• Some users have recently had their accounts hijacked. It seems that the now defunct EVGA forums might have compromised your password there and seems many are using the same PW here. We would suggest you UPDATE YOUR PASSWORD and TURN ON 2FA for your account here to further secure it. None of the compromised accounts had 2FA turned on.
    Once you have enabled 2FA, your account will be updated soon to show a badge, letting other members know that you use 2FA to protect your account. This should be beneficial for everyone that uses FSFT.

Need help w/ identify/mod part (HC13 Based):

M

macrospect

[H]ard|Gawd
2FA
Joined
Nov 22, 2004
Messages
1,711
Hello everyone,

This thread concerns ThermalTakes Hardcano 13. To make thinkg short what I am trying to do is build a 3V RECHARGABLE battery into the unit so that I dont have to keep replacing the damn battery every few weeks. It seems that the semicondustor inside it that is make by Holtek is battery hungry to keep my alarm settings and such in the memory.

So I tore the board out of the Hardcano 13 and basically heres what I have:
hc13board.JPG



Now for the technical part...

After examining the board and following the traces back to the semiconductor I came across this part. Basically what it is, is it takes the positive side of the battery and knocks the voltage down from 3V to ~2.4V. Now, when the board is recieving power it prevents the 4.65V current from flowing back into the battery.

Heres the part:
cir6.jpg


Heres a clearer picture with the same part in a different location:

cir3.jpg


Seems easy enough so far?

Now what I would like to do is take the battery/button holder off of the pcb, and take 2 wires, solder them to the PCB where the positive and negative traces begin, and run them to a unit that I am going to build that is made of 2 RECHARGABLE 1.5V AA batteries. Now to keep them charged I was going to take the 3.3V Rail off of the PSU, stick a resistor in there to knock the voltage down to 1.5V and hook it up to the 2 1.5 AA Batteries.

Does this sound like it would work? Please let me know what that part mentioned above is, as well.

Thanks for ANY help!

Heres the semiconductor that controls the LCD screen on the Hardcano 13:

semiconductor.jpg
 
Wow, that's a stupid design... if the battery is only there to preserve alarm settings (things that get programmed once) then they should have added a cheap 9346 type of eeprom to the board to hold those settings and forgot the battery.

You can probably just replace the lithium coin cell / holder with a battery - 3.6 volts (3 cells) worth of NiCd/NiMH or a single LiOn cell. Then put a high value resistor in parallel with D2 to keep the battery charged... the value will depend on the battery type you go with.

You could try a supercapacitor instead of batteries - something like http://www.panasonic.com/industrial/components/pdf/ABC0000CE2.pdf might solder directly in place of the backup battery that's already there, and you can get these from Digikey. But it might not last all that long with your computer's power off - i can't really tell from the datasheet.
 
I know its a REALLY horrible design. ThermalTake was not thinking when they made the board layout.

Anyway I am glad to see that this idea will work.

I think I found a battery I may use, its a NiCad battery pack that they commonly use in phones. Here the rating is 3.6V on it, do you think I will need a resistor to get rid of those .6 volts?

As for a resistor how would I calculate what ohm rating to get? I would rather use a battery instead of a supercap too.

Still any ideas what the part labeled D2 and D3 are? My guess is a fuse but I know thats probably wrong. I need to know what that part does before I continue.

Thanks for the reply I appreciate it!
 
As an eBay Associate, HardForum may earn from qualifying purchases.
A 3.6V battery will work... the HT48 microcontroller will handle up to 5 volts, and the series diode will drop that down to ~3V volts anyway. The top charge on the battery is

D1 through D3 are diodes. From the looks of things, D2 feeds the HT48 with 5 volts from the computer, and D3 feeds the HT48 with 3 volts from the battery. Not sure what D1 does. When the computer is on, the HT48's voltage will be ~4.3-4.4 volts due to D2's drop, and D3 will be reverse biased so no current flows into the battery. When the computer's off, D3 will conduct and the HT48 will be at ~2.4 volts, sufficient to keep it going.

A fully charged 3.6V NiCD pack will be at ~3.75 volts when it's fully charged. The pack you've linked to has three AAA's in it, so it's about 200mAh... to trickle charge it, you should stay around C/100 (2mA). At 4.35 volts (estimated HT48 voltage), the difference is 0.6V at 2mA, which is 0.6/.002 or around 300 ohms. So put a 330 ohm resistor in parallel with D3, replace the lithium battery + holder with that NiCd pack, and you're done!
 
What am I missing here?
Forgive me because I am half awake but shouldn't the stand-bye voltage from the PSU keep the battery from draining even when the machine is turned off?
 
Mister X said:
What am I missing here?
Forgive me because I am half awake but shouldn't the stand-bye voltage from the PSU keep the battery from draining even when the machine is turned off?
Click to expand...
Stand by voltage?

Im not sure what PSU you are using but when my computer is off(still plugged into the wall) The voltages on the 12V and 5V rails are 0V(well maybe a few tenths of a volt).


Gee...
Thank you so much for the info/help. Thanks for letting me know what the part(the diodes) are, as well as what type of resistor I should use.
You really helped me out on this one, I think I am going to pick up a NiCad battery at the local electronics shop tomorrow. :)

If this all works out(I am mostly sure it will) it will save me fustration and the problems with having to buy a new $5 CR2032 battery every month :mad:

Ill post back when I get all my parts and solder them together :)

PS sorry for the somewhat poor image quality, It was taken with my old (ok VERY old) Intel Play QX3 computer microscope that is at a magnification of 10X :)
 
I told you I was half awake. ;)

The purple wire on a ATX PSU (+5VSB) always has +5 volts whenever the PSU it plugged in.
 
No problem :)

I knew you were thinking about a wire on that ATX plug that is always live (just personally didnt know what color it was)

I didnt always want to keep the unit live (fans would spin(very slowly) even if the computer was off :p
So instead of using the NON RECHARGABLE CR2032 battery I was thinking of a replacment solution, the solution found above.
 
$5 for a CR2032? I get them from the dollar store for $1 canadian. There's no roman characters at all on them (entirely chinese) but damnit, they work. :D

Lemme know how it works out.
 
Gee,

Would a higher rated mAh battery last longer?

I accidentally bought a 600mAh battery, and got the 330 ohm resistors. Can I just stick 100 ohm resistors in there?
 
100 ohm? go for it.

And a higher rated battery will last longer.
 
Already went it for it.

I JUSt finished soldering everything together and IT IS HOLDING MY SETTINGS! I am SO glad to see that it worked, now its a waiting game to see if it holds my memory :)

Ill post more after I cleanup ;)

Thanks gee!
 
EDIT: There is a good amount of power drain the way I have it hooked up now. Can I get away with hooking the resistor up to a 5.2V line, without having to buy something greater than 100 ohm, (If I do no problem)?
 
You could setup a 5v line from a molex with a resistor to make it 3v and then you wouldn't even have to change it.. right?

~Adam
 
Yeah thats what I was going to do. I jsut dont know how many ohms I need to get for a resistor. I calculated ~217ohms (220 ohm resistors), thats what Ill need if Im going to run it off of a 5V line. Is this right, I tried my best to plug my values into gee's equation mentioned above.

Oh one more thing, should I be getting 1/4 or 1/2 watt resistors?

And do Zener Diodes ONLY allow current to flow in 1 direction? I want to make sure I am properly using them

As you can tell I have very little electronic knowledge. What I know is what I have learned from reading various articles on different websites.

Thanks :)
 
remove the zener diode from the above picture; it will prevent the battery from recharging :D

Zener diodes conduct voltage both ways. In one direction they're a regular diode (0.5-0.7V drop), and in the other direction they go into their zener/avalanche region at their defined voltage (5.1V, 20V, etc)
 
Gee,

Whats to stop the voltage from the battery from flowing back into the computer's 5V rail, and draining the battery, once the computers off?

EDIT:

This recharge voltage is scaring me a little, its at 3.760V and still rising!

EDIT:
Shit looks like Im not supposed to have a Zener Diode in there. Just a regular diode will do what I want. Damnit now I have to take it all apart again!

Before I go out and buy more parts I dont need, is THIS the type of diode I need?
 
The D3 diode already on the PCB prevents the computer from draining the battery when it's off.

Just do what I originally said; replace the 3V battery with your NiCd pack, and put a 100 ohm resistor in series with D2. Don't bother with any other diodes.

And 3.76V is about normal for a fully charged 3-cell NiCd pack. A fully charged NiCd cell will sit at around 1.25 volts... just keep an eye on the volage - if it goes up to something like 3.9V, you're overcharging it - in this case, you'll need to increase the value of the 100 ohm resistor to 220 ohms or something.
 
Im not charging the battery from D3, Im charging it from the 5V line on the PSU as mentioned above. THe problem is somehow the 220 ohm resistor turns out to let too much voltage through(4.6V ) which would probably eventually overcharge the battery.

Does this sound right:

5V supply is charging a 3.6V 600mAh battery pack, with a 220ohm resistor on the 5V supply voltage line. BUT the problem is, after the 220 ohm resistor my supply is still WAY above the 3.6V limit (4.6V). Where did I go wrong?

Heres a picture, it should help:

front.jpg



back.jpg




Its strange, NO MATTER what ohm resistor I put in it has ALMOST no effect on the output voltage. Hell I tried some 10 watt resistors I had laying around and it only dropped the voltage like ~.03V. Im clueless at this point.
 
Wire it up like so...

(5 volts)---->|----/\/\/\/----(battery +)

where ->|- is the diode, and /\/\/\/ is the resistor.

Due to the voltage drop across the diode, you will get around 4.4 volts at the junction of the resistor and the diode. The resistor will then drop this 4.4 volts down to 3.xxx volts, the battery voltage.

The battery has to be hooked up to measure these voltage. Without the battery, there's no current flowing through the resistor and you're gonna measure the same voltage on both sides...
 
They dont call you the electronics wizard for nothing :)


Allright gave it a try, and so far so good. The voltage dropped down to 3.1V, and its still slowly going up as the battery charges. Time will tell what happens, I have a GOOD feeling this is going to FINALLY properly work.

Heres how Ihave it setup:

5V --> Silicon Diode --> Zener Diode --> Battery --> microcontroller

The silicon diode was used to prevent battery drain from the bacflow created by the zener diode.

Ill post back in a hour or so.


Gee, thank you so much for your help, you have no idea how happy I am to see this work :)


EDIT: So far im at 3.180V and still charging I may let it go overnight to achieve a full charge, then see how it holds up once I turn it off.

EDIT 2: Well I have been sitting here for 5 hours strait now (yeah Im really bored) and the voltage has climbed up to 3.216V. I am going to unplug the PSU now and see how the battery holds up overnight, Ill post back here tomorrow with results (hope this works :) )!

EDIT 3: It seems that the battery charge maxxes out at 3.230V.
 
Well I wokeup today and went to test how the battery was holiding up, and..... it wasnt! The charge dropped down to 1.5V overnight!

It seems that I have currnet that is being rapidly consumed, my guess is at the Zener diode and resistor.

Well back to the drawing board once again!

EDIT:

I did something I didnt realize yesterday, I put the silicon diode in the wrong place. Duh, anyway I am sure it is correct now (and I have a higher charge voltages of ~3.4V)

Anyway heres how I have it:
5V --> Zener Diode --> Resistor --> Silicon DIode --> Battery


What was happening is the battery was draining back to the zener diode, so the new placment of the silicon diode should do it.
 
Dont hurt me :p

Isnt that what is knocking my voltage down??


Warming the soldering iron up again.... and taking it out to see what happens...

Edit: I jumped the zener out and the voltage jumps up past 3.8V(and keeps going) do I need another resistor or what? What ohm rating do I need if I do. I have a 220 in there right now.

PS. Not a big fan of instant messages? Never reply to any ;) Its up to you.
 
Here's what we're trying to accomplish... here's the original (lithium battery) config:



When the computer's on, D3 is conducting and the MCU voltage is equal to 5V minus the forward drop of D3 (eg, 0.6 volts), or 4.4 volts. Since 4.4V > 3V, D2 won't conduct and the battery is effectively disconnected from the circuit.

When the computer's off, D3 isn't conducting. Instead, the MCU gets its power supply from the 3V battery, and the forward drop of D2 (0.6V again, though probably less) puts the MCU at 2.4 volts.


Now, here's the original method I proposed:



Here, the operation of the circuit is exactly as before. The only difference is, when the computer is on and the MCU voltage is sitting at 4.4 volts as before, D2 isn't conducting but R1 is conducting. Since the battery is at approx. 3.6 volts, the current into the battery is (4.4-3.6)/R. As long as this current doesn't go above 1/100th of the battery's mAh rating, it's generally safe.

The only concern about this is that D3 now has to supply the charging current for the battery as well as power the microcontroller. Since the charge current is only a couple of mA, it shouldn't be a problem. But if for some reason it is, we'll just use a different diode:



Since the voltage drop of D1 (new diode) and D3 should be much the same, the circuit will behave identically.

If the battery is overcharging, you're putting too much current into it. To check the charge current, just measure the voltage across the resistor and divide by R to get your answer in amps. Again, make sure that this is less than 1/100th of the battery's mAh rating - if it's too high, just put in a bigger value resistor.

There you have it. No zener diode required...
 
Ok Ill give this a try.

Thank you so much for taking time to make me diagrams (well built ones may I add). Im a visual learner so they helped me out ALOT. Ill break out my soldering iron once again and resolder it as the diagram shows above

Ill post back when I have this completed.


Thanks!

Oh and as for the original design it goes something like this (I know the diagram is not nearly as neat as yours but it gets the basic idea across)

orig.jpg


(D1 is being used on the alarm speaker, I think.)


I soldered it together as shown in the third schmeatic. Heres my result with a 330 ohm resistor. Wouldnt this cause the battery to overcharge? Of course the end of the resistor (where the probe is touching) would go to the battery. Keep in mind that my supply is 5.29V.
5v.jpg
 
Well I keep trying with this thing with NO success at all. As of this poing I haqve made no progress since I began this project. I dont think I ever have had this much grief over a simple little resistor.

Its almost as if the resistor isnt doing its job! I stick a 200 ohm resistor onto the 5V line, its still 5V, I stick a 330 ohm resistor on there, and its still 5V (maybe .02 lower).

I dont know if I have dud resistors, a bad meter, or its some other phenomena. Maybe ill look for other advice to get a second opinion.

Gee, I KNOW your diagrams are correct, Its just I need to get to the bottom of this to figure whats going on?!

DaRkF0g
 
Do you have the battery hooked up?

If you disconnect the battery, no current will flow through the resistor so its voltage drop will be zero. And the charging diode will probably also measure zero volts of drop... so yes, you'll measure 5 volts.

But with the battery hooked up, the 5V will drop to whatever the battery's voltage is.
 
Gee,

Im sorry but the battery was not hooked up at the time. I have it hooked up now and I am measuring 3.88V and slowly rising across the resistor. I am so convinced that this will work I am going to leave the battery on there and let it charge. Everything should work out. I JUST got back from picking up some 270 ohm resistors.

At this point I am going to leave the battery on there. If it blows up then hell, Ill just get another one. These resistors should work for my application however! I used various online Resistor calculators and all gave me the result of 270 ohms.

This HAS to be it.


As always, thank you for your reply.

PS. I started with ~3.1V left on the battery, any idea as to how long it will take for a full charge?


EDIT: 1:30 PM the battery voltage is at 3.965V (charging)

EDIT: The charge seemed to be going up a bit too high at 4V, so I stuck a 330 ohm into there. Hopefully this will work.
Is it true that its not the voltage that I have to worry about, but instead the mA going into the battery?

My damn meter wont read mA at DC volts. I dont know whats wrong with it.
 
Gee,

Do you think that 4.03V is too high for a charging voltage? (Thats what it is as of 4:23PM)
 
for a NiCd battery, yes.

Measure the voltage across the resistor.
 
How do I do that? It discharges when you try to read across the resistor (Starts at ~2V(+/1V) and rapidly decends).
 
How? turn on the circuit and let it run for a while. Then take your multimeter, red test lead on one side of the resistor, black on the other. ;)

Divide the answer in volts by 270 (or whatever your resistance is) and you have the charge current in amps... at 600mAh and C/100 charging, the current should be less than 0.006 amps, which works out to 1.62 volts (.006 * 270) across the resistor.

Actually, fuck... sorry, it's a series battery. The charge current should be less than .006/3, or 2mA. Make the resistor 3 times bigger. :D
 
I still cant measure the resistance, because of when the battery charges the voltage is constantly changing. The higher the charge the lower the voltage across the resistor.

At this point in time I am at .859V across the resistor, and a 3.795 charge on the battery (and rising).

As for that last comment you made, if I tripled the resistors value, that would be 810 ohms! The highest value I can find is 680 (acutally I have a few of them right here). Can I put 1 resistor after another to get 810, or dont resistors work that way?

PS. Theres a 330 ohm resistor in there right now.

Thanks!
 
Let the circuit run for a while until it stabilizes.

Put two resistors in series - 680 + 220 is 900 ohms... which isn't C/100 exactly, but is enough to keep the battery charged.
 
Allright ill let it run a while, hell I gave the batteries 4.02V and they havent exploded yet. I know by the end of this I am going to end up getting a new battery pack :p

If I had 900 Ohms of resistance then it would probably take the battery ages to charge! First to figure out the current though.

EDIT:
Allright gee,

Looks like my current across the 330 ohm resistor is .704V(and VERY slowly dropping), with a rising charge of 3.964V on the battery.

So, that would mean I have a whimpy forward current of .0021mA.

Now...... I need to find a way to knock that 4V voltage down to 3.7-3.6V range. Ideas?
 
If the batteries are AAA, and they're trickle charging at 2.1mA, they're safe.

4.x volts seems unusually high, but I wouldn't worry. The battery pack is being charged correctly (current wise), and the microcontroller can handle 5 volts (or a battery voltage of ~5.5V including the diode drop you have).

Tape the whole mess together and call it done. Run the computer overnight and see where the battery voltage ends up.
 
Allright Ill take your word for it....

I just got done with putting it all back together, and it still runs :p Ill wait a few days and see what happens.

Thank you so much gee. You have NO IDEA how much you helped me out on this one!

DaRkF0g
 
You must log in or register to reply here.
Back
Top