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Adapting 220VAC drop in charger for 110VAC

M

Michael

Jan 1, 1970
0
Hi All,

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.

Any and all info would be appreciated!

Thanks in advance,

Mike


I
 
C

Chris

Jan 1, 1970
0
Michael said:
Hi All,

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.

Any and all info would be appreciated!

Thanks in advance,

Mike


I

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.

https://www.jameco.com
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.

Good luck
Chris
 
M

Mike

Jan 1, 1970
0
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.

--Mike
 
C

Chris

Jan 1, 1970
0
Mike said:
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.

--Mike

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):

|
| L1
| o-------. ,------------>|-----o
| )|( |
| )|( |
| .--' '---. |
| | o--. |
| )--. ,---' | |
| )|( === |
| L2 )|( GND |
| o-------' '------------>|-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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:

|
| L1
| o----o--. ,------------>|-----o
| | )|( |
| | )|( |
| .-)--' '---. |
| | | o--. |
| | '--. ,---' | |
| | )|( === |
| L2 | )|( GND |
| o--o----' '------------>|-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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.

| _
| L1 / \ |-------------------.
| o----o--. ,------------>|-----o--( A )o+ |
| | )|( | \_/ | |
| | )|( | | |
| .-)--' '---. | | |
| | | o--o--------------------o- |
| | '--. ,---' | | | |
| | )|( === | | |
| L2 | )|( GND | | |
| o--o----' '------------>|-----' | |
| '-------------------'
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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.

Good luck
Chris
 
J

James Thompson

Jan 1, 1970
0
Chris said:
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):

|
| L1
| o-------. ,------------>|-----o
| )|( |
| )|( |
| .--' '---. |
| | o--. |
| )--. ,---' | |
| )|( === |
| L2 )|( GND |
| o-------' '------------>|-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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:

|
| L1
| o----o--. ,------------>|-----o
| | )|( |
| | )|( |
| .-)--' '---. |
| | | o--. |
| | '--. ,---' | |
| | )|( === |
| L2 | )|( GND |
| o--o----' '------------>|-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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.

| _
| L1 / \ |-------------------.
| o----o--. ,------------>|-----o--( A )o+ |
| | )|( | \_/ | |
| | )|( | | |
| .-)--' '---. | | |
| | | o--o--------------------o- |
| | '--. ,---' | | | |
| | )|( === | | |
| L2 | )|( GND | | |
| o--o----' '------------>|-----' | |
| '-------------------'
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

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.

Good luck
Chris
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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