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Variable Power Regulator Mystery DAMMITDAMMITDAMMIT!!!!

I suppose I shouldn't be complaining.  After all, the circuit works:

...and more or less this (the cap straddles the gutter on my board  but Fritzing is limited...)

There's no connection to pin 3, which is "output" supposedly.  Yet it works as a voltage controller-from 18v to 3v-12v (range.)

I based this off of the old Mims circuit on p.76 of the Engineer's Notebook II but made a few adjustments, obviously. First, I barely got 1v of variation (4v-5v) with the 50kΩ pot. When I took out the 1.2kΩ resistor going from the wiper to the output rail, I started to get a little more.  With a 100KΩ pot plus a 10kΩ resistor in serial making up "R1," I get a useful range.

My gut tells me it shouldn't work, but my multimeter tells me it does.  Is it just some kind of bass-ackward voltage division at this point?  The output isn't coming from between a pair of resistors.  Is this chip designed to work this way?  I feel like this:

 

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Basile brought up a couple points in the shoutbox that I had not intended to go into.  But since it was brought up, I will go a little deeper.  What I'm going to cover is useful in a lot of contexts other than this particular circuit.

It was mentioned that the datasheet says the adjust pin current is 50 uA instead of 100.  That is true, when the chip is operating normally.  It does say the max current through the adjust pin is 100 uA.  Whatever the datasheet says, we KNOW it has to be in the range of 100 uA, otherwise Ohm's law is broken and we need to throw it out.  A 110K (100 + 10 in series) resistor with 11 volts across it HAS to have 100 uA through it.  So the question is, why does that adjust pin current go down when the regulator is working properly?  I will show a very simple example that is similar to what is buried deep inside one of these three terminal linear regulators.  They are a lot more complex, but work in a similar manner.  Take a resistor and a zener diode arranged like this:

Vin (12V)  -------------------------------wwww----------------------------|--------------------------------Vout (5V)

                                                   1000 Ohms                |

                                                                               /---/\---/

                                                                                /___\    5V zener diode

                                                                                   |

                                                                                   |

                                                                                 Ground

With nothing connected to the output terminal, 7 mA will flow through the zener diode.  That is because ther will be 7 volts across the resistor (12 volt input minus 5 volt zener drop).  The zener current would flow out through the adjust pin of a regulator.  Now if you connect a load to the output, say a 1000 ohm resistor, it will have current flowing through it.  A 1000 Ohm resistor will have 5 volts / 1000 ohms or 5 mA.  The current through the first resistor has to stay the same, because the zener keeps its voltage at 5 and the input stays at 12, leaving 7 volts across the first resistor and keeping the 7 mA.  But now that 7 mA divides between the load (5 mA) and the zener (2 mA) and the "adjust terminal" current drops because of the load.

EDIT:  I also intended to mention  that the 350 minimum load current is 5 mA instead of 10, so a 240 Ohm resistor between output and adjust will work fine as Basile pointed out.  Do keep in mind the difference between adjust pin current and load current (through the output pin).

As Oddbot pointed out, you are measuring with no load.  The circuit isn't regulating anything, but it is acting similar to a voltage divider.  The internal circuitry of the reg uses a certain amount of current, about 100 micro amps roughly, which must return to ground through the adjust pin.  That current flows through your 100K and 10K resistors, creating a voltage drop across them.  That voltage drop is, of course, E=IxR, which gives about the range you are seeing.

To make it work, place a 120 ohm resistor from the output to the adjust pins, and a 1 or 2K pot from adjust to ground.  Here's how it works.  The reg has to have a minimum load of about 10 mA (going from memory, check the datasheet) flowing out the output pin to regulate.  It regulates by making the output pin 1.2 Volts above the adjust pin.  1.2V / 120 ohms gives 10 mA output, meeting the minimum current.  That current (plus the operating current, which is nearly negligible now) flows through the pot to ground, causing a voltage drop (10 mA x 1K = 10V) so the adjust pin is now at that dropped voltage, causing the output to be 1.2V higher.  The voltage between out and adjust stays 1.2, so the current stays at 10 mA through the 120 resistor.  But as you adjust the pot the voltage dropped across it varies from 0 Volts (at 0 ohms) to 10 volts (at 1000 ohms) causing the output to be 1.2 (that's where the minimum comes from) to 11.2 Volts.  You can add a fixed resistor in series with the pot to raise the minimum (and max) voltage.  For instance, to get a 3V minimum output the adjust terminal needs to be 1.8 (3-1.2) so you need R to be R=1.8/.01 = 180 ohms to give a range of about 3 to 11.8 Volts output.  Remember, the reg can't "regulate" without the minimum load, which is provided by the 120 resistor, even though it may seem to give the proper voltage.

In the EH2 circuit, the load resistor was only 1.2k Ω so maybe that's what was failing me. Thanks bdk-I'll give it a try.

As you are testing without a load I would hardley say it's working.

I could point out the obvious and say read the datasheet rather than someone elses possibly faulty circuit diagram but in your case I think this might be more useful.

https://docs.google.com/file/d/0B__O096vyVYqTGhQNy1Jdm1Ibnc/edit?usp=sharing

In the good old days, Australian Icon Dick Smith made his money teaching Australians electronics. This is the data section from one of those catalogs. It has a lot of useful information on not only this voltage regulator but a lot of other common electronic components.

Sadly these day Dick Smith is owned by a pack of retards that completely destroyed all Dick Smith stores and turned them into a bunch of second rate consumer electronic stores where the staff couldn't tell you the time if you held a watch in front of there face.

As you can tell I have some strong feelings on the matter as Dick Smith taught me electronics, not the wankers who call themselves the Australian education department.

Thanks for that OddBot! It's almost as good as this one electronics reference poster I got somewhere...;-) Anyway, their version of the 317/350 circuit looks like some of the more complicated implementations I found. (I chose the one I did because I was too lazy to do the math in the data sheet version but I still wanted a low component count.) Dick Smith must be the Aussie version of Tandy/RadioShack when I was a kid.

In this case low component count = low reliability = dumbass!

Following a bad schematic drawn by someone who doesn't know what their doing because it looks easier is like following a guy who's bumper sticker reads "Don't follow me, I'm lost too!"