I bought some LM317LZ voltage regulators.

I think that the output voltage and the input voltage have a constant ratio defined by the user. That is, if I had a 9V supply and I want the regulator to output 3V, then if the battery is 1/2 dead, the input is 4.5V and the output is 1.5V.

What I want to do is be able to fix the voltage to 3V just for the receiver. I need the extra voltage to drive the LED output of the receiver because 3V doesn't give out much light for an LED, and I don't have ridiculously small value resistors to even out the current.

If there is a circuit that allows me to force the output of the regulator to 3V from any input voltage source above 3V (9V at most), then I would like to see it.

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an alternative to that is:
1) use a powerful op amp and as an input the constant voltage of a zenner diode or
2) a power transistor: the collector goes to Vcc. the load between the emmiter and GND a zenner diode from base to GND and a resistor that will provide enough current for the zener diode between Vcc and the transistor base. the out put voltage is Vz - 0.7 (the voltage drop of the base/emmiter)
with both these methods u will have constant voltage even if the input changes but does not go below a minimum

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Check the datasheet, you'll find the typical circuit used to maintain line regulation.

The output voltage is function of the resistors used and the resulting voltage applied on the Vref pin. It is *not* a constant ratio between input and ouput voltages.

I used one of these 317 regulators only once, but I seem to remember that I needed at least 3V more on the input to have a reliable output. So in your case, that would be at least 6V on the input (and the proper resistors) to get a constant 3V on the output. Maybe you get a fixed-ratio voltage drop when the input drops too low... Again, check the datasheet.

 

One unfortunate property of common regulators is that the supply line always needs to be 3-4 volts higher than the required regulation voltage. This means that when your supply is 9V you'll get a good 3V output but, once the supply drops below 6V you're into the danger area where the regulator may not function properly anymore. What about powering the project with two 6V batteries in series? This would give you a 12V main supply and the regulated output wouldn't be effected until the main supply dropped to half it's original output! I think it's fair to say a battery is dead by the time it's dropped half it's output voltage anyway. Or you could use two 9V batteries in series which would give you a main supply of 18V and an even better starting point.

Bear in mind that there's a maximum supply voltage for regulators. You'll have to check the datasheet to see what yours is.

Brian

 

I like Thunderchild's suggestions.

Question for everyone : could a "step-down" DC/DC be used here and would it be more efficient than a regulator?

 

Yes but, a regulator is designed to do just that, efficiently, while the methods suggested may not have as good a regulation nor be as efficient at it.

Maybe. The efficiency of the regulator depends on the voltage drop across it, there are low dropout regulators available. DC/DC converters would be more expensive too.

 

Have you ever purchased alkaline battery cells of the skinny AAAA size? They are very expensive and don't have much capacity due to their small size. There are 6 of them in series inside a little 9V battery.

A little 9V battery is considered to be dead when its voltage drops to 6V with about 10mA or more load. A 6V input to an LM317 with its output set to 3V is fine. When the 9V battery voltage drops below 6V then throw it away, it is dead!

It sounds like you need the radio's voltage to be regulated at 3V with the LM317, but need to operate the LEDs at the full supply voltage for brightness. Then use a constant current sink or source for the LEDs instead of a simple current-limiting resistor and their brightness will be the same with both a new battery and an old one. :lol:

Attached Images

led_current_sink.gif (4.3 KB, 870 views)

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Uncle $crooge

 

no, and I don't want to buy extra-expensive batteries.

In my case, no battery is dead, until it is dead, but if a 9V battery is suppsed to be dead after 1/3 of it is used, that is crazy.

My LED output comes from the receiver as well. If I could condense the circuit to just using the regulator, that would be the best. Once again, my space is small.

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mstechca : an alkalyne battery's voltage drop isn't linear!... When a 9V battery only outputs 6V under minimal load, it is near 100% used and basically dead. There isn't 2/3 of its life left because it still outputs 6V...

See Nigel's nice battery discharge graphs.

 

It is a variable voltage regulator and the circuit is in the datasheet. Only 2 resistors set the regulator's output voltage. With 160 ohms from the output to the adjust terminal and 220 ohms from the adjust terminal to ground then the output voltage is typically 2.97V. If the output voltage rises because the load current is not enough then add a 1.2k resistor from its output to ground.
If you set the output voltage to 3V, it will change typically only 0.3mV if the input voltage changes from 6V to 40V.

Maybe your lack of voltage regulation is because the LM317 needs a minimum load current or its output voltage will rise. A 10mA load on the output to ground or a 120 ohm resistor from its output to its adjust terminal will be adequate.

Maybe you wired your LM317 as a current regulator instead of as a voltage regulator. The output voltage of a current regulator varies all over the place in order to keep a constant current through the load when the load resistance changes. :lol:

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Uncle $crooge

 

Audioguru,

If that schematic you last posted is a current sink circuit, is it possible that I could rearrange it in such a way so that it works as a constant current source with only one output and +ve (and -ve/ground if required)

If I can get the same current output with voltage input in the range of 3 to 9 volts, then I'll be happy.

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Sure it can be a constant current source. Just use PNP transistors and swap around the parts like this.
The constant current sink and source circuits keep the current constant with a supply as low as 3V if the LED forward voltage is about 2.0V or less.

But your LM317 regulator isn't a low-dropout type so will dropout and lose regulation when its input voltage drops to about 5V or 6V and lower with it set for a 3V output. :roll:

Attached Images

led_current_sink_and_source.gif (5.8 KB, 770 views)

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Uncle $crooge

 

In the above schematic, what's going on between the two LEDs? What are those two dots?

 

The dots are more LEDs in series. You can put as many LEDs in series as you want. Since they are in series, their currents are all the same. The supply voltage must be at least as high as their total voltage drop, plus 1V minimum. If the voltage or current is too high then the driver transistor will melt. :lol:

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Uncle $crooge

 

why are two transistors necessary? How do the transistors make the output a constant current even though the voltages vary?

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