I'm At A Loss: Variable Resistor

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Adjustment to 0V is hard to obtain with a single supply. One way to get it is to create a 3V reference and use a pot as a voltage divider and buffer the output with an OP amp.

A true 0-3 V "power Supply" is going to be harder.
 

I realize this is hard to visualize, and I'm sorry.

I moved the VR to a separate breadboard and it is parallel to long axis of the small breadboard. I used jumpers to make the proper connections. I finally got some life out of the VR, but not enough. It's adjustable from +14 to +18 vdc.

I'm going to try to post a JPEG of the circuit I am trying to build.

Doesn't work. Windows and HP are fighting over control of the scanner. Niether will scan it.
 

It's not going as planned at all. Power input is through a 120 VAC -> 12 VAC transformer.

Rectification is by a W02M bridge rectifier. Filtering is by a 100 uF and a 0.1 uF connected in parallel between the positive rectifier output and the negative rectifier output, which is also the common ground for the remaining circuitry. The voltage rectifier is an LM317T. There is an IN4007 connected between the input and the output of the LM317T, the cathode end toward the input.

At the output, there is 1.0 uF capacitor connected to ground, and the load resistor, a 1.5 kOhm 10 Watt resistor, also connected to ground.

The circuitry attached to the ADJ terminal is a little more complicated. 240 Ohm resistor in parallel with another 1N4007 between the output terminal and the lead to the ADJ terminal. The 1 kOhm VR is connected between the the ADJ terminal and ground.

I have a VOM, on the volts scale, attached across the load resistor.

I can understand a few volts higher than 12 at the LM317 input, due to the filtering, but I didn't expect to see 18 vdc anywhere in the circuit, especially across the load.

I really wish I could send you a picture of the breadboard, but I'm afraid that all you would see would be the 36 jumpers.
 
KeepItSimpleStupid said:
Adjustment to 0V is hard to obtain with a single supply. One way to get it is to create a 3V reference and use a pot as a voltage divider and buffer the output with an OP amp.

A true 0-3 V "power Supply" is going to be harder.
Click to expand...

The TI datasheet on the LM317 shows a 0-30 vdc regulated voltage source. The trick is the ADJ pin, and what is connected to it. I've actually done it before, but not on a breadboard. That one is on display in an art museum.
 
I never said impossible, just harder. It needs a negative supply to do so.
=
Just wondering what your requirements are and whether your using the right device?
 
The LM317 is spec'd with a minimum allowed load current of 10mA. The more expensive LM117 has a minimum allowed load current of 5mA.
The LMx17 develops 1.25V between its output and its ADJ pin which is also across the resistor. Ohm's Law calculates the resistor to be 1.25V/5mA= 250 ohms so the LM117 can use 240 ohms. This resistor in series with the pot is the load when there is no additional load. The LM317 needs 120 ohms so that the output voltage does not rise without an additional load.

Assuming that you have the more expensive LM117 since you are using its 240 ohm resistor, the resistor current is 1.25V/240 ohms= 5.2mA. When the pot is at maximum it is 1k ohms so with the 5.2mA in it then its voltage is 5.2V. Then the output of the LM117 will be 1.25V + 5.2V= 6.45V.

If you replace the 240 ohms resistor with 120 ohms for an LM317 then the resistor current is 1.25V/120 ohms= 10.4mA. When the pot is at maximum it is 1k ohms so with the 10.4mA in it then its voltage is 10.4V. Then the output of the LM317 will be 1.25V + 10.4V= 11.65V.

You have a 1N4007 rectifier diode parallel with the 240 ohm resistor. Maybe its polarity is backwards? It is not needed in your circuit, it is used to discharge an additional filter capacitor sometimes used from the ADJ pin to ground. You do not have the capacitor.

TI's datasheet shows an LM117 in a 0V-30V regulated supply. It has a +35V supply and a -10V supply. The resistor is 150 ohms, not 240 ohms and the pot is 5k ohms, not 1k. It uses a 1.2V regulated reference IC connected to the lower terminal of the pot so that the output of the circuit can go down to near 0V.
 
In your post #26 you pretty much describe a textbook LM317 regulator circuit with the diode protection. Between common and the LM317 input what do you read for voltage? LM317 input to common?

Ron
 
Your 100uF main filter capacitor is MUCH TOO SMALL. For a load of 0.5A it should be 2200uF or more.
 
audioguru said:
Your 100uF main filter capacitor is MUCH TOO SMALL. For a load of 0.5A it should be 2200uF or more.
Click to expand...

I tried it with a 3300 uF cap, but, even after 15 minutes, the voltage did not stabilize. Spikes to 39 vdc sliding down to 0. With the 100 uF, it took about 45 seconds to stabilize.
 
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In general, you are correct. I was looking at figure 27 of this datasheet. https://www.ti.com/lit/ds/symlink/lm317l-n.pdf

Apparently there were a few parts, and a voltage source that I left out.

The diode in parallel with the 240 Ohm resistor, connected to the ADJ terminal is to provide protection from the 1.0 uF capacitor that is in parallel with the load resistor.

I have rebuilt my computer, and bought a software removal tool and a registry cleaner. I have a schematic on the scanner that shows what I am trying to build, but no way to scan it yet.

The final goal is a regulated voltage/current source that can produce 0.5 vdc at 100 mA, max.
 
Honduras:

A circuit I designed a number of years ago, sort of, does what you ask. It's goal, however was a little different and very complex. It was a (4-quadrant) voltage source, but without controlled current. It did auto-range in the I-V converter, so I had 4 ranges with +-10V out for current, the highest range being +-100 mA. So, you could set the voltage very accurately and measure the current very accurately over 4 decades.

The modes were zero check, zero correct, open circuit voltage and 2 terminal/4 terminal measurements (i.e. sense leads). It also had a >+-10V out detect with a 1 sec pulse and changed the color based on polarity. I wasn't allowed to fix the DC part of the design, but 40 pA of offset was extremely good. We cared about the AC performance and the sense leads. I had to handle around +-40 mA from about -1 to +0.6 volts. I forgot that the D/A would not generate 0V when commanded to output 0.

I used a Keithley 213, **broken link removed** and it outputted a little over 1 mV for 0.

It was basically an Analog Source Measure Unit (SMU), but only had modes:
1. Voltage measure
2. Source V/Measure I in 4 ranges in 2T
3, Source V/Measure I in 4 ranges and 4T mode.

The voltage source was 4 quadrant. i.e +V-I, +V+I, -V-I and -V+I

The leads drop a significant percentage voltage drop with say 6 feet of 22 AWG wire ' (low wire guage) , ~30 mA and 0.5V.

It was a cool project and expensive. Two HP 5 digit system meters were used to measure V and I.
 
I've reached the point of following advice from someone I consider innovative.

Forget the voltage regulation. Use the 1.24 volts from ADJ and a voltage divider. Use an LM317L as a current limiter. Does this seem promising?

The reason I want to start with a 9 volt regulator (thus 12 VAC input) is that another project (10 MHz QRP receiver + antenna tuner) requires two stable 9 vdc inputs.

I figure that, if I can't get a 9 volt regulator to work, there's not much chance I can get < 1 vdc regulator to work.
 
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Honduras said:
The diode in parallel with the 240 Ohm resistor, connected to the ADJ terminal is to provide protection from the 1.0 uF capacitor that is in parallel with the load resistor.
Click to expand...
Nope, the diode is not needed. A protection diode is needed from input to output when there is an output capacitor more than 25uF and the input or output is shorted.
Page 7 of your datasheet says the diode across the 240 ohm resistor of an LM117 protects it when there is a capacitor of 10uF or more from the ADJ pin to ground (that you do not have) and the output voltage is higher than 25V.
 
Almost all is clear now.

I used the wrong drawing for the design.

I did get the variable resistor to work. It even does what it is supposed to do. It's just that I set up the wrong circuit.

The circuit I built is designed to increase the output voltage of the regulator by changing the offset voltage to ground. That's why the output voltage was so high.

Sorry for making a mistake that has wasted so much time. I did learn a lot from the discussion.
 

I didn't say it is needed. I prefer to be careful, and I have about 2 pounds of them that I need to find homes for.
 
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