Im building a AC to DC battery AA charger using a full wave rectifier and a 5v regulator. Im wanting to charge 3 AA batteries. There 1.2V each with 2450mAH. So i was wanting to charge them at a rate of 245mA. The problem im having is, how would i make the output of the 5v voltage regulator output the 245mA of current. Do i put in a 20 ohm resistor (5v/20ohm=245mA) on the output and then connect the batteries to it. Any help would be appreciated. Thanks

 

What kind of batteries are these? Some battery chemistries are very particular about their charging profiles and you can damage or destroy them if you don't do it correctly (and 'destroy' can be in the catastrophic since in the form of flames and explosions). Generally you do not want to recharge them in series, but rather on a cell by cell basis.

No, you cannot just put a current limiting resistor in series with them as you describe. First, the current would be the difference between the regulator output and the battery voltage with that then being divided by the resistor value -- you can't just plug any voltage and any resistance into Ohm's Law to come up with just any current. But if you do this, the batteries will still be charged to a 5V (though they will likely fail catastrophically before they get there).

With a 10V AC source charging 3.6V worth of batteries, you are going to waste 2/3 of the power as heat and only about 1/3 is going to end up in the batteries.

Why not just purchase a cheap recharger?

 

there nimh AA batteries. Its part of a project im working on. Its for an analog class and we have to have a final project in about a month.

 

And you were thinking that a full-wave bridge, a capacitor, a single IC, and a resistor (i.e., four components) makes a reasonable final project for a class in analog electronics?

Then again, if you haven't learned Ohm's Law by this point in an analog electronics class....

Have you searched at all for how to charge NiMH cells?

 

You can modify your reg circuit to constant current by disconnecting the control terminalfrom ground and reconfiguring as below.

As Wbahn says you are wasting power with the 10v supply so you only need a half wave rectifier to reduce this.

 

And you were thinking that a full-wave bridge, a capacitor, a single IC, and a resistor (i.e., four components) makes a reasonable final project for a class in analog electronics?

Then again, if you haven't learned Ohm's Law by this point in an analog electronics class....

Have you searched at all for how to charge NiMH cells?

This is not my only part of my final project, this is just a small part....and i know ohms law just fine. I appreciate you trying to help, but you dont have to be an a** about it.

 

You can modify your reg circuit to constant current by disconnecting the control terminalfrom ground and reconfiguring as below.

As Wbahn says you are wasting power with the 10v supply so you only need a half wave rectifier to reduce this.

View attachment 75198

You can modify your reg circuit to constant current by disconnecting the control terminalfrom ground and reconfiguring as below.

As Wbahn says you are wasting power with the 10v supply so you only need a half wave rectifier to reduce this.

View attachment 75198

Thanks for the reply. Im going to be using the output of the rectifier as a rail to charge a couple other battery packs as well...one of which is close to 10v. Thats why i have a 10v supply. This is the first day this project was given, so I only worked on it for about 1 hour today. lol

 

What do you think about this. With this setup the collector and emitter current are approx. the same. The base of the transistor is will get the voltage that is across the LED ~2v. The base to emitter voltage will drop it to 1.8v. So 1.8v/245mA=7.34ohm. So ill have the batteries on the collector which I should have approx. 245mA of current to charge my batteries. Will this work?

 

Have you analyzed this circuit? As shown, what is the collector voltage? 5V - 4.2V = 0.8V. If the base and emitter voltage are, as you claim, 2 V and 1.8 V respectively, what mode is this transistor in?

Have you thought about what happens if your batteries are taken out of the circuit? I'm assuming that the 4.2V battery (V1) in your circuit represents the batteries to be charged?

If your LED is dropping 2 V, why do you think that the base-emitter voltage drop will only take it down to 1.8V? Your emitter voltage will be more like 1.4 V and your collector current will be closer to 175 mA. At your target of about 250 mA, the power in an 8Ω resistor will be around half a watt -- don't neglect thermal considerations. Also, 8 Ω is not a standard resistor size (but 7.5 Ω is).

Don't forget to consider how much voltage overhead you will have available for your battery pack. If the base is at 2 V, like you say, then even in hard saturation you will only be able to get your collector down to about 1.6 V and since you want it to stay linear you probably don't want it falling much before about 1.8 V. That only gives you 3.2 V for your batteries. Three fully charged NiMH batteries will have a terminal voltage of around 3.75 V and you will need a charging voltage of around 1.4 V to 1.6 V per cell (4.2 V to 4.8 V for the pack).

If you wanted to use a topology like this, you would be much better off using a zener diode to establish your base voltage. That would not only give you better accuracy and stability, but also let you limit the zener current and not waste so much power there.

It is still probably a bad idea to charge your cells in series, though if you limit the current to well under C/10hr you will probably be okay.

Might I recommend a current-mirror based approach?

 

Have you analyzed this circuit? As shown, what is the collector voltage? 5V - 4.2V = 0.8V. If the base and emitter voltage are, as you claim, 2 V and 1.8 V respectively, what mode is this transistor in?

Have you thought about what happens if your batteries are taken out of the circuit? I'm assuming that the 4.2V battery (V1) in your circuit represents the batteries to be charged?

If your LED is dropping 2 V, why do you think that the base-emitter voltage drop will only take it down to 1.8V? Your emitter voltage will be more like 1.4 V and your collector current will be closer to 175 mA. At your target of about 250 mA, the power in an 8Ω resistor will be around half a watt -- don't neglect thermal considerations. Also, 8 Ω is not a standard resistor size (but 7.5 Ω is).

Don't forget to consider how much voltage overhead you will have available for your battery pack. If the base is at 2 V, like you say, then even in hard saturation you will only be able to get your collector down to about 1.6 V and since you want it to stay linear you probably don't want it falling much before about 1.8 V. That only gives you 3.2 V for your batteries. Three fully charged NiMH batteries will have a terminal voltage of around 3.75 V and you will need a charging voltage of around 1.4 V to 1.6 V per cell (4.2 V to 4.8 V for the pack).

If you wanted to use a topology like this, you would be much better off using a zener diode to establish your base voltage. That would not only give you better accuracy and stability, but also let you limit the zener current and not waste so much power there.

It is still probably a bad idea to charge your cells in series, though if you limit the current to well under C/10hr you will probably be okay.

Might I recommend a current-mirror based approach?

Thanks for the reply. Yeah sorry i was doing it too fast and didnt realize I put 1.8v. Base to emitter drops 0.6V. lol sorry. Well back to the drawing board i guess. I guess i could put a higher voltage regulator....but whats the current-mirror based approach? again, thanks for the help

 

Also, i did analyze it in multisim and it puts out what I need, but i wont know till i actually build it. Also, like you said, its not the most efficient way.

 

This topic is extremely interesting Teckky than you for posting. Bahn I'm familiar with zener diode applications in this format but not sure what you mean by current mirror.