, and as far as I remember the saturation voltage was in few mV
range.
Peter

-----Original Message-----
From: Dan Michaels [EraseMEoricomspam_OUTspamTakeThisOuTLYNX.SNI.NET]
Sent: Tuesday, June 20, 2000 10:53 AM
To: PICLISTspamspam_OUTMITVMA.MIT.EDU
Subject: Re: [EE]: Analog Design Challenge - explain this


Peter Schultz wrote:
........
>Officially called inverz mode, used in early days for discrete D/A
>converters for achieving very low and stable saturation..
>Peter
>

So, Vec[sat] < Vce[sat] would apparently be due to the much
heavier doping in the emitter region w.r.t. the collector region?

2000\06\20@165627 by Dan Michaels

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John Hallam wrote:
........
>> a) When Q2 is off.
>
>        In this case, Q1 conducts in inverse mode as described by another
>poster and about 3mA of base current flows.  The load is powered by the
>10V supply and the battery charges via the 10R and via Q1 at a rate
>depending on Q1's reverse hFE.  So far so good.
>
>> b) When Q2 is on.
>
>        Now an extra ~8.5mA of base current flows in the inverse
>conducting Q1 (Q1 base is nailed at ~9V by the 1N4001 and the collector PN
>junction diode drop).  The result is that the Q1 current goes up by about
>4x.  The load is powered as before and the battery charging current is
>about 4 times what it was in case (a).  The circuit probably fails the
>nose test at this point (if it didn't before).
>

One last go at this [before dying].

I did some more looking/calculations. For part b), granted you should
have about 3+8 mA base current with Q2 on. The question here is how much
[inverse] Ie this will produce???????? It's hard to find values of
hFE[reverse]. My Millman and Grabel [circa early 90s] says "beta-forward
for IC transistors between 50 and 250, beta-reverse usually between
1 and 5". So quite low. IC transistor is basically planar.

From this, you might expect upwards to 50 mA of Ie [inverse] feeding
the battery. However, Russell said it was actually about 1/2 of 300 mA.
I guess it's possible hFE[reverse] for his transistor is closer to
15-20 than to 5.

Another possibility is that hFE[reverse] may be even more sensitive to
temperature than hFE[forward]. With say ~4v and 50+ mA, the transistor
would be dissipating ~0.2W, which might be enough to cause a temprature
rise in the emitter region [small and not designed for power dissipation],
thereby causing hFE[reverse] to increase, thereby increasing Ie and
dissipation, ....... --> positive fb + thermal runaway of hFE.
============

But what I wouldn't expect would be the VCB reverse breakdown voltage
to be exceeded in any course, since the total Vdiff between E and C
should never exceed ~4V, ie Vin=10v - Vbatt=6v, unless the battery
were very heavily discharged.

good nite, enough transistors for one day,
- DanM

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