I have a standard cheapo Radio Shack battery charger for 9 volt, D cell, C cell and AA batteries.
I opened it up when the On light started flickering out (just needed a new flashlight bulb).
I was surprised to find that there were only wires inside, no circuits.
All the power conversion was in the big block transformer that plugs into the wall.
Instead of the normal direct current output it says it has 8vac and 3vac outputs (vac=volts alternating current).
I have a digital test meter and it does actually read 8vac output, while the battery reads 9vdc.
So how does alternating current charge a direct current battery? Wouldn’t the polarity be wrong half the cycles?
And why 8 volts and not 9 volts? How can that charge it enough?
Does a DC meter really read zero volts? (Just the AC hum from a half-wave rectified waveform would read as AC on a meter. Gotta check the DC too!)
I’d put a meter in series with the battery being charged (use clip leads or something) to measure current and see if there’s a DC component. If there is, then start looking for the hidden diode, or use a scope to check whether the waveform is partly rectified.
carbon-zinc batteries can be recharged, but they will soon leak because they are clad with the zinc which gets thick and thin spots every time the metal goes into a salt and back to a metal.
But that doesn’t make any difference to the question.
And no, there is no diode. The only component in there is a single common resister near the bulb.
There has to be a diode. If there’s an LED present, they may be using that, or more likely, the diode is contained in the wall wart power supply. You cannot charge a battery with AC with no rectification.
What Q.E.D says, there’s a diode in there somewhere.
IIRC I have seen a circuit for recharging “non-rechargable” batteries which deliberately didn’t use full wave rectification. The suggestion was that the cells rechaged better if there wasn’t constant current.
Yes, but the “wrong” half-cycles are either discarded (half-wave) or inverted (full-wave) during the rectification.
As Q.E.D. says, there has to be some kind of rectification, usually a diode (or two). If the transformer unit says AC out, the rectifier is downstream from that. (A D.C. “transformer” has the rectifier inside the wall unit.) Your “single resistor” might be a diode or both in one package – I’m not familiar with current manufacturing practices.
As far as the voltages not seeming to match, we are talking oranges and apples here. Steady D.C. voltage is pretty straightforward, but powerline A.C. voltage, since it varies from zero to positive peak to zero to negative peak, is usually expressed as RMS (Root Mean Square, IIRC), which can be thought of as an average. Instantaneous voltage on the peaks is greater than the RMS value (it’s a sine wave curve).
Carbon-Zinc isn’t strictly rechargeable. You can, however, hook them up to a DC trickle and rejuvenate them to some extent.
Radio Shack has been selling Carbon-Zinc “rechargers” since I was a little kid back in 1976.
They don’t recharge all that well, but the worst that happens when one goes bad is that it leaks all over the place - rather than exploding or catching fire like the warnings on non-rechargeable alkalines warn you can happen. Rechargeable alkalines is a whole 'nother mess of fish, however.
Q.E.D., Small Clanger, Musicat -
You insist on a diode? Welcome to my problem. This is where I started, wondering how AC would charge a DC battery.
But, unfortunately, there is no diode I can see inside the box.
There is nothing hidden anywhere, as it is one of the simplest types of old fashioned slap-together Tandy construction.
And I seriously doubt that the wall connector is mislabeled, since when I tested with my meter I tested both at those wires and at the battery contacts themselves. There was no rectification between one and the other.
The meter at the point of battery contact read 8vac (for 9-volt) and 3vac (for a pair of other batteries).
But then I’ve been wrong before, so you can look for yourselves.
The charger’s picture is here (420k).
I have notated the incoming 3 wires and the position of the 9-volt bay on the other side.
The only things besides simple wires and simple crimp connectors are two things I would call resistors (circled), on ether side of the bulb on the ground line, seen in closeup here (27k)
However, it should be noted that even if those are not resistors they are not in the circuit for the 9-volt bay. That connects directly to the 8vac wire and the ground.
Herman_and_bill is (are?) right. I think those are diodes, not resistors. I can’t quite make out the circuit from your picture, since some wires look hidden.
As far as reading AC voltage where you expect DC, remember some important things. Unless you are using a 'scope, your meter is designed for accurately measuring only a simple sine wave. The voltage readout will not be accurate for all other circumstances and this is one of those. This is a very crude device (that’s not an indictment; it doesn’t need to be anything else to do the job) and no one cares if the D.C. has ripple in it. In fact, I can guarantee it has lots of ripple and your stupid meter will interpret that as A.C. voltage, and give a wrong value.
If you read D.C. volts, that won’t be too accurate, either, since you are not reading a steady direct current as you would from a battery source. A simple D.C. meter is accurate only under that condition.
A 'scope would be a very useful tool here. If you can get your hands on one, you can not only see the waveform, but measure the voltage. With a battery in place as a load, current should be flowing in only one direction, but you should see two positive “humps” multiplied by 60 per second (I assume you are in the U.S.; Canada uses 50); those are the positive half of the sine wave plus the negative half inverted, assuming this is wired as a full-wave rectifier. Due to the crudeness of the components, the waveform is probably distorted from a true sine wave, but it should still reveal its origin.
A Radio Shack voltmeter is only as good as the interpretation of it.
A power rectification circuit would never use Ge diodes, as their PIV ratings are typically too low. It’s probably a 1N400X-series Si diode, which can be had for less than $0.01 each in large quantities.
Apparently you didn’t read or understand my previous post. Limbo Donni may indeed be getting 9+ volts, but he can’t tell since he is using the wrong equipment to measure it.
The output of the AC transformer is rated at 8 volts RMS. This is about 23 volts peak-to-peak (11.4 plus, 11.4 minus). Rectify this and you have about 11 volts peak DC. Allowing for some diode loss, and you probably still have more than 8.
