12V battery recharger circuit

[boylesg]

http://www.talkingelectronics.com/projects/100 IC Circuits/1-100_IC-Ccts.html#20

Could some one explain to me why you would put the transistor on common pin of the voltage regulator as opposed to the output line, i.e. like a light switch.

I do not have enough of an understanding of how a linear voltage regulator works.

 

[betwixt]

The LM317 is an adjustable regulator. Essentially it provides the high current passing capability of a 1.25V fixed voltage regulator but is optimized so as little current as possoble flows through it's ADJ pin. What it does is try to keep 1.25V between it's ADJ and output pins (OUT = ADJ + 1.25V) so that you can lift the ADJ pin to a higher voltage with a simple circuit but still have high current capability.

The Zener diode regulates 12V so the potentiometer can lift the ADJ pin between ground and 12V, giving an output voltage between 1.25V and 13.25V. The transistor is wired to work as a current limiter. The output current flows through the 6R8/100R resistors and drops a voltage across them. Some of that is tapped off on the 100R to the base of the transistor so when Vbe is reached, the transistor starts to conduct. As it conducts, it pulls the ADJ voltage lower and hence reduces the output voltage to limit the current into the load.

There are actually two problems in that schematic, it should have a load resistor across it's output to maintain a minimum current flow and the output should also have a capacitor across it to maintain stability. As it stands, you might find the output voltage is higher than predicted and not very stable.

Brian.

 

[boylesg]

The LM317 is an adjustable regulator. Essentially it provides the high current passing capability of a 1.25V fixed voltage regulator but is optimized so as little current as possoble flows through it's ADJ pin. What it does is try to keep 1.25V between it's ADJ and output pins (OUT = ADJ + 1.25V) so that you can lift the ADJ pin to a higher voltage with a simple circuit but still have high current capability.

The Zener diode regulates 12V so the potentiometer can lift the ADJ pin between ground and 12V, giving an output voltage between 1.25V and 13.25V. The transistor is wired to work as a current limiter. The output current flows through the 6R8/100R resistors and drops a voltage across them. Some of that is tapped off on the 100R to the base of the transistor so when Vbe is reached, the transistor starts to conduct. As it conducts, it pulls the ADJ voltage lower and hence reduces the output voltage to limit the current into the load.

There are actually two problems in that schematic, it should have a load resistor across it's output to maintain a minimum current flow and the output should also have a capacitor across it to maintain stability. As it stands, you might find the output voltage is higher than predicted and not very stable.

Brian.

Sorry Brian I should have specified which circuit I was looking at.

It is the gel cell charger which does not have a zener diode nor a 6.8R/100R resistor combination.

 

[betwixt]

Sorry...!
The same explanation applies though, the output voltage is 1.25V + Vf of the LED + collector to emitter voltage of the transistor. As the voltage drop across the 1 Ohm resistor makes the transistor conduct, it pulls the collector voltage down and reduces the output voltage, hopefully making less current flow into the cell. It is therefore self limiting. Again, not a clever design because of the high dynamic resistance of the LED and it's sensitivity to light but better than nothing.

Brian.

 

[boylesg]

Sorry...!
The same explanation applies though, the output voltage is 1.25V + Vf of the LED + collector to emitter voltage of the transistor. As the voltage drop across the 1 Ohm resistor makes the transistor conduct, it pulls the collector voltage down and reduces the output voltage, hopefully making less current flow into the cell. It is therefore self limiting. Again, not a clever design because of the high dynamic resistance of the LED and it's sensitivity to light but better than nothing.

Brian.

Well I suppose you could cover the LED with duct tape to solve the light problem.

OK let be get this straight.....

I more or less get how the LM317 works:

1) Reference voltage of 1.25V between OUT and ADJ.
2) As the load draws more current less current flows through the voltage divider and so the rference voltage falls below 1.25V.
3) That results in the LM317 passing more current in order to raise the voltage at OUT.
4) That raises the current through the voltage divider and therefore the reference voltage back to 1.25V so that the current through the LM317 does not rise any further.

OK so far?

With the transistor and LED on ADJ....

1) With the battery fully charged there must be a small current flowing through 470R + 5K + 2.2K to the battery negative which means that there is a negative voltage on the transistor base.
2) The transistor must therefore be in a high resistance/high voltage state across the CE junction.
3) Hence little current flows to ground through the transistor and the reference voltage remains close to 1.25V....well perhaps not quite because it is supposed to keep trickle charging a full battery at a few tens of mA.
4) If you have a flat battery then that must mean that there is much less current flowing through 470R + 5K + 2.2K to the battery negative.
5) So that would mean a greater voltage across the 1R resistor and at the transistor base.
6) So the transistor begins conducting across its BE and CE junctions, which puts it in a low resistance/low voltage state.
7) That drops the reference voltage and causes the LM317 to conduct more current.

Does that sound about right Brian?

Why do you even need the transistor and LED etc? Why not just a stock standard LM317 circuit? Wouldn't it work just as effectively at charging the battery and reducing the current as it reaches full charge?

 

[betwixt]

Almost right.

The LM317 works like a 'floating' 1.25V regulator. Internally, it uses the voltage on the ADJ pin to set the output voltage and it is designed so as little current as possible flows through the ADJ pin. If the ADJ pin is grounded it behaves just like a conventional 7805 type of regulator but with 1.25V at the output pin, if you raise the ADJ voltage, the output voltage also rises. If the ADJ pin is shorted to the output pin it will fail to regulate but give a high voltage out.

Ignoring the LED and transistor for now, the ADJ pin is connected to a potential divider made from 470 Ohms at the top and between (2.2K + 1 Ohm) and (7.2K + 1 Ohm) at the bottom. Adjusting the variable resistor sets the ADJ pin voltage and hence the overall voltage to the battery.

When current flows in to the battery, a voltage is dropped across the 1 Ohm resistor (1V per Amp by Ohms law). The B-E junction of the transistor is across the resistor so when the current exceeds about 0.6V it starts to conduct between C-E and as it is in parallel with the lower part of the potential divider, it pulls the voltage at ADJ lower and reduces the output voltage. The LED is not essential to it's operation, in fact it reduces the effect of the current limiting because ADJ can never go completely to 0V so the output will always be at least 1.25V + LED Vf or about 3V. The designer added it just so you could see when the transistor is conducting and limiting the output, a kind of crude 'overload' indicator. The different resistor values at the bottom of the schematic are alternatives to the 1 Ohm in the main schematic, different values will drop 0.6V at different currents so you can select a value appropriate to the limit you want.

Don't expect very good results from a simple circuit like this but it will work to some degree.

Brian.

 

[boylesg]

Almost right.

The LM317 works like a 'floating' 1.25V regulator. Internally, it uses the voltage on the ADJ pin to set the output voltage and it is designed so as little current as possible flows through the ADJ pin. If the ADJ pin is grounded it behaves just like a conventional 7805 type of regulator but with 1.25V at the output pin, if you raise the ADJ voltage, the output voltage also rises. If the ADJ pin is shorted to the output pin it will fail to regulate but give a high voltage out.

Ignoring the LED and transistor for now, the ADJ pin is connected to a potential divider made from 470 Ohms at the top and between (2.2K + 1 Ohm) and (7.2K + 1 Ohm) at the bottom. Adjusting the variable resistor sets the ADJ pin voltage and hence the overall voltage to the battery.

When current flows in to the battery, a voltage is dropped across the 1 Ohm resistor (1V per Amp by Ohms law). The B-E junction of the transistor is across the resistor so when the current exceeds about 0.6V it starts to conduct between C-E and as it is in parallel with the lower part of the potential divider, it pulls the voltage at ADJ lower and reduces the output voltage. The LED is not essential to it's operation, in fact it reduces the effect of the current limiting because ADJ can never go completely to 0V so the output will always be at least 1.25V + LED Vf or about 3V. The designer added it just so you could see when the transistor is conducting and limiting the output, a kind of crude 'overload' indicator. The different resistor values at the bottom of the schematic are alternatives to the 1 Ohm in the main schematic, different values will drop 0.6V at different currents so you can select a value appropriate to the limit you want.

Don't expect very good results from a simple circuit like this but it will work to some degree.

Brian.

When I tried this circuit on my bread board I could never get the LED to light up, even if I used a 3mm red LED.

When I checked the current through the LED with my multimeter, it was only ever 1mA or less.

I suppose the BC548 could be buggered a bit, or the LM317 itself. Any other suggestions?

When I run the soldered up circuit on my 12V gel cell, which I seem to be only able to charge up to 10V with a car battery charger, I get around 50mA going into the battery.

If I connect up a flat 6V lantern battery I get 2.5A going into the battery (very briefly).
That was to simulate a flat gel cell.

Given the circuit is supposed have a limit of 650mA output, something is wrong some where.

 

[betwixt]

The LED current will be quite low, better designs use the LED to indicate an over current circuit has kicked into action rather than using the current control signal itself. If you look at the LM317 data sheets they show this exact circuit but with one extra component fitted, a 100 Ohm resistor in series with the transistor base pin. This is a good idea because normally the base curent is extremely low and the effect of the resistor is insignificant but in the case of a serious overload or shorted output, the whole output of the LM317 is connected across the base and emitter in the design you show. That would certainly fry the transistor in a fraction of a second!

Brian.

 

[boylesg]

The LED current will be quite low, better designs use the LED to indicate an over current circuit has kicked into action rather than using the current control signal itself. If you look at the LM317 data sheets they show this exact circuit but with one extra component fitted, a 100 Ohm resistor in series with the transistor base pin. This is a good idea because normally the base curent is extremely low and the effect of the resistor is insignificant but in the case of a serious overload or shorted output, the whole output of the LM317 is connected across the base and emitter in the design you show. That would certainly fry the transistor in a fraction of a second!

Brian.

If I remove the LED then how would you add it back to indicate either charging complete or charging is proceeding?

I have fiddled around with the circuit but I can't figure out how to make it work.

 

[betwixt]

You can't do it with a simple circuit like that. Removing (replacing with a link) the LED will improve the current limiting slightly although with that circuit it will never completely protect against a shorted or almost shorted battery.

A better design would still use a sense resistor in series with the output, that part is pretty much essential to check the current, but instead of using the voltge it drops to turn a transistor on, it would connect to a comparator. The comparator would compare the voltage drop (= output current) against a reference voltage to decide if too much was flowing and if necessary cut off or 'throttle back' a series pass regulator. The output of the comparator would be a logic signal, indicating whether the current limit was operating or not and so could be used to drive an LED at whatever brightness you want. The actual limiting current can be adjusted with a small potentiometer to set the reference the comparator compares against so you wouldn't have to change sense resistors.

Brian.

 

[boylesg]

You obviously mean something like this which is also in the LM317 datasheet I am looking at:

Although I don't need 5A output.

I have tried this one in multisim and I cannot get it to work. I can't adjust the voltage level of the output.

It doesn't specify what V+ should be and it does not specify what R8 should be anyway, so I am left guessing.

 

[boylesg]

With this circuit I can change it to 13.4V output by changing the 1.1k pot to a 5k pot.

And, at least in multisim, I think I have figured out how to add a charge indicator.



LED2 only comes on when the battery voltage reaches 12V

 

[boylesg]

What do you think of this design Brian?

Seems to work in multisim at least.

The current tapers off to tens of mA as the battery reaches full 12V charge and the comparator turns the LED on at this point.

 

[betwixt]

Sorry for not replying quicker - I finally got so frustrated at slow internet that I installed a direct satellite link - woo hoo! - it's 200 times faster now

The last circuit looks OK, the transistor starves it self of base current when the battery voltage approaches the LM317 voltage so it is inherently overcharge proof. The comparator does nothing except operate the LED when the current reaches a certain point so it isn't actually telling you if the circuit is limiting at all, just whether a certain current has been exceeded. I am a little concerned at the value of sense resistor being as high as 12 Ohms, I would have used something much smaller and amplified the voltage across it if necessary. I'm not sure why the 10K resistor is fitted at the ouput of the comparator either, maybe the designer had trouble with leakage current making the LED glow dimly when it should have been off completely.

Brian.

 

[boylesg]

The 10k resistor on the LM393 output is specified in the datasheet.
Apparently these comparators are current sink devices rather current source devices, so you tie the output to Vcc similar to the way you tie the base of a PNP to Vcc. I was scratching my head for a bit as to why I couldn't get it to work in multisim, then I remembered.

The purpose of the LED in this case is to come on when the battery reaches full charge rather than to indicate when charging is taking place - seemed the easiest way to use the comparator at the time. But I found that all I need to do is reverse the comparator input pins and the LED is on while the battery is charging and off when it reaches 12V.

Apart from the LM317 circuit that comes straight from the datasheet, this would be my first circuit of this type that I nutted out for myself - I did not find it on the web.

I read some where that you are supposed to charge gel cells at about 1/10 their Ahr rating so I just did V=IR to come up with the about 20R for a flat charge current of around 700mA, plus a safety margin Although I was playing around with the value and the voltage it drops.

I suppose I could just as easily play with the resistor on the BD140 base to limit the current to 700mA or so. Does it really matter which way you do it?

 

[betwixt]

Well done for experimenting and finding a solution!

A load resistance is indeed needed at the output of the LM393 because it has an open-collector output and so can only sink current to ground. I'm still not sure why the 10K is needed when the LED and R13 provide a suitable load anyway. You are quite right about swapping the comparator inputs, at least on this device it reverses the logic of the output pin making the LED light under the opposite conditions.

You should understand that although the circuit works, it doesn't provide an ideal charger circuit. What you really need is a power source that works in constant current mode (Yes, C/10 charge rate) so the supplied voltage tracks the cell voltage but switches to constant voltage mode when C/10 can no longer be maintained because the terminal voltage has maxed out. The present circuit is somewhere between the two, the voltage regulator part just keeps the supply voltage to safe level and the series transistor just reduces the current. Nothing actually changes the voltage from the regulator in sympathy with the load, even if you short the output connections it will still try to produce the same voltage.

Brian.

 

[boylesg]

Some of the LM317 datasheets there are a series of transistors and resistors for R2 such that you can TTL select the voltage - a possibility.

Any other hints?

 

[betwixt]

In those deigns the transistors are effectively uses as switches and as each has a different collector resistor they allow the lower part of the potential divider to be selected. They can be used singly or in combination to select different resistances to ground. However, the transistors have to be driven to saturation to work properly so you would have to consider driving them from a digital signal. in reality, that means using a microcontroller to produce the digial select signals so you are looking at a completely different (but far more versatile) design altogether.

Brian.

 

[boylesg]

I have another idea short of an arduino.

I could have 4 different voltage levels (3V, 6V, 9V, 13V), with 4 different transistor and resistor sets.

Control these with 4 different comparators set at different trip voltages (2 x LM393).

The transistors would have to be PNP though.

 

[boylesg]

Brian I have been fiddling with this circuit in multisim and watching the voltage dropped across the CE junction of the BD140.

With a pretend flat battery of 1V (DC source), the transistor itself drops about 10V. So the flat battery only experiences a charge voltage of about 3V.

So I am not sure that I really need any additional circuitry to vary the voltage output.

The charge current gradually decreases while the charge voltage gradually increases as the battery approaches full charge.