Moving To Another Voltage Regulator Problem

[Honduras]

Lm317-LZ 1 mA output. (Within the range I'm interested in.)

Am I going to have to power this with DC from batteries. Maybe 4.5 vdc?

 

[Ramussons]

Load voltage?

 

[Honduras]

Load voltage?

Unknown. The load will be various vintage analog current and voltmeters. Some of the ammeters measure micro Amps. No data sheets. They are not cheap.

I need to start with half the rated current, then run the voltage up until I get a half scale reading on the meter.

 

[MikeMl]

You are building a current source? The minimum input voltage must be > than the sum of three voltages:

The load (meter) V + 1.25V + Vdo

Vdo is the drop-out voltage, and is on the data sheet...

 

[audioguru]

Are you making a regulated 1mA current source?
The minimum load current for an LM317L current regulator is shown on Fairchild's datasheet as 5mA. Texas instruments show a minimum output current of 2.5mA and a minimum input voltage of 5V.

 

[KeepItSimpleStupid]

If these are current movements, then the usual technique is to get a voltage source and a potentiometer and adjust the pot for half scale.

You then measure the potentiometer resistance after making the meter read 1/2 scale. That's your internal resistance. From that and the voltage source, you know the full scale. current. Take I=V/R and 1V/(1E6+2*Rm) < 1uA minimum.

The 1E6 is a 1 M pot turned all the way up. 2*Rm is what you get at mid scale.

0 ohms for the variable pot isn't good, so you can also use a fixed resistor in series with a variable one.

I think I used a D cell. Just use the measured voltage.

 

[Honduras]

If these are current movements, then the usual technique is to get a voltage source and a potentiometer and adjust the pot for half scale.

You then measure the potentiometer resistance after making the meter read 1/2 scale. That's your internal resistance. From that and the voltage source, you know the full scale. current. Take I=V/R and 1V/(1E6+2*Rm) < 1uA minimum.

The 1E6 is a 1 M pot turned all the way up. 2*Rm is what you get at mid scale.

0 ohms for the variable pot isn't good, so you can also use a fixed resistor in series with a variable one.

I think I used a D cell. Just use the measured voltage.

This is worth putting in my little black book.

The one remaining concern is the 1.5 vdc. As I understand it, what pops meters is excess current, not excess voltage. Does this sound reasonable? I'm concerned about the voltage popping the meter, even at 1.5 vdc

 

[audioguru]

I'm concerned about the voltage popping the meter, even at 1.5 vdc

A meter movement has its current for full scale and its resistance specified, not the voltage. A voltmeter is a current meter with a series resistor that limits the current.
A 50uA meter movement might have a resistance of 3.5k ohms. Then when connected directly to a 1.5V battery, Ohm's Law calculates a current of 1.5V/3.5k= 429uA which will probably destroy it.

 

[KeepItSimpleStupid]

SO you start out with say a 10 M variable resistor at max and a 1.5 volt battery. If you need finer resolution then use a fixed resistor and a smaller potentiometer. 50 uA is the smallest I've seen for an analog meter.

You need to know the meters internal resistance to "Make a voltmeter" out of it.

 

[Honduras]

I tried the 1.5 vdc and 1 Meg variable resistor. Naturally, I picked the wrong meter first. It's designed to take +/- 1 Amp DC with no external circuitry.

The second meter is a +/- 10 milliamp meter for DC. I couldn't find any of my 1 Meg Ohm variables, so I had to use a 3 Meg single turn, but at the low resistance required, I couldn't make fine enough adjustments, so I switched to a 10 turn, 10 kOhm variable resistor when it became obvious that the required resistance was less than 1 kOhm.

I ended up with the combination of the 100 Ohm and the variable resistor equaling 330 Ohms. As I understood your instructions, this gives me a full scale meter current of 4.5 milliamps. Does that seem reasonable.

This is what it looks like . . . kind of. Oooops. How do I insert a JPG file?

 

[KeepItSimpleStupid]

OK do it again, but give me the raw values:

The instructions are:
1) Connect the fixed resistor and the variable resistor initially at maximum and then adjust for 1/2 scale.
(you might want to set the pot so that (CW decreases the resistance) so the meter moves right with CW rotation.
2) Measure the voltage of the battery.
2a) As a check, measure the voltage across the resistor and the pot combination @ 1/2 scale

3) Remove and measure the (potentiometer + variable resistance) that it took to get to 1/2 scale.

A. Battery voltage (when at 1/2 scale) = ____ Volts
B. Voltage across (fixed + variable resistance @ 1/2 scale) = ____Volts
C. Resistance of (fixed+variable) out of circuit (that was used to reach 1/2 scale) = ____ Ohms

 

[Honduras]

OK do it again, but give me the raw values:

The instructions are:
1) Connect the fixed resistor and the variable resistor initially at maximum and then adjust for 1/2 scale.
(you might want to set the pot so that (CW decreases the resistance) so the meter moves right with CW rotation.
2) Measure the voltage of the battery.
2a) As a check, measure the voltage across the resistor and the pot combination @ 1/2 scale

3) Remove and measure the (potentiometer + variable resistance) that it took to get to 1/2 scale.

A. Battery voltage (when at 1/2 scale) = ____ Volts
B. Voltage across (fixed + variable resistance @ 1/2 scale) = ____Volts
C. Resistance of (fixed+variable) out of circuit (that was used to reach 1/2 scale) = ____ Ohms

I'll put it back together tomorrow and get back to you as soon as I can.

 

[Honduras]

This more than you asked for, but . . .

Vin = 1.590 vdc, (D Cell)
The readings below are R measured, Voltage across component, measured, component resistance calculated.
Safety resistor: 99.9 Ohms, 0.526 Volts, 97.32 Ohms
Variable resistor: 161.4 Ohms, 0.868 Volts, 161.4 Ohms

Thevenin's Theorem suggests that the voltaage dropped across the meter is around 0.772 Volts.

This gives a calculated current through the whole thing of 5.377 mA based on the values for the variable resistor, which seems pretty stable.
If I base it on the safety resistor I get 5.31 mA. I've figured it three ways (Safety resistor, Variable resistor, Safety resistor + Variable resistor.) I get a slightly different set of values each time, but each one gives me a value of pretty close to 5.3 for Imeter, which is pretty much what it looked like to me, although I thought I was closer to an even 5 mA.

 

[KeepItSimpleStupid]

OK, I get: =1.59/(99.9+161.4)*2*1000 = 12.17

1/2 scale current = C3 = 0.526/99.9; Full scale current = C3*2*1000 or 10.53 mA FS
1/2 scale current = C4 = 0.868/161.4; Full scale current =C4*2*1000 or 10.76 mA FS

The meter resistance is (99.9+161.4) or 261.3

The 12.17 is off. If you measured the voltage not when the meter was at half scale might cause it. Measuring small value resistors you need to use the 4 terminal method.
You measure the current through, and the voltage across the resistor. The current is measured in nay part of the circuit. The voltage is measured
at the device terminals. Do you have a 4 or 5 wire ohmmeter?

The 2 is for half scale and the 1000 is conversion from amps/mA. C3 nd C4 are spreadsheet artifacts.

Around 10 mA nominal FS, I would believe.

Now, make a voltmeter out of it with a calculated series resistor. I wish you were more in the K range of resistors.

 

[Honduras]

OK, I get: =1.59/(99.9+161.4)*2*1000 = 12.17

1/2 scale current = C3 = 0.526/99.9; Full scale current = C3*2*1000 or 10.53 mA FS
1/2 scale current = C4 = 0.868/161.4; Full scale current =C4*2*1000 or 10.76 mA FS

The meter resistance is (99.9+161.4) or 261.3

The 12.17 is off. If you measured the voltage not when the meter was at half scale might cause it. Measuring small value resistors you need to use the 4 terminal method.
You measure the current through, and the voltage across the resistor. The current is measured in nay part of the circuit. The voltage is measured
at the device terminals. Do you have a 4 or 5 wire ohmmeter?

The 2 is for half scale and the 1000 is conversion from amps/mA. C3 nd C4 are spreadsheet artifacts.

Around 10 mA nominal FS, I would believe.

Now, make a voltmeter out of it with a calculated series resistor. I wish you were more in the K range of resistors.

Somewhere I must have got off track. I'm not trying to make a voltmeter. I'm just trying to use some old analog amp/milli-amp/micro-amp meters.

It is a +/- DC milli-ammeter with FSs of + 10 and - 10 milli-amps, so I think your method will work great for me.

I have about 20 antique x-ammeters. I just want to see if they are good, and how far out of calibration they are.

I'm not sure why I need to do the conversion from Amps to milli-amps. It measures in milli-amps.

You have to keep in mind that I am not good at explaining myself.

 

[KeepItSimpleStupid]

Yep I know tough time explaining. Numbers for 1 and 100 are easier to work with so 1e-3, e-6, e-9 get turned into mill-, micro- and nano-

One of the uses of ammeters is to change the full scale to volts or say a microamp meter to read mA. For that you need the meter resistance and the full scale current. That is the resistance of the of the "external element". The reason why it is the resistance because 1/2 the voltage is dropped across your external resistor & variable resistor and 1/2 across the meter.

Generally you can usually calibrate the zero and FS of the meter. Difficult, but possible.

If your using a traditional ohmmeter the lead resistance is probably at least 0.5 ohms *2. So, the 100.xx part of your measurements are not real. There is much to learn about accuracy and precision.

The resistor/voltage source method is pretty safe.

 

[Honduras]

Yep I know tough time explaining. Numbers for 1 and 100 are easier to work with so 1e-3, e-6, e-9 get turned into mill-, micro- and nano-

One of the uses of ammeters is to change the full scale to volts or say a microamp meter to read mA. For that you need the meter resistance and the full scale current. That is the resistance of the of the "external element". The reason why it is the resistance because 1/2 the voltage is dropped across your external resistor & variable resistor and 1/2 across the meter.

Generally you can usually calibrate the zero and FS of the meter. Difficult, but possible.

If your using a traditional ohmmeter the lead resistance is probably at least 0.5 ohms *2. So, the 100.xx part of your measurements are not real. There is much to learn about accuracy and precision.

The resistor/voltage source method is pretty safe.

I think I have meters for the full range of current that I anticipate seeing. I just wabt to know if they work and are within calibration range.

OFF ON A FLYER -
In my magic box today I found an LED ammeter that I think I paid $2.49 for. Unfortunately, it is a 0-10 Amp meter. It would be wonderful, and profitable, if I could use these instead, and sell the antiques. It's externally powered with two sense leads.

What I wonder is how to get it to show micro amps? Any ideas. One of the many things I want to do with whatever is to try to get an idea of ESR sized leakage currents for vacuum tubes.

 

[audioguru]

The 0-10A current meter must have a very low resistance shunt resistor that it measures the voltage across. If the shunt resistor is removed then the meter becomes sensitive voltmeter (most with a digital display have full-scale of 0.2V) then you can use Ohm's Law to calculate a new shunt resistance.

 

[audioguru]

The 0-10A current meter must have a very low resistance shunt resistor that it measures the voltage across. If the shunt resistor is removed then the meter becomes sensitive voltmeter (most with a digital display have full-scale of 0.2V) then you can use Ohm's Law to calculate a new shunt resistance.

 

[Honduras]

The 0-10A current meter must have a very low resistance shunt resistor that it measures the voltage across. If the shunt resistor is removed then the meter becomes sensitive voltmeter (most with a digital display have full-scale of 0.2V) then you can use Ohm's Law to calculate a new shunt resistance.

Cool. All I have to do now is learn Surface Mount soldering. These are the really small ones.