15A current limit?

[James W]

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts.. Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.

Any help would be appreciated.

- jim

p.s. When the engine is off, I intend to seperate the auxiliary battery
from the rest of the system with a relay.

 

[Thomas C. Sefranek]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.

How can yo know this?

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

Are you confusing house wiring codes for current with this DC application?

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.

7 volts?

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts..

NO, the limiting device will need to dissipate 12 volts at 15 amps
until the voltage of the battery being charged rises.

Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.

FETs have no advantage over bipolars in "typical voltage drop".
With 2 power bipolars, some resistors, and a heat sink this can be done.

Any help would be appreciated.

- jim

p.s. When the engine is off, I intend to seperate the auxiliary battery
from the rest of the system with a relay.

--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

 

[CWatters]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit.

Are you trying to charge the aux battery from the main battery or connect
both batteries to the one charger? Both are possible but require different
circuits.

The normal reason for connecting two batteries in parallel like this is
"jump starting" a car that has a flat battery (jump start = UK English
slang). In that case you need wires that can handle much more than 15A and
more like 150A.

Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.

If you are planning on using croc clips to connect the aux battery then they
WILL be connected together one day (=15V) or if you connect a duff battery
that has a short circuit it's also 15V.

If you connect the battery the wrong way around it could be 15+12=27V or if
you have the two batteries connected and the aux is reversed it's 24V.

If the aux battery is in a car and you crank the engine it's 15V with some
noise spikes.

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts.

No it's more like 225W (or 405W if battery reverse connected)

 

[James W]

Thomas C. Sefranek wrote:

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.


How can yo know this?

I know this because the aux battery will never be discharched below 11V

and the vehicles charging system will never be great that 15V, so 4V
potential difference

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.


Are you confusing house wiring codes for current with this DC application?

I used a table to automotive wiring current capacities.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.


7 volts?
SEVEN? why 7?

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts..


NO, the limiting device will need to dissipate 12 volts at 15 amps
until the voltage of the battery being charged rises.

The device will have 4V potential across it... not 12 (Vce=4)

Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.


FETs have no advantage over bipolars in "typical voltage drop".
With 2 power bipolars, some resistors, and a heat sink this can be done.

I thought fets had ~0 ohms ON resistance, whereas bipolars suffer a
diode drop.

Any help would be appreciated.

- jim

p.s. When the engine is off, I intend to seperate the auxiliary battery
from the rest of the system with a relay.

 

[James W]

CWatters wrote:

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...


This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit.



Are you trying to charge the aux battery from the main battery or connect
both batteries to the one charger? Both are possible but require different
circuits.

This is in a vehicle. The only time the batteries will be connected

together is when engine is running and alternator is charging. When
engine is running, I want to limit current to 15A to aux batt.

The normal reason for connecting two batteries in parallel like this is
"jump starting" a car that has a flat battery (jump start = UK English
slang). In that case you need wires that can handle much more than 15A and
more like 150A.

NO, I will NOT be jump starting using this aux battery.

Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.


If you are planning on using croc clips to connect the aux battery then they
WILL be connected together one day (=15V) or if you connect a duff battery
that has a short circuit it's also 15V.

The whole thing will be protected with a 20A fuse, just in case
something goes awry.

If you connect the battery the wrong way around it could be 15+12=27V or if
you have the two batteries connected and the aux is reversed it's 24V.

If I drive the vehicle into a tree at 70mph, bad things will happen
to... but since I don't plan on doing either, and theres a 20A fuse in
line, I feel pretty safe.

If the aux battery is in a car and you crank the engine it's 15V with some
noise spikes.

In addition to the 20A fuse, the whole circuit will be sitting behind an
NO-relay, that will only be closed when ignition is ON ( open while off
and while starting)

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts.


No it's more like 225W (or 405W if battery reverse connected)

P=IV?? 4V*15A=60W, no?

 

[CWatters]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvr9h9i395i1ce@corp.supernews.com...

No it's more like 225W (or 405W if battery reverse connected)

P=IV?? 4V*15A=60W, no?

Yes that's right. I guess we thought you should allow for the case when the
aux battery is completly flat or sort circuit, in which case it's....

15 x 15 = 225W

I was also allowing for reverse connection (27V)

27 x 15 = 405W

Now that you have explained the situation further I guess you safely assume
60W.

 

[CWatters]

Try a variation/cut down version of this design...

http://www.mitedu.freeserve.co.uk/Circuits/Power/13v820a.htm

 

[James W]

Yes, it's always difficult to decide what to put in the original post.
Two much detail, and nobody will read it.. To little, and everyone makes
assumptions..

Any.. I still don't know how to build the current limit device... Any ideas?


CWatters wrote:

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvr9h9i395i1ce@corp.supernews.com...


No it's more like 225W (or 405W if battery reverse connected)


P=IV?? 4V*15A=60W, no?


Yes that's right. I guess we thought you should allow for the case when the
aux battery is completly flat or sort circuit, in which case it's....

15 x 15 = 225W

I was also allowing for reverse connection (27V)

27 x 15 = 405W

Now that you have explained the situation further I guess you safely assume
60W.

 

[CWatters]

Also this app note...
http://www.vishay.com/document/70596/70596.pdf


"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:SvkLb.3765$__7.194695@phobos.telenet-ops.be...

Try a variation/cut down version of this design...

http://www.mitedu.freeserve.co.uk/Circuits/Power/13v820a.htm

 

[CWatters]

"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:8OkLb.3778$Zn1.356411@phobos.telenet-ops.be...

Also this app note...
http://www.vishay.com/document/70596/70596.pdf

Actually that's for much lower currents but it might give you the idea.

 

[Thomas C. Sefranek]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvr95hd0fl7sb1@corp.supernews.com...

Thomas C. Sefranek wrote:
"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than
5V.


How can yo know this?

I know this because the aux battery will never be discharched below 11V
and the vehicles charging system will never be great that 15V, so 4V
potential difference

O.K.... If you say so.

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.


Are you confusing house wiring codes for current with this DC
application?

I used a table to automotive wiring current capacities.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.


7 volts?
SEVEN? why 7?

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts..


NO, the limiting device will need to dissipate 12 volts at 15 amps
until the voltage of the battery being charged rises.

The device will have 4V potential across it... not 12 (Vce=4)

Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.


FETs have no advantage over bipolars in "typical voltage drop".
With 2 power bipolars, some resistors, and a heat sink this can be done.

I thought fets had ~0 ohms ON resistance, whereas bipolars suffer a
diode drop.

I WISH FETs had 0 Ohms... They are now appearing with low on resistances
in the milliohms. I have measured saturation voltages in the MICROvolts in
small signal bipolars.
Yes, MICROvolts. Anyway, power Bipolars can get down to .3 volts drop in
saturation.
If the FET had a 1 ohm saturation that would be 15 VOLTS at 15 amps.
But... they are better these days.

In any case, why do you care? Even with a bipolar burden of 1 volt, you
still
have to drop your 4 volts somewhere. So what burden voltage will you allow
the
circuit to use with your FET solution? ?The 12 to 11 volt path, leaving 1
volt for the circuit
regulating 15 amps? It can be done, but it involves a current sense
amplifier.

Any help would be appreciated.

- jim

--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

 

[James W]

The reason I care about the bipolar diode drop.. is that I'm charging a
battery, and that last 0.6 of a volt is VERY important.

Thomas C. Sefranek wrote:

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvr95hd0fl7sb1@corp.supernews.com...


Thomas C. Sefranek wrote:

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...


I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than

5V.


How can yo know this?


I know this because the aux battery will never be discharched below 11V
and the vehicles charging system will never be great that 15V, so 4V
potential difference


O.K.... If you say so.

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.


Are you confusing house wiring codes for current with this DC

application?

I used a table to automotive wiring current capacities.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.


7 volts?

SEVEN? why 7?

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts..


NO, the limiting device will need to dissipate 12 volts at 15 amps
until the voltage of the battery being charged rises.


The device will have 4V potential across it... not 12 (Vce=4)

Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.


FETs have no advantage over bipolars in "typical voltage drop".
With 2 power bipolars, some resistors, and a heat sink this can be done.

I thought fets had ~0 ohms ON resistance, whereas bipolars suffer a
diode drop.


I WISH FETs had 0 Ohms... They are now appearing with low on resistances
in the milliohms. I have measured saturation voltages in the MICROvolts in
small signal bipolars.
Yes, MICROvolts. Anyway, power Bipolars can get down to .3 volts drop in
saturation.
If the FET had a 1 ohm saturation that would be 15 VOLTS at 15 amps.
But... they are better these days.

In any case, why do you care? Even with a bipolar burden of 1 volt, you
still
have to drop your 4 volts somewhere. So what burden voltage will you allow
the
circuit to use with your FET solution? ?The 12 to 11 volt path, leaving 1
volt for the circuit
regulating 15 amps? It can be done, but it involves a current sense
amplifier.



Any help would be appreciated.

- jim

 

[Thomas C. Sefranek]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvsbnb3u1ke5ca@corp.supernews.com...

The reason I care about the bipolar diode drop.. is that I'm charging a
battery, and that last 0.6 of a volt is VERY important.

That's NOT a reason, it's a belief.
Why do you believe the .6 volts is VERY important?
Why do you believe .6 volts? (Most power bipolars can saturate at .3 volts
or less!)

If you want to use an FET, you will encounter the 3-5 VOLT gate drive
requirement.
An N-Channel low side driver would make sense, as you have 12 volts to work
with.
Conversely, a P-Channel high side driver.

SO again, you need to sense the current in order to limit the current.
(You have not said WHY you want to limit the current...)
Resistively, it makes sense to keep the burden low, as it is voltage loss
similar to your belief
about Bipolars.
Immagine a resistor in series which will give you 1 millivolt per amp.
You need an op-amp to amplify that low voltage up to the gate drive voltage
range.

But, using either an FET or a bipolar transistor you will encounter a
voltage drop in the device.
(A power FET could have a .05 ohm RDSon) [many power FETS have higher
RDSons!)
that times 15 amps still gets you back to bipolar voltage drops. .75
volts!

It sounds like your trying to design a battery charging a battery,
(A 12 volt battery charging a 12 volt battery is poor!)
this is BEST done with a DC-DC converter.

If they really are both 12 volt batteries, use a relay and forget the
current limit.

E-Mail me if you would like more details on building your solid state low
dropout FET relay.

Tom
--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

 

[Thinker]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

What made you decide on a 15 amp limit? 19ga wire will carry 30
amps without overheating and considerably more for a short time (such
as at the beginning of the charging cycle. I don't really think you will
need any elaborate current control unless your auxilary battery is very
large.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts.. Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.

Any help would be appreciated.

- jim

p.s. When the engine is off, I intend to seperate the auxiliary battery
from the rest of the system with a relay.

 

[James W]

Where can I find data in the current carrying capacity of wiring for
automotice applications. Note, this wire will be running in body panel
channels, wiring harnesses, etc... so 'free air' current capacity
numbers will not be applicable IMHO.

- jim

p.s. The 15 Amp limit is the charge rate at which I want to recharge the
batteries, per manufacturers recommendations.

Thinker wrote:

What made you decide on a 15 amp limit? 19ga wire will carry 30
amps without overheating and considerably more for a short time (such
as at the beginning of the charging cycle. I don't really think you will
need any elaborate current control unless your auxilary battery is very
large.

 

[Bill Bowden]

James W <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message news:<10036eaecnsk3e7@corp.supernews.com>...

Where can I find data in the current carrying capacity of wiring for
automotice applications. Note, this wire will be running in body panel
channels, wiring harnesses, etc... so 'free air' current capacity
numbers will not be applicable IMHO.

- jim

p.s. The 15 Amp limit is the charge rate at which I want to recharge the
batteries, per manufacturers recommendations.

Thinker wrote:

What made you decide on a 15 amp limit? 19ga wire will carry 30
amps without overheating and considerably more for a short time (such
as at the beginning of the charging cycle. I don't really think you will
need any elaborate current control unless your auxilary battery is very
large.

18ga copper wire is 5.5 milliohms per foot, so you would
lose about 0.8 volts at 15 amps on a 10 foot run, and
the wire would dissipate 12 watts which is a little high.
If you go to 12ga, the resistance is only 1.6 milliohms
per foot and voltage loss is .24 volts over 10 feet, and
the heat generated is 3.6 watts which is probably nothing
to worry about. If you don't want to lose the .24 volts,
go to 10 ga wire which is only 1 milliohm per foot.

-Bill

 

[Robert C Monsen]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.

It sounds like you are trying to build a 15A max trickle charger.

Your maximum voltage drop is 4V, and you want to limit the current to 15A.

Try this:

http://home.comcast.net/~rcmonsen/misc/charge-limit.GIF

You tune it for the particular current you want by tuning the voltage out of
the trimmer to be the same as the voltage across the current sense resistor
R1. For this design, at 15A, thats 150mV above ground. Thats a 6.2V zener, I
think. Using a different zener means different resistor values in that area.

The p-mosfet shown won't work, because it will die given the current (its
model was handy in my drafting program.) You'll have to find a power mosfet
that will support the power and current ratings you need. Also, the current
sense resistor, R1, should be at least 5W rating. You can find current sense
resistors at mouser or digikey that have this kind of rating.

The current rating for the mosfet you need is something like 25A (since you
are using a 20A fuse.) Power rating = 25A x max voltage differential between
batteries. If your battery has a short, that would be 15A x 15V = 225W, so
watch out; even with current limiting, you can still fry the mosfet and set
your RV on fire.

You also will need a big heatsink.

Regards,
Bob Monsen

 

[Nick Hull]

In article <10036eaecnsk3e7@corp.supernews.com>,
James W <change-to-jw22-at-trlp-dot-com@trlp.com> wrote:

p.s. The 15 Amp limit is the charge rate at which I want to recharge the
batteries, per manufacturers recommendations.

A crude & simple method is to run the current thru a light bulb. A
headlight takes about 4 amps, so 4 in parallel will limit the curent to
16 amps at zero batt voltage. The voltage drop will get very low at
trickle currents.

--
free men own guns - slaves don't
www.geocities.com/CapitolHill/5357/

 

[Ross Mac]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:vvpo81eef7pde8@corp.supernews.com...

I need to build a circuit that can limit current to ~15A. This will be
used in an automotive setting. Maximum voltage drop will be less than 5V.

This is essentially a DC battery charger operation. I want to be able to
charge an auxiliary battery that is located quite some distance
(~20feet) from the main battery and charging circuit. Other
considerations limit wire size to 10Ga, hence the requirement for
current control.

The auxiliary battery will (obviously) never be drawn down to lower than
~11V, so we'll have less than a 4V differential between the charging
circuit and the auxiliary battery.

Assuming for the moment, that the auxiliary batteries series resistance
in ZERO, then the current limiting device/circuit will be dissipating
~60watts.. Further, since I want the auxiliary battery to eventually
become fully charged, I need to avoid the typical voltage drop of a
bipolar transistor.. Hence, I'm thinking FET.

Any help would be appreciated.

- jim

p.s. When the engine is off, I intend to seperate the auxiliary battery
from the rest of the system with a relay.

Usually if you are going to charge a second battery in an RV or towed
vehicle you would want a diode isolator such as this one...
http://www.hellroaring.com/nodiode.htm
This way you won't need a solenoid and there would be no voltage drop from
the diode isolator.....There are a bunch of similar solutions if you do a
google search.....
Hope this helps.........My 2 cents on the subject....Ross