Will a dimmer switch work instead of an incandescent light bulb in asimple battery charger?

[Martin Brown]

Kanon Kubose formulated the question :

Application of a sledgehammer first to the device itself and then to the
head of anyone dumb enough to build it would seem the best solution.

Why don't you drop this stupid idea and get a real battery charger? :-?

Or an old scrapbox laptop PSU and a chunky power resistor.

Dirt cheap lead acid battery chargers can be bought off the shelf for
not much more than the cost of building this death trap.

 

[[email protected]]

I was thinking about this more, if there was a way to make it safer. Probably the breaker, being 20 amp or so, wouldn't help, right, given that 25 mA is lethal? What about a GFI, though?

The whole idea of this " battery charger " is low cost. Spending any moneyto make it safer makes no sense. If you want to make it safer the thing to do is to add an isolation step down transformer. To do that cheap get a free microwave oven and remove the secondary winding. Then rewind a secondary on it. Do a search on the internet on microwave oven transformers for how to do that.

If you use common sense and make the connections to the battery without anypower applied, the thing is not dangerous. But if there is any possibility of children getting access to it, then you ought to have it and the battery in a box with a lock on it.

Dan

 

[George Herold]

I was thinking about this more, if there was a way to make it safer. Probably the breaker, being 20 amp or so, wouldn't help, right, given that 25 mA is lethal? What about a GFI, though?

So we all agree the $3 charger is no good. But hidden in this thread is another question. Can the OP take his 2/4/6 amp Schummacher(sp) charger and reduce the minimum 2 amps to maybe 1 amp or 0.5 amp. (I think that's what he wants anyway.)
So a current splitter.
Like maybe a diode and resistor... maybe a zener diode?
(Hmm maybe easier to get the Schumacher schematic and hack it for 1 amp?)

George H.

 

[Kanon Kubose]

Yes, if I could do that, I'd be a happy camper. What's a good circuit to use?

Kanon

 

[ehsjr]

On Monday, July 8, 2013 1:00:25 PM UTC-7, George Herold wrote: ....
Can the OP take his 2/4/6 amp Schummacher(sp) charger and reduce
the minimum 2 amps to maybe 1 amp or 0.5 amp.

....

Yes, if I could do that, I'd be a happy camper. What's a good circuit to use?

Kanon

View in fixed font:

--------------- -----
| +|------Vin|LM317|Vout---+
| Schumacher | ----- |
| | Adj [R1]
| | | | D1
| | +----------+--->|---+
| | |
| | [Battery]
| | |
| -|--------------------------------+
---------------

An LM317 can handle up to 1.5 amps when mounted on a heatsink.
Current is set by R1 using the formula R = 1.25/I where I is
the desired current. So for one amp use a 1.25 ohm resistance,
for 1/2 amp use a 3 ohm resistor and so forth. The voltage
from the Schumacher needs to be a minimum of about 3 volts higher
than the battery rating (eg 15 volts to charge a 12 volt battery).
Assuming 15 volts from the Schumacher, you can charge lead acid,
NiCd and NiMh batteries rated from 1.5 volts up to 12 volts. This
is a current control circuit, so there is no need to set the
voltage to the battery rating.

Diode D1 (1N5404) prevents current flowing backwards from
the battery through the LM317.

Parts (below) at http://www.allelectronics.com
heatsink cat # HS-7019 $0.50
..22 ohm 3 watt resistor CAT# .22-3 3 for $1.00
..5 ohm 5 watt resistor CAT# .5-5 3 for $1.00
1N5404 diode CAT# 1N5404 3 for $0.80
LM317 CAT# LM317T $0.70

Ed

 

[Martin Brown]

So we all agree the $3 charger is no good. But hidden in this thread is another question. Can the OP take his 2/4/6 amp Schummacher(sp) charger and reduce the minimum 2 amps to maybe 1 amp or 0.5 amp. (I think that's what he wants anyway.)

He managed to disguise that requirement very well!

So a current splitter.
Like maybe a diode and resistor... maybe a zener diode?
(Hmm maybe easier to get the Schumacher schematic and hack it for 1 amp?)

Assuming it is a cheap and cheerful charger and not some clever constant
current into any load device then simply adding a small series resistor
of an appropriate wattage to one lead ought to do the job.

Try 1 ohm 2W in series as a rough empirical starting guess and then
adjust up or down to obtain the intended lower charge rate(s).

 

[Jasen Betts]

So we all agree the $3 charger is no good. But hidden in this
thread is another question. Can the OP take his 2/4/6 amp
Schummacher(sp) charger and reduce the minimum 2 amps to maybe 1 amp
or 0.5 amp. (I think that's what he wants anyway.)

add a series resistor of 1 ohm or so, a 15W 12V brake lamp is probably
about right.

 

[Kanon Kubose]

Thanks, it worked! I had to connect a 12V ceramic heater and a headlight inseries with the Schumacher battery charger to get the current down to 600 mA. With either one by itself, it was still too high, and with a 12V fan only, current got so low (around 200 mA) that the charger shut off thinking it was done.

I don't understand the LM317 circuit. Isn't the LM317 a voltage regulator, not a current limiter? The charger already limits current in stage one, then limits voltage in stage two.

Kanon

 

[[email protected]]

I don't understand the LM317 circuit. Isn't the LM317 a voltage regulator, not a current limiter? The charger already limits current in stage one, then limits voltage in stage two.



Kanon

The LM 317 is a voltage regulator , but can be used to limit current. What it does depends on how it is used.

Dan

 

[ehsjr]

Thanks, it worked! I had to connect a 12V ceramic heater and a headlight in series with the Schumacher battery charger to get the current down to 600 mA. With either one by itself, it was still too high, and with a 12V fan only, current got so low (around 200 mA) that the charger shut off thinking it was done.

I don't understand the LM317 circuit. Isn't the LM317 a voltage regulator, not a current limiter? The charger already limits current in stage one, then limits voltage in stage two.

Kanon

The LM317 regulates the voltage at the Vout pin at ~1.25
volts higher than the voltage on the Adj pin. If you set
the voltage at the Adj pin to some specific level, Vout
will be 1.25 volts higher than that.

View in fixed width font:

LM317 Voltage Regulator

Vin--[Lm317]--Vout-+----+
Adj | |
| [R1] |
| | |
+----------+ [Load]
| |
[R2] |
| |
Gnd----------------+----+

Note that in the above voltage regulator circuit, the voltage
at the Adj pin is set by voltage divider R1 and R2. Current
drawn by the load does not flow through those resistors, so
the voltage at the Adj pin is not affected by load current.
Now look at the current regulator configuration:

LM317 Current Regulator

Vin--[Lm317]--Vout-+
Adj |
| [R1]
| |
+----------+
|
[Load]
|
Gnd----------------+


The LM317 works the same as it did before - it makes the
Vout voltage 1.25 volts higher than the Adj pin. But
this time the voltage divider that sets that Adj voltage
is R1 and the _load_, not R1 and R2. R2 is gone. Note also
that this time current drawn by the load is drawn through
R1. Thus the voltage at the Adj pin depends on the current
drawn by the load. Since the LM317 holds the voltage between
between Vout and Adj to 1.25, the voltage drop across R1 is
1.25. The current is found by ohms law: I = V/R which in this
case is I = 1.25/R1 Notice that you don't need to consider the
load resistance in that equation.

See the LM317 datasheet for specs and examples.

Ed

 

[Kanon Kubose]

Thanks for the great explanation, Ed! One question, though: How is this circuit better than merely sticking different loads in series with the batterybeing charged to adjust current? It seems somewhat the same, just that there's an LM317 present. What does the LM317 do that the load by itself wouldn't?

Kanon

 

[rickman]

Thanks for the great explanation, Ed! One question, though: How is this circuit better than merely sticking different loads in series with the battery being charged to adjust current? It seems somewhat the same, just that there's an LM317 present. What does the LM317 do that the load by itself wouldn't?

Kanon

The LM317 *regulates* the current to the set value until the battery
voltage rises to a level that prevents the regulator from working.

Using a resistor will result in different current with different
batteries or with different states of discharge (different battery
voltages).

 

[ehsjr]

Thanks for the great explanation, Ed! One question, though:
How is this circuit better than merely sticking different loads
in series with the battery being charged to adjust current?
It seems somewhat the same, just that there's an LM317 present.
What does the LM317 do that the load by itself wouldn't?

Kanon

The LM317 circuit provides constant current while using different
"loads" (actually current limiting resistances) in series provides
a varying current to the battery being charged.

When using the LM317 circuit to charge the battery,
it looks like this:

Vin--[Lm317]--Vout-+
Adj |
| [R1]
| |
+----------+
|
[Battery]
|
Gnd----------------+

The battery _is_ the load. As the battery charges, the voltage
across the battery increases. However, the current remains the
same - it is whatever value R1 is divided into 1.25, regardless
of the battery voltage. Let's say R1 is 10 ohms, and the battery
voltage is 10.5 when you start charging. The current will be
1.25/10 or .125 amps. Now at the end of charge say the battery is
charged to 13.8 volts. The current will _still_ be .125 amps.
The LM317 circuit acts as a constant current source.

With the alternative you mention (using only a resistor as the
current limiter), the circuit looks like this:

Vin---[R1]---+
|
[Battery]
|
Gnd----------+

The current depends on two variables, Vin and the battery voltage.
The formula is I = (Vin - Vbattery)/R1. But the vbattery voltage
is continually increasing, meaning the current is continually
decreasing. If Vin is 15 volts and Vbattery is 10.5 at the start
of charge, and R1 is the same 10 ohms we used in the constant
current circuit, the initial charge current will be (15 - 10.5)/10
or .450 amps. When the battery voltage reaches 13.8 at the end of
charge, the current will be (15 - 13.8)/10 or .120 amps. That is
not a constant current - the LM317 circuit is.

You can choose whatever approach you want. With the LM317 approach,
you can set the exact current you want mathematically. With the
approach of putting various different things into the circuit to
act as limiting resistors, it's hit or miss. It's also a kluge
of jumpers and devices, whereas the LM317 is straightforward.

If you use the LM317 circuit I would recommend you add a diode
rated at 3 amps, like a 1N5404:

Vin--[Lm317]--Vout-+
Adj |
| [R1]
| | D1
+----------+--->|---+
|
[Battery]
|
Gnd-------------------------+

The diode prevents the battery from discharging through the
LM317 in the event that Vin is not present (eg you turn off
the Schumacher without disconnecting the battery.) The LM317
will not be "happy" with current going through it backwards,
and will let the magic smoke out.

Also, pay attention to the wattage rating of R1. The minimum
wattage is found by I^2*R, and it's best to use higher
wattage. So, for example, say R1 is 1 ohm to give you 1.25
amps charging current. That resistor will need to dissipate
1.5625 watts - I'd use a 5 watt resistor.

Ed