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12V lead acid charger with LM317 not charging
Need help with LM317 voltage regulator circuitLM317 Constant Current Battery Charger Questioncharging a lead acid battery bank with a 19v laptop chargerHow to make 5v DC UPS Circuit?LM317 output voltage to 0 V with transformerless negative voltage created through a buck converterLM317 Sealed lead acid chargerCurrent booster with LM317 not working properly in LTspice simulation?Power supply beginner errorLead acid chargingCharge a 12V marine/boat battery with AC
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have this circuit built and tested. I think this should work and charge the battery with 0.4A to about 13.5V.
The Problem: it doesn't charge and between U1 and U3 (Battery) is a current of about 0.03A, but why?
The LM2596 is a Buck converter Breakout board and that works flawlessly.
R1 is a 4W resistor so that won't be the problem
Why doesn't it charge?
simulate this circuit – Schematic created using CircuitLab
12v lm317 lead
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
$endgroup$
add a comment |
$begingroup$
I have this circuit built and tested. I think this should work and charge the battery with 0.4A to about 13.5V.
The Problem: it doesn't charge and between U1 and U3 (Battery) is a current of about 0.03A, but why?
The LM2596 is a Buck converter Breakout board and that works flawlessly.
R1 is a 4W resistor so that won't be the problem
Why doesn't it charge?
simulate this circuit – Schematic created using CircuitLab
12v lm317 lead
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
$endgroup$
4
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23
add a comment |
$begingroup$
I have this circuit built and tested. I think this should work and charge the battery with 0.4A to about 13.5V.
The Problem: it doesn't charge and between U1 and U3 (Battery) is a current of about 0.03A, but why?
The LM2596 is a Buck converter Breakout board and that works flawlessly.
R1 is a 4W resistor so that won't be the problem
Why doesn't it charge?
simulate this circuit – Schematic created using CircuitLab
12v lm317 lead
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
$endgroup$
I have this circuit built and tested. I think this should work and charge the battery with 0.4A to about 13.5V.
The Problem: it doesn't charge and between U1 and U3 (Battery) is a current of about 0.03A, but why?
The LM2596 is a Buck converter Breakout board and that works flawlessly.
R1 is a 4W resistor so that won't be the problem
Why doesn't it charge?
simulate this circuit – Schematic created using CircuitLab
12v lm317 lead
12v lm317 lead
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
edited Aug 7 at 11:17
brhans
10.2k2 gold badges24 silver badges32 bronze badges
10.2k2 gold badges24 silver badges32 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
asked Aug 7 at 11:13
Mark OEMark OE
243 bronze badges
243 bronze badges
4
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23
add a comment |
4
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23
4
4
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Read the datasheet of the LM317, on page 9 it states:
So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V.
Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317.
The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 converter in place so remove it.
You will need a power source with a higher voltage than ~ 15 V.
As the LM317 will drop 3 V or more at a significant current, it will get hot so use a heatsink! If the LM317 gets too hot it lowers the current to lower its power dissipation (and allow it to cool down).
Note that your circuit does not have a well defined "stop charging" voltage, current will keep flowing and your battery might over charge!
I have built an LM317 based battery charger for my 12 V car battery. I use a 19 V laptop power supply I had lying around to power it. In that design I do not use the LM317 as a current source, instead I use it as a voltage regulator set to 13.5 V. Then when the battery has a lower voltage, the LM317 will hit its build-in current limit (< 2.2 A). For a car battery 2.2 A or less is fine. As the battery charges an the voltage reaches 13.5 V, the current gets smaller and smaller until only a leakage current is left.
If that 2.2 A is too much for your battery, use this circuit instead:
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
$endgroup$
add a comment |
$begingroup$
You need more than 14V at input if you want to charge a battery to 13.5V. About 18V should be enough. LM317 needs about 3V between Vin and Vout to work. And there will be 1.25V between Vout and Adj. And then the battery will have 13.5V. 13.5+1.25V+3V is 17.75V.
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
The problem you have is caused by the LM317 and the voltage you have available to it.
The LM317 requires a difference of 3V between the input and output voltage. You have 15V from your powersupply (which you drop to 14 with the LM2596.) That's only 1.5V above the output voltage (13.5V) that you need for the battery.
You really only have two choices:
- Use a higher input voltage (at least 17V)
- Use a different current regulator that can work with the available voltage.
In any case, the LM2596 is only making things worse.
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
$endgroup$
add a comment |
$begingroup$
Also, if the power supply is switched off and the 12V battery is still attached, there's a negative voltage difference (Vin-Vout) on the LM317, a situation not mentioned in the data sheet as far as I can see. If that causes a leak current, it would drain the battery.
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
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votes
active
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votes
$begingroup$
Read the datasheet of the LM317, on page 9 it states:
So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V.
Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317.
The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 converter in place so remove it.
You will need a power source with a higher voltage than ~ 15 V.
As the LM317 will drop 3 V or more at a significant current, it will get hot so use a heatsink! If the LM317 gets too hot it lowers the current to lower its power dissipation (and allow it to cool down).
Note that your circuit does not have a well defined "stop charging" voltage, current will keep flowing and your battery might over charge!
I have built an LM317 based battery charger for my 12 V car battery. I use a 19 V laptop power supply I had lying around to power it. In that design I do not use the LM317 as a current source, instead I use it as a voltage regulator set to 13.5 V. Then when the battery has a lower voltage, the LM317 will hit its build-in current limit (< 2.2 A). For a car battery 2.2 A or less is fine. As the battery charges an the voltage reaches 13.5 V, the current gets smaller and smaller until only a leakage current is left.
If that 2.2 A is too much for your battery, use this circuit instead:
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
$endgroup$
add a comment |
$begingroup$
Read the datasheet of the LM317, on page 9 it states:
So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V.
Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317.
The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 converter in place so remove it.
You will need a power source with a higher voltage than ~ 15 V.
As the LM317 will drop 3 V or more at a significant current, it will get hot so use a heatsink! If the LM317 gets too hot it lowers the current to lower its power dissipation (and allow it to cool down).
Note that your circuit does not have a well defined "stop charging" voltage, current will keep flowing and your battery might over charge!
I have built an LM317 based battery charger for my 12 V car battery. I use a 19 V laptop power supply I had lying around to power it. In that design I do not use the LM317 as a current source, instead I use it as a voltage regulator set to 13.5 V. Then when the battery has a lower voltage, the LM317 will hit its build-in current limit (< 2.2 A). For a car battery 2.2 A or less is fine. As the battery charges an the voltage reaches 13.5 V, the current gets smaller and smaller until only a leakage current is left.
If that 2.2 A is too much for your battery, use this circuit instead:
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
$endgroup$
add a comment |
$begingroup$
Read the datasheet of the LM317, on page 9 it states:
So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V.
Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317.
The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 converter in place so remove it.
You will need a power source with a higher voltage than ~ 15 V.
As the LM317 will drop 3 V or more at a significant current, it will get hot so use a heatsink! If the LM317 gets too hot it lowers the current to lower its power dissipation (and allow it to cool down).
Note that your circuit does not have a well defined "stop charging" voltage, current will keep flowing and your battery might over charge!
I have built an LM317 based battery charger for my 12 V car battery. I use a 19 V laptop power supply I had lying around to power it. In that design I do not use the LM317 as a current source, instead I use it as a voltage regulator set to 13.5 V. Then when the battery has a lower voltage, the LM317 will hit its build-in current limit (< 2.2 A). For a car battery 2.2 A or less is fine. As the battery charges an the voltage reaches 13.5 V, the current gets smaller and smaller until only a leakage current is left.
If that 2.2 A is too much for your battery, use this circuit instead:
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
$endgroup$
Read the datasheet of the LM317, on page 9 it states:
So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V.
Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317.
The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 converter in place so remove it.
You will need a power source with a higher voltage than ~ 15 V.
As the LM317 will drop 3 V or more at a significant current, it will get hot so use a heatsink! If the LM317 gets too hot it lowers the current to lower its power dissipation (and allow it to cool down).
Note that your circuit does not have a well defined "stop charging" voltage, current will keep flowing and your battery might over charge!
I have built an LM317 based battery charger for my 12 V car battery. I use a 19 V laptop power supply I had lying around to power it. In that design I do not use the LM317 as a current source, instead I use it as a voltage regulator set to 13.5 V. Then when the battery has a lower voltage, the LM317 will hit its build-in current limit (< 2.2 A). For a car battery 2.2 A or less is fine. As the battery charges an the voltage reaches 13.5 V, the current gets smaller and smaller until only a leakage current is left.
If that 2.2 A is too much for your battery, use this circuit instead:
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
edited Aug 7 at 11:39
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
answered Aug 7 at 11:25
BimpelrekkieBimpelrekkie
56.8k2 gold badges55 silver badges130 bronze badges
56.8k2 gold badges55 silver badges130 bronze badges
add a comment |
add a comment |
$begingroup$
You need more than 14V at input if you want to charge a battery to 13.5V. About 18V should be enough. LM317 needs about 3V between Vin and Vout to work. And there will be 1.25V between Vout and Adj. And then the battery will have 13.5V. 13.5+1.25V+3V is 17.75V.
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
You need more than 14V at input if you want to charge a battery to 13.5V. About 18V should be enough. LM317 needs about 3V between Vin and Vout to work. And there will be 1.25V between Vout and Adj. And then the battery will have 13.5V. 13.5+1.25V+3V is 17.75V.
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
You need more than 14V at input if you want to charge a battery to 13.5V. About 18V should be enough. LM317 needs about 3V between Vin and Vout to work. And there will be 1.25V between Vout and Adj. And then the battery will have 13.5V. 13.5+1.25V+3V is 17.75V.
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
$endgroup$
You need more than 14V at input if you want to charge a battery to 13.5V. About 18V should be enough. LM317 needs about 3V between Vin and Vout to work. And there will be 1.25V between Vout and Adj. And then the battery will have 13.5V. 13.5+1.25V+3V is 17.75V.
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
edited Aug 7 at 11:25
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
answered Aug 7 at 11:17
JustmeJustme
6,4412 gold badges6 silver badges18 bronze badges
6,4412 gold badges6 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
The problem you have is caused by the LM317 and the voltage you have available to it.
The LM317 requires a difference of 3V between the input and output voltage. You have 15V from your powersupply (which you drop to 14 with the LM2596.) That's only 1.5V above the output voltage (13.5V) that you need for the battery.
You really only have two choices:
- Use a higher input voltage (at least 17V)
- Use a different current regulator that can work with the available voltage.
In any case, the LM2596 is only making things worse.
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
$endgroup$
add a comment |
$begingroup$
The problem you have is caused by the LM317 and the voltage you have available to it.
The LM317 requires a difference of 3V between the input and output voltage. You have 15V from your powersupply (which you drop to 14 with the LM2596.) That's only 1.5V above the output voltage (13.5V) that you need for the battery.
You really only have two choices:
- Use a higher input voltage (at least 17V)
- Use a different current regulator that can work with the available voltage.
In any case, the LM2596 is only making things worse.
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
$endgroup$
add a comment |
$begingroup$
The problem you have is caused by the LM317 and the voltage you have available to it.
The LM317 requires a difference of 3V between the input and output voltage. You have 15V from your powersupply (which you drop to 14 with the LM2596.) That's only 1.5V above the output voltage (13.5V) that you need for the battery.
You really only have two choices:
- Use a higher input voltage (at least 17V)
- Use a different current regulator that can work with the available voltage.
In any case, the LM2596 is only making things worse.
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
$endgroup$
The problem you have is caused by the LM317 and the voltage you have available to it.
The LM317 requires a difference of 3V between the input and output voltage. You have 15V from your powersupply (which you drop to 14 with the LM2596.) That's only 1.5V above the output voltage (13.5V) that you need for the battery.
You really only have two choices:
- Use a higher input voltage (at least 17V)
- Use a different current regulator that can work with the available voltage.
In any case, the LM2596 is only making things worse.
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
answered Aug 7 at 11:31
JREJRE
28k7 gold badges53 silver badges91 bronze badges
28k7 gold badges53 silver badges91 bronze badges
add a comment |
add a comment |
$begingroup$
Also, if the power supply is switched off and the 12V battery is still attached, there's a negative voltage difference (Vin-Vout) on the LM317, a situation not mentioned in the data sheet as far as I can see. If that causes a leak current, it would drain the battery.
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
$endgroup$
add a comment |
$begingroup$
Also, if the power supply is switched off and the 12V battery is still attached, there's a negative voltage difference (Vin-Vout) on the LM317, a situation not mentioned in the data sheet as far as I can see. If that causes a leak current, it would drain the battery.
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
$endgroup$
add a comment |
$begingroup$
Also, if the power supply is switched off and the 12V battery is still attached, there's a negative voltage difference (Vin-Vout) on the LM317, a situation not mentioned in the data sheet as far as I can see. If that causes a leak current, it would drain the battery.
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
$endgroup$
Also, if the power supply is switched off and the 12V battery is still attached, there's a negative voltage difference (Vin-Vout) on the LM317, a situation not mentioned in the data sheet as far as I can see. If that causes a leak current, it would drain the battery.
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
answered Aug 8 at 8:02
Arthur HoornwegArthur Hoornweg
1
1
add a comment |
add a comment |
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4
$begingroup$
Why do you put the Buck converter in between? Just to get rid of that one volt?
$endgroup$
– jusaca
Aug 7 at 11:23