Charging voltage on garage door

Viva

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About a year ago, I replaced the battery (3.5Ah/24V) on my garage door (Digidoor One). I wasn't paying attention, plugged the cable in the wrong way around, and blew a resistor on the board.

Since then, the board couldn't charge the battery and after a couple of months that battery was destroyed.

I recently checked out a neighbour's working Digidoor One and was able to determine that the resistor I need to replace is 20 Ohm (5 watt). I visited three shops today (Yebo Electronics, Cape Town Components and Mantech). None had a 20 Ohm resistor, so I settled for a 22 Ohm (5 watt) resistor.

I soldered it in place this afternoon and the multimeter reports 36V on the cable that connects to the battery. This seemed high to charge a 24V battery, so I checked the voltage on the neighbour's garage door and measured 30V.

Questions:
  • Can the 6V difference as measured be explained by the 2 ohm difference between the resistors?
  • I have not yet connected the terminal to the new batteries, as I don't want to damage the new batteries. What is the risk?
Lastly, since daily loadshedding is not kind to the rather small 3.5Ah (24V) batteries that garage doors typically use, I bought 2 x 12V/8Ah gel batteries instead and connected them in series. The idea is that the depth of discharge wouldn't be as deep and the batteries should last longer. My concern is that the Digidoor One's circuitry won't charge the 8Ah battery fast enough if loadshedding sessions occurs in quick succession. This is not a major issue, but I'm kind of wondering if a higher voltage from the board (36V instead of 30V) won't compensate for the larger battery capacity.

I basically need to know if I can plug my 2x12V batteries in to be charged by 36V, or should I instead continue the search for the correct resistor in order to charge at 30V, assuming the the 6V difference is indeed caused by the 22 ohm resistor.

I'd appreciate any input from knowledgable members.
 
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i gather the 36 volt is with no load on the circuit.with batteries connected max is +25 % so no higher than 30 to 31 volts .measure the other one with the battery disconnected .does the board have a voltage pot on it that you can set as well as amps.that resistor is maybe only a drop for amps .where you were at 29 watts charge you are now at around 26 watts charge (only possibly )cant say without seeing the circuit .
 
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22 ohm at1,5A is 33 volt. 20ohm is 30V. Connect it, Switch off AC and check volt and operate door. Make sure of the polarity, mark your wire with black and or red tape or cutex.
 
22 ohm at1,5A is 33 volt. 20ohm is 30V. Connect it, Switch off AC and check volt and operate door. Make sure of the polarity, mark your wire with black and or red tape or cutex.
do those boards charge at 1.5 amp ?mmm (edit)2.0 amp will push 40v edit.1.65 amp at 22ohm gives 36 volt .which is what the voltage is so 20ohm will be 33 v without a load .?
 
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i gather the 36 volt is with no load on the circuit.with batteries connected max is +25 % so no higher than 30 to 31 volts .measure the other one with the battery disconnected .does the board have a voltage pot on it that you can set as well as amps.that resistor is maybe only a drop for amps .where you were at 29 watts charge you are now at around 26 watts charge (only possibly )cant say without seeing the circuit .
I measured both the 36V on my door, and the 30V on my neighbour's door, with the battery disconnected. Those are the voltages measured on the wires which connect to the battery terminals.

If by "load" you mean that a measurement must be taken while the door is opening or closing, you're correct, I measured without any load.

It doesn't seem like my model has a voltage pot, no.
 
I measured both the 36V on my door, and the 30V on my neighbour's door, with the battery disconnected. Those are the voltages measured on the wires which connect to the battery terminals.

If by "load" you mean that a measurement must be taken while the door is opening or closing, you're correct, I measured without any load.

It doesn't seem like my model has a voltage pot, no.
no (load)as in connected to a discharged battery .that voltage is high enough to cook a battery .look for a 20 ohm .is it one of the big green resistors
 
do those boards charge at 1.5 amp ?mmm (edit)2.0 amp will push 40v edit.1.65 amp at 22ohm gives 36 volt .which is what the voltage is so 20ohm will be 33 v without a load .?
The original resistor (see attached) is color coded: red, black, black, gold. Gold means a tolerance of 5% from what I can gather.

My new resister is color coded: red, red, black, gold, black. I'm unable to determine what a tolerance color of black means. Maybe it means 0% tolerance?

When I just took another reading, it measured 37.4V.

Anyway, it seems like I should rather find a 20 ohm resistor, and get the voltage down to 30V, otherwise I'm going to damage my new batteries.
 

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yes you need a red .2.black .0. black .0 .gold .you get different colour resistors as in main body colour brown or blue etc .i presume yours was brown
 
i think you need to look for a 1/4 watt resistor .blue in colour body RES 20 OHM 1% 1/4W AXIAL security product manufacturers dont usually use standard resistors .
 
i think you need to look for a 1/4 watt resistor .blue in colour body RES 20 OHM 1% 1/4W AXIAL security product manufacturers dont usually use standard resistors .
The picture attached previously is of the original resistor (on my neighbour's door opener). It has a white/beige body. Judging by its size, it is a 5W resistor. From what I understand, a resistor rated for a higher wattage than necessary won't do any harm. So I'm pretty confident I'm looking at the correct type of resistor. The only issue is that I unable to find a 20 ohm. 18 ohm and 22 ohm are available. My only other option is to use two 10 ohm resistors in series.
 
if you dint come right i can pudo it to you
 

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The picture attached previously is of the original resistor (on my neighbour's door opener). It has a white/beige body. Judging by its size, it is a 5W resistor. From what I understand, a resistor rated for a higher wattage than necessary won't do any harm. So I'm pretty confident I'm looking at the correct type of resistor. The only issue is that I unable to find a 20 ohm. 18 ohm and 22 ohm are available. My only other option is to use two 10 ohm resistors in series.
yip its a blue grey body
 
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if you dint come right i can pudo it to you
That is very kind to offer! I'll make a couple of calls tomorrow to try and find it locally. If I don't come right, I'll let you know. Thanks for the help!
 
That is very kind to offer! I'll make a couple of calls tomorrow to try and find it locally. If I don't come right, I'll let you know. Thanks for the help!
have you tried RS components in cape town ? and communica ?
 
Did you measure the other door ac input, etc. What power supply does it use, switching or normal transformator. Normally their would be a diode as well, is it okay.
 
Did you measure the other door ac input, etc. What power supply does it use, switching or normal transformator. Normally their would be a diode as well, is it okay.
I didn't measure AC input on either door. I'm sure there is no issue on this front. Both doors are functioning just fine (in terms of being able to open and close) while being powered by AC only, or the battery only, or with both connected.

The only trouble is that my board couldn't charge the battery after I burned the resistor.
 
The charging circuit usually also has an LM317 somewhere (3pin voltage regulator). I suspect that went first which is why the resistor blew. A 22Ohm should also work in that position, but I think your best bet is to source a replacement board.
 
The charging circuit usually also has an LM317 somewhere (3pin voltage regulator). I suspect that went first which is why the resistor blew. A 22Ohm should also work in that position, but I think your best bet is to source a replacement board.
This is interesting. I'll try to see if I can find a LM317. If you're right, I can simply replace the LM317?
 
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