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Today I was checking the voltage between charging terminals going to the battery of emergency lantern using a multimeter.

The multimeter showed a DC voltage of 9V and an AC voltage of 4V.

Could this AC voltage have damaged by old battery?

The DC voltage while being measured was not constant and the decimal part was changing in a cyclic pattern.

I only saw 3 diodes on the board. Is this a half-wave rectifier and is the multimeter mistakingly showing the ripple of half-wave as an AC voltage.

Please see the image of the circuit.

enter image description here

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    \$\begingroup\$ I actually found same circuit, but without the unneeded third diode, in a car battery charger. Here's a question about that. \$\endgroup\$ – AndrejaKo Jun 15 '11 at 19:34
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It's fairly common to see a lead-acid battery charged using rectified AC. As long as the charging current isn't beyond the capability of the battery, it will 'work'. If there isn't a series resistor somewhere, or some primary-side limiter, the winding resistance of the transformer could be what's limiting the charging current.

A handheld multimeter is sensitive to 60Hz AC, so yes, your DC reading was likely skewed by the low-frequency ripple.

The circuit you have drawn is a full-wave rectifier.

If there isn't any explicit current-limiting protection in the charger, it is possible that the charger can become damaged if subjected to long-term overload. Wall-wart adapters have similar failure modes (usually the transformer goes high-impedance).

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The circuit you show is a full wave rectifier, not half wave. Half of the transformer secondary will be conducting each cycle. However, the third diode on the right doesn't make sense. There is no need for it since there are already diodes in series with each of the two paths current can come from to get there.

The meter didn't mistakenly show you AC. The voltage coming out of the rectifier circuit has a AC component to it. The DC reading was jumping around a little probably because of the meter's sampling interval beating against the 2x power line frequency of the AC component. A simple mechanical meter would probably have shown you a steady DC reading.

If this voltage is being used to charge a lead-acid battery, then probably there is no problem, assuming the voltage is in the reasonable range. There is nothing explicit limiting the current in your schematic, but the transformer will have some internal resistance. This is probably good enough, especially if the charger is intended for that unit.

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  • \$\begingroup\$ Thanks for the answer. Does all rectifier's has a AC component? Because when I have used my multimeter for similar charing circuits I haven't detected AC before.Also what would happen if an AC voltage is directly used to charge a battery without using a rectifier ? \$\endgroup\$ – Johnes Thomas Jun 15 '11 at 19:07
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    \$\begingroup\$ Rectifiers either pass only the positive half of the AC waveform (half wave) or take the absolute value of the AC waveform (full wave). If the AC waveform is a sine, like the power line is, then there will be a AC component in the result. This can be smoothed or reduced in various ways, most commonly by putting a large capacitor on the output of the rectifier. \$\endgroup\$ – Olin Lathrop Jun 16 '11 at 12:18
  • \$\begingroup\$ No, AC can't be used to charge a battery. A chemical battery is inherently DC, and must have a net DC to current to charge it. If the peaks vary too much from the average DC, then the battery can be damaged. Negative current will discharge instead of the charge the battery. Excessive current in either direction will damage it. \$\endgroup\$ – Olin Lathrop Jun 16 '11 at 12:19
  • \$\begingroup\$ Yes, all rectifiers convert AC input to an unwanted AC component, called "ripple", in addition to the desired DC. The exact size of the ripple (as measured by a voltmeter set to AC) depends on the output capacitor and the load. Most digital electronics devices have a relatively large capacitor and a relatively small load, resulting in an AC ripple that may be too small for a multimeter to measure. A car battery charger typically has no capacitor and a relatively heavy load, resulting in an easily-measured volt or two of AC ripple. \$\endgroup\$ – davidcary Jun 19 '11 at 1:08
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The extra diode may be a voltage drop mechanism, to limit supplied voltage to the battery. Could be that they had a transformer they needed to use for whatever reason and the output voltage was just a tad too high, or it could have been a design tweak.

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  • \$\begingroup\$ A Resistor could also be used to drop the voltage right? What advantage would a diode provide over a resistor? \$\endgroup\$ – Johnes Thomas Jun 17 '11 at 6:14
  • \$\begingroup\$ @JohnesThomas A diode would give a fairly constant 0.6-0.7 V drop over a wide range of currents, a resistor would give a drop proportional to the current. \$\endgroup\$ – Andrew Jun 15 '17 at 10:44
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At best this looks to me like it would produce a pulsed DC current at twice the frequency of the mains (50Hzx2 or 60Hzx2) with the multimeter measuring the 1/2 wave AC at the higher frequency. If that third diode had a breakthrough voltage, meaning that even in the correct direction the lower voltage wasn't passing, this would turn the current into something more square waved...really noisy on an AM radio.

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