hi, i have a 24v 3amp. power supply which uses tl431 and pc817 for voltage control. I added a trimpot and replaced output caps for output voltage adjustment from 15v-35v output. Now , i want to use this power supply as a lithium charger for my 18650 7s pack. For that i need to add a cheap and simple constant current feature. I have some lm358 and tl431 ICs laying around. IF these can be used in the circuit that would be great. I don't need an accurate current control, just need a simple and dirt cheap circuit for CC feature.
The feedback circuit looks something like this
(Attachment Link)
I have a question will the output current vary with variation in voltage or will be stable at minimum 1A if a 1.2ohm resistor is used? Sorry if I am wrong but I think this circuit is more of a current regulator circuit which will only regulate the output current
Hey, thanks for a detailed yet simple explanation. I have a question will the output current vary with variation in voltage or will be stable at minimum 1A if a 1.2ohm resistor is used? Sorry if I am wrong but I think this circuit is more of a current regulator circuit which will only regulate the output current and the extra power will be wasted through the power resistor in form of heat. Am I right?
I think I need Constant Current source circuit which can adjust (lower) the output voltage in order to output the SET current. I have read online about LM358 op-amp CC circuit but failed to implement in my current power supply.
A guy at youtube showed this circuit but it didnt worked. (I am using 29 volts for the 7S pack)
(Attachment Link)
And yes, I am using protected cells (BMS) with balancer.
Hello again,
Here are two circuits.
One is the original with the added opto but drawn more clear so you can see the right way to connect it. Notice that the Vout has moved. Vout Old is where the rectifiers typically connect, and Vout New is the actual new Vout to connect to the load.
The other is almost the same but uses a PNP transistor to sense current. 0.6 to 0.7 Ohms will provide 1 amp output current limit.
Hello again,
Here are two circuits.
One is the original with the added opto but drawn more clear so you can see the right way to connect it. Notice that the Vout has moved. Vout Old is where the rectifiers typically connect, and Vout New is the actual new Vout to connect to the load.
The other is almost the same but uses a PNP transistor to sense current. 0.6 to 0.7 Ohms will provide 1 amp output current limit.
Sharing you the exact circuit design of my power supply. Where to put the transistor?
Hello again,
Here are two circuits.
One is the original with the added opto but drawn more clear so you can see the right way to connect it. Notice that the Vout has moved. Vout Old is where the rectifiers typically connect, and Vout New is the actual new Vout to connect to the load.
The other is almost the same but uses a PNP transistor to sense current. 0.6 to 0.7 Ohms will provide 1 amp output current limit.
Sharing you the exact circuit design of my power supply. Where to put the transistor?
The base emitter resistor should be a power resistor, the one that drives the second opto LED can be 1/4 watt.
Looking at your drawing, is the output at the point right at the right side of that inductor?
LATER:
Assuming that is the output as it was, here is a drawing that makes the addition even simpler yet. Note no second opto needed the feedback can be tapped into at the IC chip feedback node.
Note i need to know the value of R3 and the lower resistor also to set R7 value.
The 100 Ohm and R7 can be 1/4 watt, RCL has to be a power resistor.
If RCL is 0.6 to 0.7 Ohms (for 1 amp output) then it has to be rated for at least 2 watts but 5 watts is better.
Diode is any Si like 1N4002 or similar.
I'll check out your transistor to see if it will work. [looks like it should work]
Hello again,
Here are two circuits.
One is the original with the added opto but drawn more clear so you can see the right way to connect it. Notice that the Vout has moved. Vout Old is where the rectifiers typically connect, and Vout New is the actual new Vout to connect to the load.
The other is almost the same but uses a PNP transistor to sense current. 0.6 to 0.7 Ohms will provide 1 amp output current limit.
Sharing you the exact circuit design of my power supply. Where to put the transistor?
The base emitter resistor should be a power resistor, the one that drives the second opto LED can be 1/4 watt.
Looking at your drawing, is the output at the point right at the right side of that inductor?
LATER:
Assuming that is the output as it was, here is a drawing that makes the addition even simpler yet. Note no second opto needed the feedback can be tapped into at the IC chip feedback node.
Note i need to know the value of R3 and the lower resistor also to set R7 value.
The 100 Ohm and R7 can be 1/4 watt, RCL has to be a power resistor.
If RCL is 0.6 to 0.7 Ohms (for 1 amp output) then it has to be rated for at least 2 watts but 5 watts is better.
Diode is any Si like 1N4002 or similar.
I'll check out your transistor to see if it will work. [looks like it should work]
Hi, i only have 10 ohm and 0.1 ohm power cement resistor readily available. Can the circuit be modified for 1-2 amp output.
R3 is a 10k resistor and below that is a 10k pot for variable voltage.
Hello again,
Here are two circuits.
One is the original with the added opto but drawn more clear so you can see the right way to connect it. Notice that the Vout has moved. Vout Old is where the rectifiers typically connect, and Vout New is the actual new Vout to connect to the load.
The other is almost the same but uses a PNP transistor to sense current. 0.6 to 0.7 Ohms will provide 1 amp output current limit.
Sharing you the exact circuit design of my power supply. Where to put the transistor?
The base emitter resistor should be a power resistor, the one that drives the second opto LED can be 1/4 watt.
Looking at your drawing, is the output at the point right at the right side of that inductor?
LATER:
Assuming that is the output as it was, here is a drawing that makes the addition even simpler yet. Note no second opto needed the feedback can be tapped into at the IC chip feedback node.
Note i need to know the value of R3 and the lower resistor also to set R7 value.
The 100 Ohm and R7 can be 1/4 watt, RCL has to be a power resistor.
If RCL is 0.6 to 0.7 Ohms (for 1 amp output) then it has to be rated for at least 2 watts but 5 watts is better.
Diode is any Si like 1N4002 or similar.
I'll check out your transistor to see if it will work. [looks like it should work]
Hi, i only have 10 ohm and 0.1 ohm power cement resistor readily available. Can the circuit be modified for 1-2 amp output.
R3 is a 10k resistor and below that is a 10k pot for variable voltage.
Hello again,
Oh well 10 Ohm is way too large and 0.1 Ohm is way too small.
The way this works is that the resistor RCL drops some voltage as the output current increases. As the voltage across RCL raches near 0.6v the transistor begins to conduct emitter to collector and that drives the node between the two original feedback resistors as shown on the drawing. In order to get to 0.6v and drop that much at 1 amp output, the resistor has to be sized according to Ohm's Law which says:
V=I*R
where I is the current and R is the resistance and V the voltage.
Solving for R we get:
R=V/I
and since we need 0.6v at 1 amp this is very simply:
R=0.6/1
which of course equals 0.6 Ohms, although something close to that would be good enough like 0.7 Ohms possibly even 0.5 Ohms for a little more current output.
But 10 Ohms and 0.1 Ohms is just too far off, With 0.1 Ohms the current would not begin to limit until about 6 amps, and with 10 Ohms the current would limit around 60 milliamps.
Even 1 Ohm would allow 600 milliamps.
So as is, the circuit would not work with 0.1 Ohms or 10 Ohms. I am not sure if we can do it that simply with either of those two. I would suggest you fine a source for electronic parts if you want to build circuits yourself.
I dont know where you can get them in India but i know there are a lot of professors that teach there so they must be getting parts from somewhere. You could check with one of the universities if you cant find any places else maybe they would give you one or charge a small fee.
There are online outlets like Mouser and Digikey but there are others too maybe you could check around the web.
In the mean time i will give that 0.1 Ohm some thought but the 10 Ohm passing 1 amp would dissipate 10 watts i doubt you want to use that. The problem with a 0.1 Ohm is that we would have to incorporate a differential amplifier which would take another nearly identical transistor.
What other parts do you have on hand or can get if you need them?
I have many old circuit boards from which I can salvage transistors, I do have 2pcs. same B772D PNP transistor. Have many type of power MOSFETS (N-channel). Working Telecom rectifiers of 48v output DC voltage and many other faulty equipments. One more thing I noticed is you removed the second optocoupler for Current feedback in the last circuit. Will the circuit work without it?
QuoteI have many old circuit boards from which I can salvage transistors, I do have 2pcs. same B772D PNP transistor. Have many type of power MOSFETS (N-channel). Working Telecom rectifiers of 48v output DC voltage and many other faulty equipments. One more thing I noticed is you removed the second optocoupler for Current feedback in the last circuit. Will the circuit work without it?
Yes the idea was to provide a second feedback path to the secondary side regulator IC itself rather than to the primary side of the transformer with the AC regulating chip. That means it controls the secondary side regulator instead of the primary side AC regulator and the secondary side regulator controls the primary side regulator so it limits the current just the same.
I had contemplated using a custom differential amplifier made of transistors but it gets a little messy and you'd probably have to buy some parts anyway.
An op amp might be used but it would have to be a low voltage one im not sure if you have that either and not sure if you would get short circuit protection. That's unless you can provide a second DC power supply such as a DC wall wart to provide the power for the op amp. Then you can use a 0.1 Ohm resistor for current sense and it shouldnt get hot either.
They make special chips for sensing current but again you'd have to buy that.
There are a lot of options but they are all more involved than just getting a 0.6 Ohm power resistor :-)
However, see if you can use a second DC wall wart without too much trouble. That opens the door to a lot of other options.
If you think you can guarantee that the battery voltage never drops below 5 volts an op amp may be an option too, but the voltage would have to always be present when charging. A variation on this idea is to use two diodes and a second battery that could help provide power to the op amp in the event of a short circuit or very low output voltage.
Im not sure if you have any op amps though.
QuoteI have many old circuit boards from which I can salvage transistors, I do have 2pcs. same B772D PNP transistor. Have many type of power MOSFETS (N-channel). Working Telecom rectifiers of 48v output DC voltage and many other faulty equipments. One more thing I noticed is you removed the second optocoupler for Current feedback in the last circuit. Will the circuit work without it?
Yes the idea was to provide a second feedback path to the secondary side regulator IC itself rather than to the primary side of the transformer with the AC regulating chip. That means it controls the secondary side regulator instead of the primary side AC regulator and the secondary side regulator controls the primary side regulator so it limits the current just the same.
I had contemplated using a custom differential amplifier made of transistors but it gets a little messy and you'd probably have to buy some parts anyway.
An op amp might be used but it would have to be a low voltage one im not sure if you have that either and not sure if you would get short circuit protection. That's unless you can provide a second DC power supply such as a DC wall wart to provide the power for the op amp. Then you can use a 0.1 Ohm resistor for current sense and it shouldnt get hot either.
They make special chips for sensing current but again you'd have to buy that.
There are a lot of options but they are all more involved than just getting a 0.6 Ohm power resistor :-)
However, see if you can use a second DC wall wart without too much trouble. That opens the door to a lot of other options.
If you think you can guarantee that the battery voltage never drops below 5 volts an op amp may be an option too, but the voltage would have to always be present when charging. A variation on this idea is to use two diodes and a second battery that could help provide power to the op amp in the event of a short circuit or very low output voltage.
Im not sure if you have any op amps though.
Ok then, I will find a 0.5, 0.6 power resistor. In the meantime, can you clarify:
a). Value of resistor R7?
b). Is the (100) resistor on the transistor base pin a 100ohm?
c). You didn't include the other Optocoupler at the transistor collector pin as used in previous circuits, or you forget to include one?
QuoteI have many old circuit boards from which I can salvage transistors, I do have 2pcs. same B772D PNP transistor. Have many type of power MOSFETS (N-channel). Working Telecom rectifiers of 48v output DC voltage and many other faulty equipments. One more thing I noticed is you removed the second optocoupler for Current feedback in the last circuit. Will the circuit work without it?
Yes the idea was to provide a second feedback path to the secondary side regulator IC itself rather than to the primary side of the transformer with the AC regulating chip. That means it controls the secondary side regulator instead of the primary side AC regulator and the secondary side regulator controls the primary side regulator so it limits the current just the same.
I had contemplated using a custom differential amplifier made of transistors but it gets a little messy and you'd probably have to buy some parts anyway.
An op amp might be used but it would have to be a low voltage one im not sure if you have that either and not sure if you would get short circuit protection. That's unless you can provide a second DC power supply such as a DC wall wart to provide the power for the op amp. Then you can use a 0.1 Ohm resistor for current sense and it shouldnt get hot either.
They make special chips for sensing current but again you'd have to buy that.
There are a lot of options but they are all more involved than just getting a 0.6 Ohm power resistor :-)
However, see if you can use a second DC wall wart without too much trouble. That opens the door to a lot of other options.
If you think you can guarantee that the battery voltage never drops below 5 volts an op amp may be an option too, but the voltage would have to always be present when charging. A variation on this idea is to use two diodes and a second battery that could help provide power to the op amp in the event of a short circuit or very low output voltage.
Im not sure if you have any op amps though.
Ok then, I will find a 0.5, 0.6 power resistor. In the meantime, can you clarify:
a). Value of resistor R7?
b). Is the (100) resistor on the transistor base pin a 100ohm?
c). You didn't include the other Optocoupler at the transistor collector pin as used in previous circuits, or you forget to include one?
Hi,
R7 can be 1k but you can try higher too like 2.2k first. You need to test it for current limit too before you use it.
Yes 100 Ohms 1/4 watt or 1/2 watt.
I thought i told you this new circuit doesnt need an extra opto because the feedback on the 431 chip is already isolated from the mains line.
Yeah 0.6, 0.7, even 1 Ohm will give you about 600ma.
I have assembled the circuit and used a 0.5ohm resistor and 1k at R7, however I am getting only 600ma at output. How to increase the output? The circuit is fine now but charging takes much time for huge battery (100ah)
I have assembled the circuit and used a 0.5ohm resistor and 1k at R7, however I am getting only 600ma at output. How to increase the output? The circuit is fine now but charging takes much time for huge battery (100ah)0.5 Ohms would provide a bit over 1 amp but that's if everything is perfect. Circuits are not always perfect because of various reasons including component variations.
If you need more current (like 1 amp) then you would have to decrease the 0.5 Ohm resistor.
If you are getting 600ma and you are sure you are getting that, then if you parallel another 0.5 Ohm resistor you would get twice that or 1.2 amps, roughly.
I have assembled the circuit and used a 0.5ohm resistor and 1k at R7, however I am getting only 600ma at output. How to increase the output? The circuit is fine now but charging takes much time for huge battery (100ah)Keep in mind that if the transistor shorts collector to emitter there will be no current limit and no voltage regulation. The voltage regulation is extremely important with these batteries because over charging could cause fire or explosion, and those kinds of fire are hard to put out.
I have assembled the circuit and used a 0.5ohm resistor and 1k at R7, however I am getting only 600ma at output. How to increase the output? The circuit is fine now but charging takes much time for huge battery (100ah)100AH battery is a pretty hefty size. In theory at 1 amp that would take 100 hours to charge, which is more than 4 earth days. It may be safer than high current though, but the other thing to check is that ti actually charges with as little as 1 amp. It may take more because 1 amp is pretty low for a 100AH battery. That would be C/100 which could be too low to properly charge this battery.
But wow C/100 that must have cost quite a bit to purchase.
Wow you got four 200AHr batteries for free?
The regulator chip handles the voltage regulation along with those two resistors. If you want to use a second independent circuit then more parts will have to be added. This would include a voltage reference IC chip, and a comparator and a few resistors. The reference voltage connects to one pin of the comparator, the other input pin goes to a set of resistors that drop the output voltage down so it matches the reference voltage almost perfectly but will sense a voltage that is slightly higher than the output. If the output goes too high, the comparator trips and its output goes to a low (or high) state. That can be used to either:
*cut the voltage off
*just light an LED or buzz a buzzer.
*both of those above
So can you get a voltage reference IC chip somewhere? You might get away with another 431 chip but a regular voltage reference IC is easier to use and you can get some really really stable types.
You also need a comparator such as the LM339 or equivalent (they make a 2 unit model also, that's a 4 unit model meaning there are 4 comparators in it. A search would turn up these parts. Obviously an 8 pin part will be smaller and easier to use. I suggest getting all 'through hole' parts though unless you are very familiar with SMD soldering and probably have to make a circuit board that takes those parts too.