voltage regulator

[mstechca]

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.

 

[Thunderchild]

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

 

[Joel Rainville]

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.

 

[ThermalRunaway]

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

 

[Joel Rainville]

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?

 

[Klaus]

I like Thunderchild's suggestions.

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.

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

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.

 

[audioguru]

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:

 

[mstechca]

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.

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

A little 9V battery is considered to be dead when its voltage drops to 6V with about 10mA or more load.

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.

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.

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.

 

[Joel Rainville]

A little 9V battery is considered to be dead when its voltage drops to 6V with about 10mA or more load.

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.

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 .

 

[audioguru]

I bought some LM317LZ voltage regulators.

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.

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:

 

[mstechca]

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.

 

[audioguru]

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:

 

[Joel Rainville]

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

 

[audioguru]

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:

 

[mstechca]

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

 

[audioguru]

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

He, he. They do it with majic! :lol: :lol:
Remember how transistors work? Look at the current sink circuit:
1) The upper transistor conducts because the resistor R supplies it with base current. The lower transistor doesn't conduct because its base voltage is very low.
2) When the current in the upper transistor exceeds about 20mA, then Ohm's Law says that about 0.54V or more is developed across the 27 ohm resistor.
3) With about 0.54V between its base and emitter, the lower transistor conducts just enough to shunt to ground some of the base current of the upper transistor, so it conducts less.
The circuit will stabilise at about 20mA regardless of the supply voltage if the rules are followed.
Analog ICs are full of constant current sinks and sources. :lol:

 

[eemage21]

I'll explain based on the circuit on the left.

Theoretically, you could get away using only one transistor, but you wouldn't be making a very good current source, because it would vary greatly with the load you'd be applying to it.

When you analyze transistor circuits, one of the main things you need to remember is that in the active region the voltage drop from base to emitter (output with an arrow) is around 0.6V. What happens in the circuit is the following:

1) the transistor on the lower left (the one with grounded emitter), causes 0.6 drop accross the 27 ohm resistor, and that drop is relatively constant (this gives you the stability), the second transistor (the one with collector connected right to the LED), isolates your current source (the 27 Ohm resistor with 0.6 V accross it), from the load (LED's). It does it thanks to low impedance looking into the emitter, and high impedance looking into the collector.

Another way of looking at this is that the transistor on the left, provides you with a stable biasing for the transistor on the right. You could get away with replacing the transistor on the left with another resistor such that the Voltage at the gate of the first transistor due to the resulting voltage divider, would 1.2 V, and you'd still end up with 0.6 V accross the 27 Ohm resistor.

 

[eemage21]

Regarding LM317's - as long as you pick the right ratio for your resistors, and keep your input voltage above 1.25V above your output voltage (in yourcase you need to make sure your input doesn't go below 4.25V if you want 3V output), then you will be fine. If you want to go even lower, go with a low-dropout voltage regulator, and if you don't want to worry about picking resistors, pick a constant voltage type (as opposed to the adjustable version). In either case, your assumption in the original post that the output voltage drops with the input voltage, is wrong. Your output voltage will stay constant over entire range, down to 4.25V.

 

[audioguru]

the transistor on the lower left (the one with grounded emitter), causes 0.6 drop accross the 27 ohm resistor, and that drop is relatively constant

No it doesn't. Current doesn't come out of the transistor's base and go down into the 27 ohm resistor.

the second transistor (the one with collector connected right to the LED), isolates your current source (the 27 ohm resistor with 0.6 V accross it), from the load (LED's). It does it thanks to low impedance looking into the emitter, and high impedance looking into the collector.

Huh? This transistor supplies the LED current, not the 27 ohm resistor.
Impedances have nothing to do with it.

Another way of looking at this is that the transistor on the left, provides you with a stable biasing for the transistor on the right.

You are sort-of correct. :lol:

 

[eemage21]

I never said the current comes out of the base. The current comes out of the collector of the transistor on the right side.

Also - you were wrong in stating that the transistor on the left is not conducting - it is.

Impedances have everything to do with it - even more so, because you are using such a small resistance in your degenerative emitter feedback.