Hi Everyone, I have what I hope to be a simple problem. I'm working on modifying a battery pack for an old video game console (Sega Genesis Nomad to be exact) and I'm having some issues getting the charging circuit to work correctly. Below is a simple diagram of the circuit. In order to use the OEM wall wart, I decided to use a DC-DC buck converter. Everything seems to work as expected, with the exception of the DC-DC converter's LED being powered on all the time, even when removed from the power source. I don't want this, as it would drain the battery over time. I thought simply putting a diode (1n4007) between the buck converter and the battery's positive terminal would prevent current from flowing back in the converter and powering the LED, but that appears to have prevented the battery from charging correctly (Though, it did shut off the LED). Is there anything simple I am missing here?

Previously, I simply did these battery packs by installing a new DC jack and wiring it directly to the battery and load. That works fine, but requires a 7.4v power supply rather than the Sega OEM. Thanks in advance for any help, and sorry for the simple question. Please let me know if I need to provide more information.



EDIT: To clarify, the LED pictured is not the LED I'm having issues with. I installed that LED as a status light for when the power supply was plugged in. The LED in question is located on the buck converter just before the output.

 

Maybe I am missing the obvious, but could you just remove the LED?

 

Maybe it is that simple. I'm still trying to learn =)

My initial thought was that if the LED is still being powered only when the input voltage is removed, that must mean there is some drain from the battery side. Removing the LED would remove some of that drain, but then I would think the circuit is still draining the battery some. Maybe this is incorrect?

I guess my main question should be, how should I design this circuit to ensure that the charging circuit isn't draining the battery when idle?

 

Set the converter to about 8.1 volts and add a diode in series with the output.

 

So my original circuit design + the diode in series with the output was correct. When I tried that and it didn't work, is it simply because the 7.4v output was too low to charge the battery, since it is also 7.4v? I was thinking that as the batter drained, voltage would drop, and the charger would then charge the battery, but even when the battery was completely drained, it wouldn't charge with the converter set to 7.4v. Thanks for your help and any explanation you can provide.

 

Hmmm is there passive drain besides the LED. I can't say for sure but I'm thinking no? You can test it though. Between Battery + and Output+ insert and ampmeter on the mA settings. With the LED I would assume to see something in the low mA range, maybe 10 mA?. If you remove I would expect to see very low, either reading as nothing or if the meter can go low enough something in the uA range.

#12's suggestion makes sense to me. Between battery + and Output + put a regular diode (typically a 0.7V drop) with the cathode on Battery + side and anode on Output + side. Almost any diode would work, but best would be one with a voltage rating of maybe 25 or so. More is fine, but I believe it is preferred to be about double the maximum voltage expected or so before you get some minor efficiency losses or something like that. Then adjust your pot to 8.1 volts. So in one direction 8.1V flows through the diode, drops to 7.4V, and charges the battery/powers the device. In the other direction the battery can't flow reverse through the diode (well technically some uA or nA or leakage current will flow I think).

 

So with a diode you were probably giving the battery pack 6.7 volts instead of 7.4. Hence the reason for adjusting it to 8.1.

Lets go back a step. What is the battery in question and how are you charging it exactly? Is it nickel or lithium? Just hard wiring it from Output of converter to battery terminals?

 

is it simply because the 7.4v output was too low to charge the battery, since it is also 7.4v?

That's a good point. You can't charge a battery with the same voltage it's already supposed to be. For instance, a car battery is really 12.6 volts when it's fully charged, but you typically hit it with 14.4 volts to charge it.

Then we are on to, what kind of battery is it? Lead-acid and ni-cad are pretty dumb batteries, easy to charge. Lithium and NiMH are a lot pickier about their charging profile and can get frankly dangerous if you do it wrong.

 

Hah I did the same thing. Finished answering the question then realized wait a minute...what exactly is going on with the battery charging part of this idea? Something is a little off there!

 

The battery in question is a Lithium Ion 7.4v 2600mAh PCM protected battery. For the charger, it is a 10v .85a wall wart (Which outputs around 12v). I've done a similar setup, using the same battery and a 7.4v smart charger without the DC buck converter and everything worked as expected, which was why I didn't realize I would actually need to be hitting the battery with more than 7.4 volts. I'm guessing the 7.4v smart charger I used previously actually output higher than 7.4v.

With all that said, it sounds like what I tried to do the first time, with the diode in series with the output, should work fine assuming I up the output voltage of the converter?

 

I put the diode back in the circuit and set the converter to 8.1v output. That seems to be working fine, and output is right around 7.4v after the diode. What worries me, and why I orginally turned the converter down to 7.4v when I first tried it with a diode inline, is just how hot the diode and the IC on the converter is getting. That doesn't seem right to me (hot enough to burn, if you hold onto it for a few seconds) but maybe I'm underestimating how much heat is normal. Still, something doesn't seem right.