quick voltage divider question

[kevilay]

Hey guys im just working on a pedal and im trying to setup a power circuit. I need 9V and a 5v source. I have a 24v 60watt regulated supply. Here is what my setup is:


---Input-------25k trim pot------10k resistor-----ground


Now I adjusted the trims  so reading between the top of the 10k resistor and ground is 9v. Now when I attached this 9v to the input of my l7805 (5v regulator) My voltage at the resistor is dropping from 9v to 3v, and my regulator output is 2. Am I pushing to much current through it? What would be proper values.

Thanks
Kevin

[kevilay]

Does anyone have any suggestions ? Do I need a regulator if I have a regulated supply. As long as I devide it properly?

[slacker]

You can't really do this with a voltage divider, whatever you connect to the divider acts like another resistor in parallel with the bottom one, so will change what voltage you get. If what you attach only pulls a constant small amount of current then it might work, but if it pulls a larger current or the current changes it won't work and the voltage will fluctaute all over the place.

The 5 volt regulator will run straight off the 24 volts, so no need to drop the voltage to 9 volts. If you want 9 volts for something else the easiest thing to do is use a 9 volt regulator.

[kevilay]

Crap I don't have a 9v regulator. I only have my 24v regulated supply  and a 12v wall plug for pedals. I also have 9v batteries. Is there a way to do this without batteries ?

[CynicalMan]

You can make a transistor regulator. It's not as good as an IC, and you'll need a 9-10V zener, but it'll work much better than a voltage divider.
http://www.satcure-focus.com/tutor/page5.htm

That said, what do you need to power with the 9V supply? If the current demands are low enough, you might be able to get away with a voltage divider. Also, if you power supply can handle it, you can try reducing the resistors' values to increase the available current. If there's AC involved, make sure to throw in a big capacitor from 9V to ground to make sure that there's no signal going through the power line.

[slacker]

If you have another 7805 you can use it to make a variable voltage regulator and get 9 volts from that, see page 8 of the datasheet.

http://html.alldatasheet.com/html-pdf/9047/NSC/78L05/36/1/78L05.html

[Gurner]

Crap I don't have a 9v regulator. I only have my 24v regulated supply  and a 12v wall plug for pedals. I also have 9v batteries. Is there a way to do this without batteries ?

If your anticipated current 'draw' for the 9V supply is low, you might get a way with a suitable opamp configured as a voltage follower - so pretty much your original divider with the output into this....



(careful, because not all opamps will operate up to 24V...therefore maybe press your 12V wall wart into play with a voltage divider ...also most opamps won't supply a lot of current, but if it's for a simple low parts count circuit your ought to be alright).

[phector2004]

that's a really cool idea, Gurner!

How much current can it supply? Is this calculated by taking a specific chip's power dissipation and dividing by output voltage?

[R.G.]

How much current can it supply? Is this calculated by taking a specific chip's power dissipation and dividing by output voltage?

As shown, it's limited to the current the opamp can put out before it goes into current limiting, which can be read from the part datasheet, which is available from google, like all other human knowledge.

However, you'll find that for most opamps this is only about 10-20ma. One solution for this is to insert a transistor after the opamp. The transistor is connected with collector to the power supply, and the output is its emitter. The opamp feedback is taken from the emitter of the transistor. This lets the opamp supply up to its maximum current as base current to the transistor. Now you can get as much as HFE times the opamp output as emitter current. For small signal devices like the 2N3904, this can be as much as 100-200ma.

There are, predictably, gotchas. First, the opamp almost certainly self-limits it output current to keep itself from dying on shorts. The new transistor does not. Short its output and it will burn up. The transistor also self-heats. With its collector to the power supply and its emitter at a fixed voltage, there is a constant voltage across it. This can be multiplied by the current through it and produce the amount of heat it will make internally. For 24V in and 9V out, this is 15V. The TO-92 plastic transistor package can dissipate 625mW before burning out, so the transistor can supply .625W/15V = 41.2ma before burning out. This is worse for 5V, of course.

Hmmm. Need less volts or more transistor dissipation. Ahah! Use a TO-220! That can dissipate 2W before burning up! So we can get ... 133ma. This isn't working all that well.    Ah. The problem is the large voltage across the device. If I know I only need, say 200ma, I can put a resistor in series with the transistor collector that drops the voltage down as current increases. Then the resistor is sharing the power wasting with the transistor. If we have 200ma and 15V, that's 3W. Something's going to waste 3W, so let's make it be mostly resistor. We'll use a resistor of 13V/0.2 = 65 ohms. At 200ma, it drops 13V, and the transistor only sees 2V and 200ma, or 0.4W. OK, now I can use a TO92 device and a 3W power resistor.

One could also use a darlington transistor and cut the output current down from the opamp, or up the output current to amperes. And, one could add some transistor sensors of current and shutdown, etc. For even higher currents, one could compute the heat developed and design heat sinks, etc.

[PRR]

> How much current...?

Indeed. Without that info it's all "you could do this, or you could do that, or you could do another thing...."

Like: I need a 100 foot hose. Well, how big? To dampen seedlings? To wash the elephant? To put out a burning building?

> You can't really do this with a voltage divider

I bet he "could". 24V 60 Watts will support a BIG voltage divider. Hang a 10 ohm resistor across it. This will dissipate 57.6 Watts heat. It can be a big 100W ceramic job a foot long. File through the ceramic at the 9/24 and 5/24 points, solder leads. (You can buy big power resistors with slide-tap.) The 9V tap shows about 2.4 ohms to the load. 100mA 0.1A load droops about a quarter volt, which should be close enough.

I assume Kevin does not have a 10 ohm 100W power resistor handy. The cost to buy and ship will exceed the cost of a regulator chip. It will eat electric power at the rate of a penny an hour, not large but not negligible (but might reduce home heating costs a half-cent). It will set fire to paper or drapes unless in a well ventilated box, more expense.

OTOH, if you need 5V to bias a MOSFET Gate, any dang voltage-divider will do the deed. (It did not wok for the 7805 because the 7805's self-idle current exceeds the <1mA available in his 25K+10K divider.)

[kevilay]

Sorry for not replying until now. I did see I had so many responces. Thanks for all the great information! I ended up just using a 9v battery and my5v regulator I had. But this information is great to have!