I have a 220VAC drop in charger for a Motorola GP68 two-way radio. The bottom of the charger indicates 220VAC input and 10VDC output at 100mA and
170mA in rapid charge mode. I would like to replace the transformer to use this charger on 120V mains.
Looking at the discrete 12 component circuit, it looks a fullwave rectifier circuit using a 220V CT step down transformer. The CT is tied to the gnd bus and both ends of the secondary winding are connected to a 1N4004 diode. There is a 5 resistor series-parallel network (22 / 82 / 33K / 220 / 10 Ohms) used to switch between normal and rapid charge modes using one side of a DPDT slide switch. Along with 2 LEDs: one for charge mode indication and the other for charging status. The negative DC charging lead is tied to gnd via a 1N4004. The positive DC charging lead is picked off at a point that is separated from the gnd via 33K resistor and a LED.
I've got a Radioshack power transformer (part # 2731365) with the following specs:
Input: 120VAC / 60 Hz
Output (load): 12VAC C.T. @ 450mA
Output (no load): 14VAC
Current (load): 80mA max
Current (no load): 45mA max
This 120VAC transformer's loaded current rating is only 80mA yet the loaded output rating indicates 12VAC @ 450mA. How do I interpret these ratings correctly? The chargers DC output current is 100mA/170mA Knowing nothing else about this circuit (and no schematics or further info online that I could find) and being a newbie, what is the best approach to tackling this project while ensuring a safe solution.
Hi, Mike. At a suggested retail of about $400 for a pair, I'm not sure exactly why you'd try doing it this way. I'd really recommend getting a small international step-up autotransformer, and just plug your existing wall wart in to the adapter to avoid the chance of letting the smoke out.
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Jameco P/N 99477CK , $8.95 USD
They do ship to Canada, but you might want to call to get shipping charges straight, or go with a local source.
Considering you don't have any further information on the charger, your admitted relative inexperience in electronics, and the replacement cost in case of a mistake, it might be better just to get an autotransformer. If you do decide to go with a local source, make sure it's an autotransformer and is rated for at least 50 watts or so. The cheapie international voltage adapters with just a diode or SCR will definitely smoke your adapter.
Thanks for your reply and suggestion regarding a step-up transformer. The thought did cross my mind but it seems too easy and doesn't advance my understanding of electronics much.
The GP68 was a $60 purchase off eBay so I am willing to take the risk which I believe is small nonetheless.
What I am after is someone who can provide some guidelines on how to determine an equivalent 120VAC transformer and which parameters to consider. Since the transformer itself is a passive device, and as I understand it, and correct me if I am wrong, the primary characteristics of this device is its step-down ratio between the primary and secondary coils which determines the voltage and current ratios and the winding's current handling capability dictated by the wire gauge of the windings. I recognize there are other parameters including intra and inter winding capacitance, the magnetic core and its associated permeability and resulting hysteresis, copper losses, etc. including the whole theory behind Maxwells equations etc however these parameters aren't spec'd in the manufacturers datasheet for such devices so I don't believe they are going to play a significant role in this limited application.
Part of this replacement is to make a little project out of it. Since I believe the current requirements are going to be determined by the parallel resistor networks on the secondary side along with parallel charging battery load, I don't see why this is a complicated project.
I should be able to experimentally determine some of the primary characteristics of the 220VAC transformer by applying an AC source (non-lethal voltage levels) to the primary winding and measure the unloaded output off the secondary to determine the step down ratio. I suspect I could also play it extra safe and put in 1A transformer which would yield at least a 5:1 safety margin in terms of the 100mA/170mA current requirements.
Feel free to shoot holes in my theory with plausible justifications. I would just really like to understand how to make this work and if it won't work, the *specific* technical reasons why it won't work.
Hi, Mike. Please bottom post (place your answers and questions underneath what you're responding to). It makes answering easier.
Congratulations on having the gumption to make a go of it, and turning your problem into a project. Good attitude.
If I were doing this, the first thing I'd look at is whether your transformer has a dual primary. The solution might be right there before your eyes. Since you've got the charger open, check to see if the primary side has some extra pins or wires like this (view in fixed font ot M$ Notepad):
In this configuration, each half of the primary winding has 120VAC across it. By jumpering the primary so the halves are in parallel, you can accomplish the same thing:
If you're that lucky, you're done. But probably not.
If you have to do it the long way, you'll have to find out some of the characteristics of the transformer, as you mentioned. Disconnect the transformer secondary from the circuit, plug it in, and measure the no-load AC voltage (Vnl). Then reattach the transformer secondary, and put an ammeter between the junction of the two 1N4004 diodes and the rest of the circuit. Turn it on, and simultaneously measure the transformer secondary full load AC voltage (Vfl) and the AC current I.
Now that you have this information, you can figure out the internal resistance of the transformer secondary. Internal resistance (Rint) is shorthand for a lot of the losses in the transformer under load, but it's a good first cut approximation. You can use Ohm's Law to determine this resistance. The difference in AC voltages is assumed to be the result of the current flowing through the internal resistance:
( Vnl - Vfl ) / I = Rint.
Now you'll want to find a transformer with approximately (within a couple of tenths of a volt) the same no load voltage at 120VAC that the original has at 240VAC, and that also has approximately (within 10%) the same internal resistance. This is important, because you want the transformer voltage to be essentially the same under no load, half-load (100mA) and full load (170mA). If it's too low, it won't charge properly. If too high, you'll stress out the battery.
Well, I guess what this comes down to is that there is a way to do this, but I'd still recommend against it. The replacement transformer you pick will probably be an almost right compromise, and I guess replacement batteries are pretty expensive. Again, if it were my call, I'd say just go with the autotransformer adapter.
But it *is* your call. If you want to try it, you've got some information to start with.
What chris said about looking for the taps on the primary side is best. If there is none ( only 2 connections ) , and I'm not misreading this :). The unit normally runs on 220, then I would connect it to the 120 temp, and read the secondary voltage with no load. Then look for a new transformer that has near same current capability and double the output voltage as you just read from the old unit. Of course also with a center tap. My logic is if the 220 unit is running at half power, the output voltage would be half power etc. I hope that helps.
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