Charging circuit Lead acid batteries with burned resistor

[tru3533]

Hi,
I'm looking to repair an older power supply with battery backup and charger for Lead Acid batteries

Power supply consist of LM317 adjusted to 28,5V
2 x 12V Sealed Lead Acid Batteries - 9Ah

Charging circuit is defective.
Resistor R1 is burned and not possible to measure the value.

Batteries was old/damaged and reason for the resistor to burn.

Question is how to calculate the resistor?

[abdulbadii]

commonly R.b hfe times R.e as R.c ~ 0.

and commonly Darlington Tr  has hi hfe. roughly Rb ~ hfe . (R3 || R2)
scheme seems bit weird to me though (D2 , D3-D4, R || why not one R?)

[tru3533]

From what I understand
LM317 provide 28,5 V when on to the circuit. Also charges the batteries via TIP122 - R2 II R3 D3 D4 if needed
When LM317 is off, batteries are supplying the circuit via D2

As you can see R2 R3 is around 1 Ohm,
Due to high current when charging, 2 resistors divide the current flowing to the batteries

[abdulbadii]

When LM317 is off, batteries are supplying the circuit via D2

Elaborate what actually the system is, is not a simple charger ?

so check Tr Dar high Hfe, for 1000,  R1 ~ 1 K

[tru3533]

Sorry, modified my text.
Complete circuit is a power supply with battery backup.

Power supply "works" but don't charge the batteries.

Without knowing, I think R1 is at a low ohm value.
The size (1w?) and that it is burned, shows a lot of current most have been flowing through BE

Max Ib is 0,12A for TIP122, so very strange that the resistor burned and the transistor survived.

[CatalinaWOW]

Strange circuit. 

R2 and R3 having different values means current sharing for power dissipation can't be purpose, and whatever that purpose is isn't obvious to me. 

Badly depleted batteries would cause max current from LM317 to flow, limited by 1.6 A fuse.

Current in R1 will roughly be total current flowing in R2 R3 pair divided by hfe of Q1.  The hfe of the TIP-122 is stated at 1000 so current in R1 will be so much less than the current in either R2 or R3 so it shouldn't burn at all, let alone burn badly.

Vce sat of the TIP122 is stated at 2 volts given 3 A, so won't be much less than a volt at the max operating current as set by the fuse.

Since this charger is operating as a float charger the current supplied should be relatively small.  It is a compromise between battery life and recovery time when batteries are discharged.  A 0.1 C charging rate is often used.  For your 6 AH batteries this would be 0.6 Amps. 

At 0.6 A the drop across the R2 R3 pair is about a quarter volt.  Add to that four diode drops (two diodes and the two BE drops in the Darlington transistor. ) and you have about 3.2 volts subtracted from the difference between the LM317 and the battery nominal voltage.   About 0.3 V across R1.  This voltage needs to drive 0.6A/hfe or 0.6 mA through R1.  So R1 would nominally be 500 ohms.

Higher current could flow in R1 if Q1 has failed open.  Have you checked this?

But with a 500 ohm R1, even if Q1 failed open R1 would not burn without burning R2 and R3.

So something doesn't make sense.  Changing the recharge rate to 1 C only drops the value of R1 to 50 ohms, still making it burning without burning R2 and R3 unlikely.

In spite of this confusion on explaining the fault, I suspect the circuit will work well enough with a resistor in the 50-500 ohm range.  If you use 50 battery life will be compromised a bit, but it may be worth it for the faster recovery.

[floobydust]

The schematic is wrong, the circuit makes no sense. Too many diode drops.
TIP122 might be part of the output switch? The world's cheapest float charger 42292 from Harbor Freight uses TIP41 power transistor for reverse-polarity protection. It's a mind warp to figure it out.
But here the Darlington TIP122 has a built in C-E diode... so that would not work. And you don't need a power resistor for base drive.
schematic taken from http://www.desert-home.com/2012/02/battery-charging-harbor-freight-item.html

[tru3533]

@CatalinaWOW
Thank you so much for the very useful explanation, strange circuit indeed.
This power supply is coming from a  20 year old commercial product made in a low scale production.
R2 and R3 Can be that they needed 1ohm and 2w, and used what they had?
Amazingly Q1 is still working perfectly.

@floobydust
I was thinking the same way, this must be wrong! But no, I have quadruple checked the circuit many times.
After all this is a simple 2 sided print and very easy to trace. So circuit is correct.

And you don't need a power resistor for base drive.
I totally agree, but they have used it, and it even states in their user manual
If R1 is miscolored you have a problem with the batteries.

So question remains, why do they use a power resistor? and why is it burned so badly?

[floobydust]

Can you post more pictures? Who has the big heatsink? LM317 or TIP122 or both? I still say the schematic is wrong.
There are four diode drops (Q1 VBEx2, D3, D4)  so around 2.8V lost with 28.5V in, for 25.7V output- which is too low to charge the batteries.

[tru3533]

LM317 and TIP122 are both mounted on the same large heatsink

[CatalinaWOW]

A bit off topic, but the reverse polarity protection on the worlds cheapest float charger doesn't work.  I hate to say how I know, but it does involve haste and poor lighting.  Totally smokes the charger.  Fortunately they are cheap.