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'[EE:] Battery Charger Question'
2003\10\25@151055 by Olin Lathrop

face picon face
kben@DCA.NET wrote:

Topic tag changed to EE:, which it should have been in the first place.

> I am building a battery charger for 4AA NiMh batteries.
> I know they are charged by current not voltage.
> I have read quite a bit on charging, but no one mentions
> the voltage for charging. Can I ignore this ?

Not if you want the best quality charge with the least damage to the
batteries.

> I plan on
> using a 9V 1A wall wart with a buck converter circuit.
> This allows me to control the current nicely. Since MAX
> voltage for each battery charged should be < 1.8 volts.
> This seems to be OK, 1.8 * 4 = 8.2 Volts.

This implies you are putting the 4 separate AA batteries in series.  Bad
idea.  I wouldn't do this unless the cells came manufactured in one
indivisible pack.  Different cells will be at different state of charge.
This will make it difficult to charge the lowest cell fully without
overcharging the fullest one.  Doing a discharge cycle before charging can
be a good idea with NiMH batteries, but is out of the question with 4
separate cells in series.  One cell could get discharged too deep or, worse
yet, have its polarity reversed.  NiMH cells are easily killed by
overdischarge.

> My question is
> can I use a 12V or 16V wall wart, as long as I charge
> with proper current does the Voltage matter at all ?

Only the voltage at the cells matter.  The voltage at the other end of your
buck converter only matters to the buck converter.  Since it is a buck
converter, higher voltage should be fine.  It will just draw less current
from the higher voltage.

> Next question, when the 4AA pack is connected, I plan on
> discharging the pack to 4V, before starting the charge
> cycle.

You shouldn't do this with separate cells that are not in a pack.  See
above.

> I plan on discharging by using the circuit below.

The discharge current will only be about 160mA given an 8.2V pack.  That
will definitely discharge those batteries, but typical AA NiMH cells can
stand significantly more than that without damage.

> Question when the +Batt drops lower than the 5V PIN,
> the collector will be less positive than the emitter,

No, it won't.  The emitter is at 0V by definition (that's what GND means).
Your circuit has no means of generating a negative voltage, so nothing can
be at a lower potential than the emitter (GND).

The collector should be at about 200-300mV because the transistor should be
saturated.

> will this turn the transistor off ?

No, since it won't happen.

{Quote hidden}

All in all, I recommend a separate circuit for each separate AA cell.


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2003\10\25@153833 by kben

picon face
Olin Lathrop <spam_OUTolin_piclistTakeThisOuTspamEMBEDINC.COM> said:

> .....kbenKILLspamspam@spam@DCA.NET wrote:
>
> Topic tag changed to EE:, which it should have been in
the first place.

Noted, I tend to use PIC since it is a PIC project.

>
> > I am building a battery charger for 4AA NiMh
batteries.
> > I know they are charged by current not voltage.
> > I have read quite a bit on charging, but no one
mentions
> > the voltage for charging. Can I ignore this ?
>
> Not if you want the best quality charge with the least
damage to the
> batteries.
>
> > I plan on
> > using a 9V 1A wall wart with a buck converter circuit.
> > This allows me to control the current nicely. Since
MAX
> > voltage for each battery charged should be < 1.8
volts.
> > This seems to be OK, 1.8 * 4 = 8.2 Volts.
>
> This implies you are putting the 4 separate AA
batteries in series.  Bad
> idea.  I wouldn't do this unless the cells came
manufactured in one
> indivisible pack.  Different cells will be at different
state of charge.
> This will make it difficult to charge the lowest cell
fully without
> overcharging the fullest one.  Doing a discharge cycle
before charging can
> be a good idea with NiMH batteries, but is out of the
question with 4
> separate cells in series.  One cell could get
discharged too deep or, worse
> yet, have its polarity reversed.  NiMH cells are easily
killed by
> overdischarge.

OK, should have been 1.8V * 4 = 7.2v.
I want to bury the 4 pack of AA batteries in the robot,
and not have to dig it out to charge it. So, I thought
I would make a charger designed for 4AA batteries.

> > My question is
> > can I use a 12V or 16V wall wart, as long as I charge
> > with proper current does the Voltage matter at all ?
>
> Only the voltage at the cells matter.  The voltage at
the other end of your
> buck converter only matters to the buck converter.
Since it is a buck
> converter, higher voltage should be fine.  It will just
draw less current
> from the higher voltage.

Good, thanks.

>
> > Next question, when the 4AA pack is connected, I plan
on
> > discharging the pack to 4V, before starting the charge
> > cycle.
>
> You shouldn't do this with separate cells that are not
in a pack.  See
> above.
>
> > I plan on discharging by using the circuit below.
>
> The discharge current will only be about 160mA given an
8.2V pack.  That
> will definitely discharge those batteries, but typical
AA NiMH cells can
> stand significantly more than that without damage.
>

OK, I will use a smaller resistor, with higher power
rating.

> > Question when the +Batt drops lower than the 5V PIN,
> > the collector will be less positive than the emitter,
>
TYPO, I meant to say Collector will be at 4V according
to discharge routine. Base will be at 5V from PIC
pin. I realize Emitter will be at GND 0V,
so my question about collector and base still stands.
I thought I read collector has to be more positive than
base and base has to be more positive than emmitter,
or transistor would be off. Apparently I am mistaken.

> No, it won't.  The emitter is at 0V by definition
(that's what GND means).
> Your circuit has no means of generating a negative
voltage, so nothing can
> be at a lower potential than the emitter (GND).
>
> The collector should be at about 200-300mV because the
transistor should be
{Quote hidden}

separate AA cell.
>

Since power is drawn from the pack, I assumed the
batteries would drain somewhat equally. Since I only plan
on draining to 4v and batteries are from same
manufacturer I thought this would be ok.

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2003\10\25@160818 by Eric Christensen

picon face
On Sat, 25 Oct 2003 15:38:36 -0400
kbenspamKILLspamDCA.NET wrote:
>
> Since power is drawn from the pack, I assumed the
> batteries would drain somewhat equally. Since I only plan
> on draining to 4v and batteries are from same
> manufacturer I thought this would be ok.
>

It might be okay for a little while, but the batteries will eventually
get out of balance.

On the ISU solar car project, we had 26 batteries in series.  We
monitored the voltage and temperature of each battery and controlled our
charging/discharging accordingly.  If the batteries became too out of
balance, we would charge/burn off cells to bring the pack back into
balance.

If the batteries are out of balance, you can only charge the entire pack
until the 'fullest' battery is full and you can only discharge until the
'emptiest' battery is empty.  This can eat into the usable battery power
very quickly.

This is also a common problem with laptop batteries.  The batteries
don't necessarily wear out so much as they get horribly out of balance
and can't be used.

With only 4 batteries, it wouldn't be too bad to build a system to
monitor and charge each battery individually.  Or maybe just warn you so
that you can pull the battery and balance it every so often.

Eric

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2003\10\25@201742 by Larry Bradley

flavicon
face
>I'm no expert, but from what I've read, NiMH cells don't suffer from the
>same symptoms as NiCad's. There does not appear to be the same need to
>discharge them before charging that is sometimes recommended for NiCads.


But in either case, you have to be VERY careful when "fully" discharging,
since if one of the cells happens to be in somewhat worse shape than the
others, you can actually end up reversing the voltage on the cell, which is
game over. As Olin said, discharge cells individually, or at least monitor
the voltage across each cell while discharging, and quit if the voltage on
any of the cells gets too low.


Larry Bradley
Orleans (Ottawa), Ontario, CANADA

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2003\10\25@205552 by Olin Lathrop

face picon face
kben@DCA.NET wrote:
>>> Question when the +Batt drops lower than the 5V PIN,
>>> the collector will be less positive than the emitter,
>>
> TYPO, I meant to say Collector will be at 4V according
> to discharge routine.

No.  The transistor will be (at least should be) saturated, so the
collector will be around 200-300mV.  The other side of the collector
resistor may be 4V when the pack is low, but the transistor only "sees"
the base, emitter, and collector voltages.

> Base will be at 5V from PIC pin.

No again.  The other side of the base resistor from the transistor may be
at 5V, but the transistor base will be one diode drop above ground, about
700mV.

> I thought I read collector has to be more positive than
> base and base has to be more positive than emmitter,
> or transistor would be off. Apparently I am mistaken.

A saturated transistor's E-C voltage is typically 200mV, while the E-B
voltage of an "on" transistor is one diode drop, around 700mV.  When the
transistor is on but not saturated, the collector can be much higher.  In
any case, it's the base current that decides whether the transistor is
"on", not the collector current or voltage.



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2003\10\25@211045 by Russell McMahon

face
flavicon
face
> > I thought I read collector has to be more positive than
> > base and base has to be more positive than emmitter,
> > or transistor would be off. Apparently I am mistaken.
>
> A saturated transistor's E-C voltage is typically 200mV, while the E-B
> voltage of an "on" transistor is one diode drop, around 700mV.  When the
> transistor is on but not saturated, the collector can be much higher.  In
> any case, it's the base current that decides whether the transistor is
> "on", not the collector current or voltage.

Olin is, of course, correct. But a point of clarification.
When the transistor is on, the collector can be LOWER than the base voltage
under the on conditions mentioned above.
This can be deduced from the above but is somewhat unintuitive and some
people do not realise that this can happen.

As Olin says, when the base is driven on it rises to around 0.6 to 0.7 volts
relative to emitter.
The collector can go essentially to emitter voltage plus saturation voltage.
If the saturation voltage is less than Vbe then collector voltage will be
below base voltage.


       RM

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2003\10\26@091624 by kben

picon face
Thanks to all who replied, that clears up my confusion
and I have what I need to proceed.

Thanks again,
          Kevin

Russell McMahon <.....apptechKILLspamspam.....PARADISE.NET.NZ> said:

> > > I thought I read collector has to be more positive
than
> > > base and base has to be more positive than emmitter,
> > > or transistor would be off. Apparently I am
mistaken.
> >
> > A saturated transistor's E-C voltage is typically
200mV, while the E-B
> > voltage of an "on" transistor is one diode drop,
around 700mV.  When the
> > transistor is on but not saturated, the collector can
be much higher.  In
> > any case, it's the base current that decides whether
the transistor is
> > "on", not the collector current or voltage.
>
> Olin is, of course, correct. But a point of
clarification.
> When the transistor is on, the collector can be LOWER
than the base voltage
> under the on conditions mentioned above.
> This can be deduced from the above but is somewhat
unintuitive and some
> people do not realise that this can happen.
>
> As Olin says, when the base is driven on it rises to
around 0.6 to 0.7 volts
> relative to emitter.
> The collector can go essentially to emitter voltage
plus saturation voltage.
> If the saturation voltage is less than Vbe then
collector voltage will be
> below base voltage.
>
>
>         RM
>

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