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

Ok, thanks, guys. Good stuff to know. If this thread helps prevent even one accident from someone trying to implement that $3 battery charger, it'll be worth it.

Yes and very likely dead soon afterwards. As will be the person who tries to rescue him without first switching off the lethal contraption.

The 60W light bulb limits the *current* on 120v to about 500mA. I am not absolutely sure what it takes to kill a human but 10mA is enough to prevent you ever letting go and above 25mA can kill particularly if the current is flowing across your chest.

Until the light bulb heats up it will actually pass a *much* higher current than its nominal operating current and will not drop many volts doing so. More than enough to kill you. 120v US mains is slightly more forgiving than 240 in the UK but building this is utter madness.

I have had to administer first aid to US installation engineers that checked British 240v mains wiring for being live with a wet finger.

It would be hard to imagine a more effective death trap for the unwary!

Definitely you are Darwin award material.

--
Regards, 
Martin Brown

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?

"Kanon Kubose"

** No help as the user is handling both AC supply conductors - ie active and neutral.

... Phil

Kanon Kubose formulated the question :

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

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John G

Right. Power-mains circuit breakers are *not* there to prevent shocks. They're there to prevent fires caused by overloads.

GFI won't solve the basic problem. If you grab both connectors in separate hands, you can get a thoroughly lethal dose of current right through your heart, travelling through the hot-to-neutral pathway. No leakage to ground, no imbalance in the hot/neutral wires, hence a GFI won't detect a problem and won't trip.

As they say, an effort to polish a turd is not a good investment of time. This circuit simply should not be built. Doing it *right* (with a proper transformer and good isolation) isn't either difficult or expensive.

Ouch!

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.

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.

--
Regards, 
Martin Brown

ably the breaker, being 20 amp or so, wouldn't help, right, given that 25 m A is lethal? What about a GFI, though?

The whole idea of this " battery charger " is low cost. Spending any money to make it safer makes no sense. If you want to make it safer the thing t o 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 second ary on it. Do a search on the internet on microwave oven transformers for h ow to do that.

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

Dan

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.

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

Kan> >

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

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

He managed to disguise that requirement very well!

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).

--
Regards, 
Martin Brown

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

--
?? 100% natural 

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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 on ly, current got so low (around 200 mA) that the charger shut off thinking i t 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, the n 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

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

Thanks for the great explanation, Ed! One question, though: How is this cir cuit better than merely sticking different loads in series with the battery being charged to adjust current? It seems somewhat the same, just that the re's an LM317 present. What does the LM317 do that the load by itself would n'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).

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Rick