"9V" Battery Recharger

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bountyhunter

Joined Sep 7, 2009
2,512
I see a lot of threads about how to recharge the Ni-CD or Ni-MH "9V" batteries. I designed a simple one many years back, schematic attached. It is a shunt regulator that charges with approximately 10 mA and will not overcharge because it locks into a constant voltage mode when the battery is fully charged.

The LED is an indicator of charge because as the charging current through the battery diminishes, the LED gets brighter indicating closer to full charge.

NOTE: Ni-Cd and Ni-MH "9V" batteries are typically six-cell which means they are actually 7.5V (nominal) and 8.4V when fully charged. This circuit is designed for these 8.4V batteries. It is not a fast charger. For a typical 9V battery (between 100 and 150 mA-hr) it will take 10 - 20 hours to recharge fully.

There are some "9V" rechargeables with seven cells which have 8.75V nominal and 9.8V full charge. This circuit would have to be tweaked up to 9.8V to work with those.

To calibrate the circuit: disconnect battery and adjust VSET pot to make the voltage across the resistor string (where it connects to the battery terminals) read 8.4V with the battery disconnected.
 

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adam555

Joined Aug 17, 2013
858
That looks great.

I also just designed a very crude charger for 2 x AAA 1.2V batteries with the 5V from a USB port. I would be really interested in adapting your design and use it instead, so i was wondering if you could explain exactly how it works (just the part surrounding the transistor).
 

Thread Starter

bountyhunter

Joined Sep 7, 2009
2,512
Sure. It's a shunt regulator. R1 feeds current that everything runs on. Q1 and the rest are set up as a DC voltage regulator which forces a fixed voltage across the resistive divider. Voltage reference U2 forces 2.5V across R4, then R2 and R3 "scale up" that voltage to the regulated output which R2 sets at 8.4V. If no battery is connected, all the current from R1 flows down through Q1 (less the 0.5 mA through the resistive divider). Q1 sucks up as much current as necessary to keep the voltage at 8.4V.

When a battery is attached, it "drags" the voltage down below 8.4V which reduces the current draw through Q1 and the current goes through the battery. When the batt voltage reaches 8.4V (full charge) Q1 starts sucking up the current so the battery gets only a very small amount and is not overcharged.

The key to the design is that the roughly 10mA coming down divides between Q1 and the battery based on the battery voltage. As it rises, the current shifts back to Q1.
 

adam555

Joined Aug 17, 2013
858
Thanks for the explanation.

I tried to adapt your design to charge 2 AAA batteries with the 5V from the USB and these are the values I came up with:

Untitled-1.jpg

I couldn't try it for real yet (I have to buy some components) but it seems to work in a simulator. It didn't work in LTSpice though.

Any suggestions will be welcome.
 

#12

Joined Nov 30, 2010
18,223
Slick design using a 336 instead of 2 transistors, and temperature compensated, too!
My compliments.
 

Thread Starter

bountyhunter

Joined Sep 7, 2009
2,512
Thanks for the explanation.

I tried to adapt your design to charge 2 AAA batteries with the 5V from the USB and these are the values I came up with:

View attachment 61190

I couldn't try it for real yet (I have to buy some components) but it seems to work in a simulator. It didn't work in LTSpice though.

Any suggestions will be welcome.
Will probably work as long as the "set voltage" is regulated to about 1.4V/cell on the NiCad or NiMH battery. The only downside is using a shunt, you only have about 10 mA max to charge the battery so it will take a while to charge AAA batteries.
 

adam555

Joined Aug 17, 2013
858
Will probably work as long as the "set voltage" is regulated to about 1.4V/cell on the NiCad or NiMH battery. The only downside is using a shunt, you only have about 10 mA max to charge the battery so it will take a while to charge AAA batteries.
Yes, I was aiming for 50mA but I couldn't get that high.
 

#12

Joined Nov 30, 2010
18,223
I like the design with 5V and 100 ohms because the maximum power is 1/4 watt. That makes power calculations pretty much unnecessary because a quarter watt distributed among that many components will not put any of them in danger of smoking.
 

Thread Starter

bountyhunter

Joined Sep 7, 2009
2,512
Yeah I saw that mistake too but didn't want to criticise Bountyhunter as he has done such a superbly neat job hand drawing the schematic. :)
The 9V battery has a polarity connector so it only goes on one way. I suppose somebody could try to force it on reverse, never worried about that.

I suppose you could add a 1N4148 blocking diode between the right end of R1 and the top of R2 to replace that trace? That sound right?

It would reduce the charge current slightly.

ADD: it may not need it. If the battery is reversed it forces about 8V forward across diode U2 and reverse across E-B of Q1, which will Zener at about 6V with current limited by about 2V across R2 and R3 (about 10k) which limits current to about 200 uA. I don't think there is enough current or voltage to hurt anything in that path.

The other paths don't have any way of hurting anything either. D1 will be reverse biased and LED 1 will have about 2V reverse across it at most.

I think the circuit should be reverse proof if I haven't forgotten something. I have been using one of these for 30 years and have probably reversed the terminals before but the unit never went down.
 
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#12

Joined Nov 30, 2010
18,223
Bounty's circuit does look idiot proof.

As for tracecom...
Look at the 5k pot in series with a 4.99k resistor that has 2.5 volts across it.
Therefore the 5k pot has 2.5 volts across it.
A typical 1 turn pot has 270 degrees of rotation, so
you have 10.8 degrees per tenth of a volt.
If this is going to be a tight adjustment, you would try for .05 volts accuracy.
The trick is to imagine your hand and your eye trying to set a pot within 5 degrees of rotation.
Look at the pot in your imagination. A quarter turn is 90 degrees. Too easy!
Now cut that in half. Eighth of a turn. Still a piece of cake.
Now cut that in half. Sixteenth of a turn. Getting tight, but you can see it and you can hit it in one try.
You need to hit a 4th of that.

By now you can see that a free hand adjust with only one try probably won't work with a 5 degree sweet spot, but you have a volt meter, all day to adjust it, and it only needs to be set correctly once. So...it might take 5 little nudges, but it's achievable.

That's how you decide on a turns ratio. Figure how many degrees you have to hit your target and visualize it. If it's tighter than 5 degrees, reconsider which pot to buy.
 

Ramussons

Joined May 3, 2013
1,082
Yeah I saw that mistake too but didn't want to criticise Bountyhunter as he has done such a superbly neat job hand drawing the schematic. :)

It was not to criticise. It was just to correct the drawing, mainly, to avoid any future reference to an error.

Ramesh
 

tracecom

Joined Apr 16, 2010
3,944
Bounty's circuit does look idiot proof.
I'll be the test of that. :D

I want to lay out a small PCB for the circuit, and I want to use a wall-wart to provide the DC input. To that end, I have redrawn Bounty Hunter's schematic with some minor changes to the regulator circuit. Hopefully, I haven't introduced any problems, but if I have, I will correct them.

Comments are welcome from all. Thanks.
 

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