Re: DC Battery Charger

[Collin Kidder]

(hopefully this threads right. I'm posting from the forum interface and it didn't have my thread in it for some reason...)

Thanks for the suggestions. I thought about going that way. Maybe I still will use a dedicated 12V charger off of the AC. A potential problem is that this car is dual 110/220 on the charger so I cannot just hook up a 110 charger unless it auto adjusts for 220V. However, I don't think 12V chargers normally do that.

Instead, I still kind of want to do what I originally thought of. Here is a schematic I quick drew up. Please warn me if it wouldn't work for some reason.

[Lee Hart]

Collin Kidder wrote:

> (hopefully this threads right. I'm posting from the forum interface and it
> didn't have my thread in it for some reason...)
>
> Thanks for the suggestions. I thought about going that way. Maybe I still
> will use a dedicated 12V charger off of the AC. A potential problem is that
> this car is dual 110/220 on the charger so I cannot just hook up a 110
> charger unless it auto adjusts for 220V. However, I don't think 12V chargers
> normally do that.
>
> Instead, I still kind of want to do what I originally thought of. Here is a
> schematic I quick drew up. Please warn me if it wouldn't work for some
> reason.
>
> http://n4.nabble.com/file/n607107/Batt_scheme.png 

This will work, though it has some limitations.

  - Resistor R1 limits the current that the DC/DC can source, both for
    the battery and for the car's 12v system.

  - R1 and D1 cause the battery voltage to be poorly regulated even if
    the DC/DC output is well regulated.

  - You have two diodes carrying high currents. They'll get hot, and
    so will need heatsinking. They also cause two forward voltage drops,
    so you need (say) 15v from the DC/DC to get 14v on your 12v system.

Instead, I think I would use one diode and one power resistor. Wire the
DC/DC straight to your 12v accessory loads. Thus, they will get a
constant voltage (set by the DC/DC).

Wire the diode and resistor from the DC/DC output to the battery. Orient
the diode so when the DC/DC voltage is higher than the battery, the
resistor carries all the current. Pick the resistance to limit the peak
charging current that the DC/DC can supply. When the battery is fully
charged, it will draw essentially no current and so be "floating" at the
DC/DC's output voltage.

The diode has its anode to battery+, cathode to the DC/DC output and the
rest of the 12v system. If something draws more current than the DC/DC
can deliver, its output should go into current limit and its output
voltage falls. The battery now picks up the peak load via this diode.
The diode only carries current briefly during peaks, so the heatsink can
be much smaller.

--
Lee A. Hart | Ring the bells that still can ring
814 8th Ave N | Forget the perfect offering
Sartell MN 56377 | There is a crack in everything
leeahart earthlink.net | That's how the light gets in -- Leonard Cohen

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[Collin Kidder]

Oops! Yes, I had planned to directly hook up the DC/DC converter to load
but that was not shown in the schematic I posted. Adding that
essentially creates the situation you outlined and which I had actually
hoped to create. Thanks.

On 11/12/2009 12:15 AM, Lee Hart wrote:

>
> This will work, though it has some limitations.
>
>    - Resistor R1 limits the current that the DC/DC can source, both for
>      the battery and for the car's 12v system.
>
>    - R1 and D1 cause the battery voltage to be poorly regulated even if
>      the DC/DC output is well regulated.
>
>    - You have two diodes carrying high currents. They'll get hot, and
>      so will need heatsinking. They also cause two forward voltage drops,
>      so you need (say) 15v from the DC/DC to get 14v on your 12v system.
>
> Instead, I think I would use one diode and one power resistor. Wire the
> DC/DC straight to your 12v accessory loads. Thus, they will get a
> constant voltage (set by the DC/DC).
>
> Wire the diode and resistor from the DC/DC output to the battery. Orient
> the diode so when the DC/DC voltage is higher than the battery, the
> resistor carries all the current. Pick the resistance to limit the peak
> charging current that the DC/DC can supply. When the battery is fully
> charged, it will draw essentially no current and so be "floating" at the
> DC/DC's output voltage.
>
> The diode has its anode to battery+, cathode to the DC/DC output and the
> rest of the 12v system. If something draws more current than the DC/DC
> can deliver, its output should go into current limit and its output
> voltage falls. The battery now picks up the peak load via this diode.
> The diode only carries current briefly during peaks, so the heatsink can
> be much smaller.
>
>    

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[Dennis Miles-2]

On Wed, Nov 11, 2009 at 9:05 PM, Collin Kidder <[hidden email]> wrote:

>
> (hopefully this threads right. I'm posting from the forum interface and it
> didn't have my thread in it for some reason...)
>
> Thanks for the suggestions. I thought about going that way. Maybe I still
> will use a dedicated 12V charger off of the AC. A potential problem is that
> this car is dual 110/220 on the charger so I cannot just hook up a 110
> charger unless it auto adjusts for 220V. However, I don't think 12V
> chargers
> normally do that.
>
> Instead, I still kind of want to do what I originally thought of. Here is a
> schematic I quick drew up. Please warn me if it wouldn't work for some
> reason.
>
> http://n4.nabble.com/file/n607107/Batt_scheme.png
> --
> View this message in context:
> http://n4.nabble.com/Re-DC-Battery-Charger-tp607107p607107.html
> Sent from the Electric Vehicle Discussion List mailing list archive at
> Nabble.com.
>
> _______________________________________________
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> =======================revision successful ! ===================
>

Collin, with the added connection from DC/DC to Load The system will work.
You
will need to recalculate the Resistor value, with a DC/DC output at 13.8 V.
and a diode dropping 0.3 V. and a 12 V battery needing to be charged to 13.2
v. for full charge, the resistor will be only 0.075 ohms for 4 A. charge and
0.15 ohms for 2 A....
That is why I suggested a 12 V. charger, but, you will need to work out the
110/220
problem, there are many possible solutions but choosing the best is
dependent upon
 the particular connections you are making to the 110/220 ACV. sources.

Regards,

    Dennis Miles,    (Director / CEO)
Electric Vehicle Technical Institute Inc.
Tampa Bay Region, Florida, USA
       Phone (863) 289 - 0690
"Mentoring, Small Group Instruction,
and Hands-On training
for the EV Technicians:
Building, Converting, Repairing,
and Servicing the
Electric Vehicles of Today,
and the Future."
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[Collin Kidder]

On 11/12/2009 11:15 AM, Dennis Miles wrote:

>
>> Instead, I still kind of want to do what I originally thought of. Here is a
>> schematic I quick drew up. Please warn me if it wouldn't work for some
>> reason.
>>
>>
>>
>>      
> Collin, with the added connection from DC/DC to Load The system will work.
> You
> will need to recalculate the Resistor value, with a DC/DC output at 13.8 V.
> and a diode dropping 0.3 V. and a 12 V battery needing to be charged to 13.2
> v. for full charge, the resistor will be only 0.075 ohms for 4 A. charge and
> 0.15 ohms for 2 A....
> That is why I suggested a 12 V. charger, but, you will need to work out the
> 110/220
> problem, there are many possible solutions but choosing the best is
> dependent upon
>   the particular connections you are making to the 110/220 ACV. sources.
>
>    

Actually that's a good point. I realize that the amperage depends on
resistance and voltage differential. My previous numbers were not taking
the fact that the battery has voltage into consideration. I guess that I
can see that your numbers come from this: The battery is full at 13.2 so
if the output is 13.8 and there is a .3V drop then the output to the
battery is 13.5. That's only .3V over the full battery. And 0.3 / 0.075
= 4A. That's the proper way to look at it right?

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[Collin Kidder]

On 11/12/2009 12:53 PM, Collin Kidder wrote:

>
> Actually that's a good point. I realize that the amperage depends on
> resistance and voltage differential. My previous numbers were not taking
> the fact that the battery has voltage into consideration. I guess that I
> can see that your numbers come from this: The battery is full at 13.2 so
> if the output is 13.8 and there is a .3V drop then the output to the
> battery is 13.5. That's only .3V over the full battery. And 0.3 / 0.075
> = 4A. That's the proper way to look at it right?
>
>
>    

Replying to myself just to clarify this in case anyone else wants to try
it to make sure I'm not totally crazy.

The truth seems to be in between what I originally wanted to use (6 ohm
resistor) and what was suggested to me (0.075 ohm). In reality the goal
is to find a voltage on the lead acid cell which is to be considered
totally drained. I'll say that's 10.5V. Now, what I want to do is charge
at a certain amperage at total discharge (10.5V) and taper off as the
battery comes up. So if I set my desired charge to 8A then the
calculations are as follows:

13.5V output from DC/DC
.3V drop through a shottky diode
13.2V gets to the battery
10.5V is considered dead for my purposes
13.2 - 10.5 = 2.7V difference
R =  V/A so R = 2.7 / 8 = 0.338 ohms

Now, that will taper the charge from 8A down to zero as the battery
charges. There is a potential issue though. If the battery were allowed
to get REALLY dead (say, down to 4V) then 27A would pass through the
resistor. Chances are I wouldn't have a resistor rated for that sort of
wattage so I could kiss the resistor and/or the diode goodbye. However,
that's solvable (like a lot of chargers I could just refuse to charge at
that point) and all of this allows me to spend about $10 on parts
instead of having to buy another charger. Of course, it's not quite as
cool looking or fancy...

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[Evan Tuer]

On Thu, Nov 12, 2009 at 9:15 PM, Collin Kidder <[hidden email]> wrote:

> Now, that will taper the charge from 8A down to zero as the battery
> charges. There is a potential issue though. If the battery were allowed
> to get REALLY dead (say, down to 4V) then 27A would pass through the
> resistor. Chances are I wouldn't have a resistor rated for that sort of
> wattage so I could kiss the resistor and/or the diode goodbye.

You could add a fuse as well.
But I think you'd be as well just buying a suitable power supply with
output current limit, to be honest.

I have one similar to this:
http://www.vxipower.com/product_pdetail_12V-40A-OUTPUT---BCH-CHARGER_150.htm

That should be OK up to 400V (you might want to check with them first
of course).

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[Lee Hart]

Collin Kidder wrote:
> Replying to myself just to clarify this in case anyone else wants to try
> it to make sure I'm not totally crazy.
>
> The truth seems to be in between what I originally wanted to use (6 ohm
> resistor) and what was suggested to me (0.075 ohm).

The solution for this is to use a light bulb in place of the fixed
resistor. Light bulbs behave like a crude constant-current source. Their
cold resistance is about 1/10th of their hot (rated voltage) resistance.

So for example, you can use a car headlight as your resistor. It will
limit the battery charging current to about 4-5 amps, for any battery
voltage from 0v to 13v.

--
Lee A. Hart | Ring the bells that still can ring
814 8th Ave N | Forget the perfect offering
Sartell MN 56377 | There is a crack in everything
leeahart earthlink.net | That's how the light gets in -- Leonard Cohen

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