boylesg
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Sorry Brian I should have specified which circuit I was looking at.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...!
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.
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.
If I remove the LED then how would you add it back to indicate either charging complete or charging is proceeding?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.
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