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Extremely simple overnight AAA Ni-MH charger

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neazoi

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Hello,
I want to build an ultra simple charger, to charge a 750mAh 1.2V AAA Ni-MH battery.
The charger must be very simple, but it has to have a series diode, because it is needed for my circuit, so as to prevent referse power flow.

Can you please do some calculations for me, for a simple overnight (slow) charger (diode-resistor?). I do not need it to be automatic and I do not need it to fully charge the battery, since my circuit works down to 0.6v. I do not want the battery to be heated much, but I read that modern batteries, when charged with 1/10 of the current, have an internal chemistry that automatically handle overcharge situations.
 

You really need a series pass ckt to limit the max output volts, and current, with a timer, then switch over to constant current via a resistor, say 30mA in this case for overnight complete charging. Ideally some sort of temp sense (ntc) to limit the batt temp to 35 deg C while charging for long battery life
 
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    neazoi

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You really need a series pass ckt to limit the max output volts, and current, with a timer, then switch over to constant current via a resistor, say 30mA in this case for overnight complete charging. Ideally some sort of temp sense (ntc) to limit the batt temp to 35 deg C while charging for long battery life

Thanks but I said ultra simple. Quick and dirty charger, a diode and a resistor or something similar. The space is extremely limited, since it is for a handheld device. Performance is not of interest neither battery life, as long as it does not explode
 

Simplest safe solution is just to limit the current but it will not give fastest charge time or guarantee best battery life. Just assume the battery voltage is zero and calculate the resistance that allows C/10 to flow from the source voltage. The current can never exceed that amount but will reduce as the cell voltage increases. The main drawback is that the charging current drops as the cell voltage increases so you get a boost at the beginning but the charging slows down as the cell 'tops up'.

If you have sufficient overhead voltage, a better way is to use a small voltage regulator (78L05 for example) configured as a constant current source then to shunt the battery with 2 x 1N4001 diodes in series. The constant current maintains the charge rate and the diodes prevent over voltage. It isn't particularly efficient but it only uses four components.

Brian.
 
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    neazoi

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I want to build an ultra simple charger, to charge a 750mAh 1.2V AAA Ni-MH battery...

Leave it in trickle charge mode; use a zener and a resistor to limit the voltage and current. Select the resistor to get max current of 75mA and the zener to limit voltage to 1.2-1.3V.

If you want to have a series diode, select a low drop one. Or perhaps a JFET?

You are lucky that NiMH cells are pretty robust.
 
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    neazoi

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If you have sufficient overhead voltage, a better way is to use a small voltage regulator (78L05 for example) configured as a constant current source then to shunt the battery with 2 x 1N4001 diodes in series. The constant current maintains the charge rate and the diodes prevent over voltage. It isn't particularly efficient but it only uses four components.

Brian.

Yes this would be great. I have 13.8v available, from the handheld device charger.
Can you please help me with this schematic?
The battery is 750mAH 1.2v eneloop brand cell AAA, and I can spare space for a 7805 regulator (what is the smd equivalent of it and how much current can it provide in that package?)
Can you draw a rough schematic for me with the values of the components, this would help a lot really

Here is the current schematic with a similar charger but I am not sure if the charger is ok, so I prefer your idea.
 

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Not quite what I had in mind.

Ohms Law says if you apply a constant voltage across a fixed resistance, a fixed current should flow. So if you load the regulator with a fixed resistor, it must pass a fixed current. All you do is calculate the resistance to pass C/10 current. For example, if you use a 5V regulator and want 75mA to flow, use R=V/I to get ~68 Ohms. The resistor goes across the output of the regulator, (Vout to GND) and as you want the total current to flow through the battery, take the output from the GND pin. Nothing other than the load resistor connects to the output pin.

The drawback to that arrangement is you drop a little over 5V in the regulator so you need a minimum of say 7V above battery voltage for it to work. it also dissipates some heat, W=V*I = 5 *0.075 = 0.375W in the resistor and (supply voltage - 7 - battery voltage) * 0.075 in the regulator. (about 0.3W from a 12V supply)

Brian.
 
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    neazoi

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Not quite what I had in mind.

Ohms Law says if you apply a constant voltage across a fixed resistance, a fixed current should flow. So if you load the regulator with a fixed resistor, it must pass a fixed current. All you do is calculate the resistance to pass C/10 current. For example, if you use a 5V regulator and want 75mA to flow, use R=V/I to get ~68 Ohms. The resistor goes across the output of the regulator, (Vout to GND) and as you want the total current to flow through the battery, take the output from the GND pin. Nothing other than the load resistor connects to the output pin.

The drawback to that arrangement is you drop a little over 5V in the regulator so you need a minimum of say 7V above battery voltage for it to work. it also dissipates some heat, W=V*I = 5 *0.075 = 0.375W in the resistor and (supply voltage - 7 - battery voltage) * 0.075 in the regulator. (about 0.3W from a 12V supply)

Brian.

Like the attached schematic?
 

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Like the attached schematic? ...

It is often wise to consider the end user. If you remove the battery but put on the charger, the voltage on the terminals (battery connections) may rise beyond expected values.

But I agree that it is difficult to make something fool proof because fools are quite smart!
 

It is often wise to consider the end user. If you remove the battery but put on the charger, the voltage on the terminals (battery connections) may rise beyond expected values.

But I agree that it is difficult to make something fool proof because fools are quite smart!

What, do you mean with this charger the battery has to be charged alone (without connected to any circuit?
Please explain
 

A misunderstanding - the diodes go across the battery. Two in series will start to conduct at around 1.3 V so they hold the voltage down by diverting the charging current. Something more like 1.4 - 1.5V would be better but not as easy as it is below normal Zener voltage range.

What c_mitra was referring to is a consequence of using constant current, it adjusts the voltage to keep the current steady, if you remove the battery so there is no load, it maximizes the voltage which could potentially damage the remaining circuitry. Adding the voltage clamp makes sure it can't rise high enough to do any damage.

Brian.
 

A misunderstanding - the diodes go across the battery. Two in series will start to conduct at around 1.3 V so they hold the voltage down by diverting the charging current. Something more like 1.4 - 1.5V would be better but not as easy as it is below normal Zener voltage range.

What c_mitra was referring to is a consequence of using constant current, it adjusts the voltage to keep the current steady, if you remove the battery so there is no load, it maximizes the voltage which could potentially damage the remaining circuitry. Adding the voltage clamp makes sure it can't rise high enough to do any damage.

Brian.

Like this one?
 

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Thats correct.

If the battery voltage is lower than the combined Vf of the diodes, all the current goes through it, as it rises, the diodes start to conduct and divert the current to ground. It is safe if the battery is removed because the voltage still can't rise above 2 x Vf.

The ideal diode drop is just slightly higher than the voltage across a fully charged battery, it may be a little more than two 1N4001 diodes which could limit the depth of charge a little - but you did want the simplest solution. If you can find a way to clamp at 1.4 or 1.5V it would be better but Zener diodes normally only go down to 1.7V. Possibly several Schottky diodes in series might do it as they tend to have a lower Vf at low current.

Brian.
 

Thats correct.

If the battery voltage is lower than the combined Vf of the diodes, all the current goes through it, as it rises, the diodes start to conduct and divert the current to ground. It is safe if the battery is removed because the voltage still can't rise above 2 x Vf.

The ideal diode drop is just slightly higher than the voltage across a fully charged battery, it may be a little more than two 1N4001 diodes which could limit the depth of charge a little - but you did want the simplest solution. If you can find a way to clamp at 1.4 or 1.5V it would be better but Zener diodes normally only go down to 1.7V. Possibly several Schottky diodes in series might do it as they tend to have a lower Vf at low current.

Brian.

So you say this will prevent overcharging of the battery as well in a sense?
About the second paragraph, maybe if I use a 3mm 1.5v LED instead of the diodes, would be better?
 

maybe if I use a 3mm 1.5v LED instead of the diodes, would be better????

Yes (but); can the 3mm LED take 75mA full current? Common 3mm LEDs are best used around 10mA.
 

Fairchild 1N400x Vf at 75mA is quoted at ~0.72V and Diodes Inc. show it as ~0.75V so you might get away with two of those. It would only take a minute to do a test to find out. I wouldn't use an LED, a small one wouldn't handle the current and bigger ones tend to have higher Vf anyway.

Brian.
 
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Fairchild 1N400x Vf at 75mA is quoted at ~0.72V and Diodes Inc. show it as ~0.75V so you might get away with two of those. It would only take a minute to do a test to find out. I wouldn't use an LED, a small one wouldn't handle the current and bigger ones tend to have higher Vf anyway.

Brian.

Thank you
I need to ask a final thing because I am not sure. In the schematic of the preamplifier attached, I like the spst switch for practical reasons. Will there be a problem from the "backwards" power flown from the battery into the output of the regulator, when no INPUT voltage is applied to the regulator?

In other words will the regulator itself prevent power from the battery to flow back to the "in/out" point in the circuit?

- - - Updated - - -

Also, in the ON position, won't the battery be discharged through the diodes as well? (apart from the PGA amplifier)
 

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Fairchild 1N400x Vf at 75mA is quoted at ~0.72V and Diodes Inc. show it as ~0.75V so you might get away with two of those....

True, but...

I am worried how much the diodes will drain the battery. If the diode drops 0.72V at 75mA it will certainly take 10-20mA at 0.6V (two makes 1.2V - the nominal voltage)- too much to sacrifice in my opinion...
 
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In other words will the regulator itself prevent power from the battery to flow back to the "in/out" point in the circuit?
Not impossible but unlikely. I doubt 1.2V is sufficient to push current backwards through the regulator and bear in mind that even if it does, the remainder of the supply line is also in series with the discharge path. If in doubt and if you have enough supply voltage, add a series diode as well. As it is a constant current circuit, adding another diode doesn't change the operation except for making the input overhead slightly higher.
Also, in the ON position, won't the battery be discharged through the diodes as well? (apart from the PGA amplifier)
As c_mitra points out, there is a possibility that a fully charged battery and diodes with low Vf will create a discharge path. That's why I was cautious about the simplistic two diode approach. What you need really is something that clamps just a little higher than the voltage of a fully charged battery so it protects the PGA amplifier but doesn't provide a discharge path. A 'nominal' 1.2V battery may actually go up to more than 1.4V when fully charged so ideally a clamp at say 1.5V would be best but building one isn't as easy as it sounds. Two 1N400x give around 1.3V which may be too low but adding a third makes it almost 2V which is too high. You might try an active clamp, an NPN transistor across the battery with forward conducting diode between collector and base so the voltage is Vf +Vbe, it will have a lower dynamic resistance but I'm not sure the voltage will be any more precise.

Brian.
 
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    neazoi

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If in doubt and if you have enough supply voltage, add a series diode as well. As it is a constant current circuit, adding another diode doesn't change the operation except for making the input overhead slightly higher.

Just a series diode from the output of the 78M05 to the battery? (the input to the 78M05 is 13.8V)
But how the "comparison" of the series discharge diodes and the battery would be done then? (since there is an isolating diode between them)

As c_mitra points out, there is a possibility that a fully charged battery and diodes with low Vf will create a discharge path. That's why I was cautious about the simplistic two diode approach. What you need really is something that clamps just a little higher than the voltage of a fully charged battery so it protects the PGA amplifier but doesn't provide a discharge path. A 'nominal' 1.2V battery may actually go up to more than 1.4V when fully charged so ideally a clamp at say 1.5V would be best but building one isn't as easy as it sounds. Two 1N400x give around 1.3V which may be too low but adding a third makes it almost 2V which is too high. You might try an active clamp, an NPN transistor across the battery with forward conducting diode between collector and base so the voltage is Vf +Vbe, it will have a lower dynamic resistance but I'm not sure the voltage will be any more precise.

Brian.
I would not worry too much, because the preamplifier can work down to 0.65v or so, so there is enough headroom, even if the battery is not kept fully charged.
However, couldn't I just use an extra germanium diode in series with the 1n4001 diodes? (I worry about the current that a 1n34a can pass through it, can it be as high as 75mA?)


A final thing that worries me, is that ok to charge the battery that way, when the preamplifier is being powered at the same time as well?
 
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