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What did I do wrong?

john monks

Mar 9, 2012
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You need to look at the capacity of the battery. It may be 1AH or 1 ampere hour. It may be 10AH. You have to look it up. It may be written on the side of the battery.
But let's just say that the battery is 1AH. From your schematic it looks like the charge current is about 0.7 amperes or 700ma. To bring up a dead battery to full charge is about 1.6/0.7 or about 2.3 hours. If this is the case I would suggest changing the circuit to reduce the charge current.
 

1-3-2-4

May 7, 2010
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It's a 17 Ah battery I think most SLA batteries say to limit the charge under 1.5 or 1.3 A anyways I wanted to do a real test but need something to drain the battery at a quicker rate.

When everything is done I'm going to only be pulling 350 mA from the battery so that should give me around 48 hours of run time if I have that right.
 

john monks

Mar 9, 2012
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I don't know what you mean by the other transistor. Do you mean the P2N2222A?
By the top half of this circuit do you mean the part with the relay?
But anyway I don't see any problem with testing the two halves separately.
 

1-3-2-4

May 7, 2010
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I don't know what you mean by the other transistor. Do you mean the P2N2222A?
By the top half of this circuit do you mean the part with the relay?
But anyway I don't see any problem with testing the two halves separately.

No, what I meant to say was the two 4v7 zeners, I don't have those yet. and the top half I mean everything above the LED module and mains input. reason why I said because say I have a extended outage and longer then 48 hours then I need some type of low voltage cutoff so I wont drain my SLA battery too low.
 

TedA

Sep 26, 2011
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As already mentioned, an LED without a current limiting resistor is not a suitable load for the charger circuit.

You really need a lower resistance trimpot, but the 100k one might work. You might disconnect the trimpot from the circuit, and see if you can set it to values around 2k to 3k ohms. Sometimes pots jump from some small value to zero, and small for a 100k pot may be higher than the desired set resistance.

You can put a resistor network around the 100k pot to give a useful adjustment range. You might give us a list of resistors that you have and we can suggest what to use.

The trimpot adjusts the maximum output voltage at a moderate charging current. It does not adjust the current limit. The current limit is fixed, and determined by R1 and the characteristics of Q1. ( Down to some non-zero output voltage. Pull the output too low, and the current limit goes up to whatever the LM317 will provide.)

The symptom of poor voltage regulation between the no-load and loaded conditions may be the result of providing too large a bleeder resistance on the LM317 output. This may not really matter in the application, as a few mA of "float" current may not harm the battery.

See the "Minimum Load Current" spec on the data sheet. This current, plus a bit more, is usually absorbed by a bleeder resistance that also forms a voltage divider for the Adjustment Pin of the LM317.

The 470 ohm value of R3, and thus of R2 and R4, may be too large for the LM317 to regulate with no external load. The LM317 idle current may be as high as 5mA. The data sheet application circuits all suggest a 240 ohm value for R3. There is a reason that this value is not higher. The LM317 idle current varies from one unit to another, with temperature, and possibly the phase of the moon.

It is possible that when you adjust the output with no load, you are adjusting the load resistance to give the correct output voltage based on the LM317 idle output current and the total resistance of R2 + R3 + R4. At this point, the voltage across R3 can be much higher than the 1.25V Reference Voltage of the LM317.

Increasing the external load current will then pull down the output voltage until the LM317 begins to regulate, when the load current finally exceeds the idle current. At this voltage, the output becomes much stiffer, so the output voltage drops very little with further increases in load current.

The answer may be to just adjust the output voltage with some load, say a 1k resistor.

As already suggested, you should add a series resistor to the base of Q1. Even 100 ohms might save the transistor.

If you connect a fully discharged battery, the current limit should function, and the LED, D1, should light, if perhaps dimly. There should be only about 2.6mA to drive this LED. And if the charger output is pulled too low, the LED may not light, and the current limit will be higher, as well. Possibly over 3A, until the LM317 warms-up.

This charger circuit will drain the battery if the power source fails. Often a diode is used to isolate the battery, with the regulator output voltage increased to compensate for the diode drop.

I hope that some of this makes sense to you. I'm sure you can make this work. Just keep on asking questions.

Ted
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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It might help if you gave us an overview of what you want to DO with this project. I see so far there is a power supply which is actually a battery charger, and some LEDs, and possibly a battery somewhere.
I had to trade in my crystal ball, so he'p me out here!
 

1-3-2-4

May 7, 2010
53
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The battery charger part is done pretty much, all that's left for me to do is to put on the 5K trim pots and the relay, I do have diodes, they are way overkill since I was doing a project on a car and I ended up finding a simpler way to do it, so my diodes are rated for 600V.

having said that looking at the 2nd picture on here I can do something like this.

http://homemadecircuitsandschematic...2/how-to-make-simple-low-battery-voltage.html

Just put it on the battery and the load goes to the relay that way I wont totally drain my battery.
 

1-3-2-4

May 7, 2010
53
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Messages
53
As already mentioned, an LED without a current limiting resistor is not a suitable load for the charger circuit.

You really need a lower resistance trimpot, but the 100k one might work. You might disconnect the trimpot from the circuit, and see if you can set it to values around 2k to 3k ohms. Sometimes pots jump from some small value to zero, and small for a 100k pot may be higher than the desired set resistance.

You can put a resistor network around the 100k pot to give a useful adjustment range. You might give us a list of resistors that you have and we can suggest what to use.

The trimpot adjusts the maximum output voltage at a moderate charging current. It does not adjust the current limit. The current limit is fixed, and determined by R1 and the characteristics of Q1. ( Down to some non-zero output voltage. Pull the output too low, and the current limit goes up to whatever the LM317 will provide.)

The symptom of poor voltage regulation between the no-load and loaded conditions may be the result of providing too large a bleeder resistance on the LM317 output. This may not really matter in the application, as a few mA of "float" current may not harm the battery.

See the "Minimum Load Current" spec on the data sheet. This current, plus a bit more, is usually absorbed by a bleeder resistance that also forms a voltage divider for the Adjustment Pin of the LM317.

The 470 ohm value of R3, and thus of R2 and R4, may be too large for the LM317 to regulate with no external load. The LM317 idle current may be as high as 5mA. The data sheet application circuits all suggest a 240 ohm value for R3. There is a reason that this value is not higher. The LM317 idle current varies from one unit to another, with temperature, and possibly the phase of the moon.

It is possible that when you adjust the output with no load, you are adjusting the load resistance to give the correct output voltage based on the LM317 idle output current and the total resistance of R2 + R3 + R4. At this point, the voltage across R3 can be much higher than the 1.25V Reference Voltage of the LM317.

Increasing the external load current will then pull down the output voltage until the LM317 begins to regulate, when the load current finally exceeds the idle current. At this voltage, the output becomes much stiffer, so the output voltage drops very little with further increases in load current.

The answer may be to just adjust the output voltage with some load, say a 1k resistor.

As already suggested, you should add a series resistor to the base of Q1. Even 100 ohms might save the transistor.

If you connect a fully discharged battery, the current limit should function, and the LED, D1, should light, if perhaps dimly. There should be only about 2.6mA to drive this LED. And if the charger output is pulled too low, the LED may not light, and the current limit will be higher, as well. Possibly over 3A, until the LM317 warms-up.

This charger circuit will drain the battery if the power source fails. Often a diode is used to isolate the battery, with the regulator output voltage increased to compensate for the diode drop.

I hope that some of this makes sense to you. I'm sure you can make this work. Just keep on asking questions.

Ted


I switched from the 12V adapter to one that is 18.6V 3.6A I noticed the charger is quicker to charge, it's been on for maybe 2 hours now and the led is still glowing but the voltage has slowly went to to 13V the heatsink for the LM317 is around 66 C

What I do wonder is what determines when the led stops glowing? I ran the battery pretty hard yesterday testing taking it down to around 11.6 V
 

1-3-2-4

May 7, 2010
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I put the trimpot in early this morning going to place the diode inline on the output line.

birfa.jpg
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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I think the LED is there to act as a voltage drop, not to indicate anything.

However the LED will probably go off when the relay is turned off.
 

TedA

Sep 26, 2011
156
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The LED indicates that the charger is in the constant current mode. This is the first part of the battery charging sequence, when the battery is being charged rapidly, and its voltage is relatively low. The battery voltage gradually rises during this part of the charging sequence.

Once the battery reaches the voltage set point for the charger ( what the trimpot sets ), the LED will dim, and go out. Once the LED goes out, the charger is in the constant voltage portion of the charging sequence. While the voltage is held constant by the charger, the charging current gradually tapers to a low value.

Once the charging current falls to a very low value, the battery is fully charged.

Ted
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Clearly we're talking about different circuits. I looked at the last one posted (silly me)
 

TedA

Sep 26, 2011
156
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Steve,

I thought I was answering the OP's last LED question, which appeared to me to concern the LED in the original circuit he linked early-on.

However, I must admit that there is an LED in the low battery detector circuit as well.

I have to say that this low battery detector appears to be designed to finish off the battery, once it gets low enough. Given enough time, it can discharge a gel cell far more than is wise. Ideally, a low battery indicator should have a zero power indicator.

Ted
 

1-3-2-4

May 7, 2010
53
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Yeah that's why I'm going to go with the max8212cpa instead of that lm741..

Now one issue I'm dealing with now is the fan and my input voltage.. so my input is 18.6 V and my fans are 12 V rated..

just how do I go about hooking the fan up without blowing it out?
 

bbrummett

Mar 7, 2013
2
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Mar 7, 2013
Messages
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What is the current limit set to on the power supply to your circuit? good place to start looking when large Vdrops occur asumming no shorts in the circuit design.
 

1-3-2-4

May 7, 2010
53
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The limit on the supply is 3.5 A I think I can just do with a IRF510 because I will be making a temp controlled fan that will be using a 10k trim pot and a 10k thermistor.
 
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