Hello everyone. I have a question that I would appreciate your help with. I built a variable voltage power supply a while ago using an LM317. I am just getting around to putting it into an enclosure that will have a digital voltmeter mounted on the outside so that I can quickly see what voltage is being output. The voltmeter I bought when I initially started the project says that it requires an independent 9V power supply. What I think that means is that the 9V for the voltmeter and the measured voltage cannot have the same ground. I attached a schematic of how I plan to do this. The ground for the LM317 portion of the circuit is simply the "-" pole of the bridge rectifier B1 while the ground for the 9V is connected to the 3rd prong of the AC outlet. The voltmeter I will be using in the PM-129A (link to datasheet below). So here are my questions:

1. Is my assumption in this statement, "What I think that means is that the 9V for the voltmeter and the measured voltage cannot have the same ground.", correct?

2. Will the method I am using in the schematic to seperate the ground paths work?

3. Other than using a 9V battery, is there an easier way to acheive an indepent power source for the digital voltmeter other than how I have shown it in the schematic?

4. This question is not related to the design but to information in the datasheet. This voltmeter is supposed to measure 0 - 500 V DC but on the datasheet under the specifications section the maximum input is listed as 199.9 mV DC. How can the voltmeter measure up to 500 V DC if I can't apply anymore than 199.9 mV DC to it? I think I'm misinterpreting the datasheet but can some explain to me what the 199.9 mV DC limit applies to?

Thanks in advance for your help.


Voltmeter datasheet:

http://www.circuitspecialists.com/im...29A&Bspecs.pdf


Schematic:

Attached Thumbnails

 

 

No. Independent means no electrical connectivity what so ever. It also means there must be practically infinite resistance between the two power supply.

No. The two circuits are still connected somehow.

a) You should have bought the PM-129B instead. That model is designed to work with a common ground with your LM317 supply.

b) If you are stuck with the PM-129A, then either use another transformer with two 12V windings or add another small power transformer and rectifier circuit is required to power the DVM.

You drop the high voltage to within 200mV using resistive voltage divider. The datasheet will show you how and what values of resistors to use.

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L.Chung

 

OK, I understand. Thank you.

 

The datasheet for the meter does not say its max supply current but since it uses many LEDs then its max current would quickly kill a little 9V battery.

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Uncle $crooge

 

Per the suggestion by eblc1388 I have switched gears and am now going to go with the PM-129B which does not require an independent power source. I do have another question however. The instructions for the meter say that I need to provide the meter with a voltage half of full scale to calibrate it. In my implimentation I am going to set the meter for the 200V range. So that means I need to supply it with 100V to calibrate it. One way I thought I could do this would be to use the 24VAC output of my transformer and build a 4X voltage multiplier. I could have a pot in series with a 100K resistor to fine tune the output to 100V and use that (see attached schematic). Does anyone have a better idea? Is there an easier and better was to obtain a 100V reference for calibration? Thanks.

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If you wire the voltmeter to the basic measuring range (199.9mV) you require 100mV for calibration.

You might as well calibrate it leaving it connected for the range of 200V and apply any known voltage within range (measured with a DMM).

If it has a zero point calibration pot perform the offset calibration before you calibrate for range.

Boncuk

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Proper Planning Prevents Piss Poor Performance

 

All the literature says it's garunteed to read 0 volts with no voltage applied to it. I interpreted that to mean there is no zero point calibration necessary. There is only one pot on the meter so I'm assuming that's for calibrating the measured voltage. I can do what you say and apply any known voltage but I'm curious as to why the instructions say to use a voltage half of the range. Maybe it is non-linear and the midpoint of the range is most accurate? Any thoughts?

 

Yes. If the IC used is ICL7106, then this is taken care of via the intrinsic features of the IC itself which includes a "zeroing phase" in its operating cycle.

You can apply any known voltage less than the range maximum to calibrate the range. The higher the voltage the better the accuracy. In this context, I would use 199.85V to get the best accuracy. I know my meter is calibrated if I get a changing 199.8 and 199.9 display roughly half the time. So you'll need to use a 4-1/2 digit or even a 5-1/2 digit DVM if you really want to calibrate your meter.

I don't know why the instruction suggests half the voltage as it does not make sense to me. The IC operates on a charge balanced principle and there is no way half the charge is a better way to calibrate it.

Perhaps someone would tell me something I'm not aware.

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L.Chung

 

Hi LC,
I suspect the reason its non linear over its full range.

So when calibrated at the v/2 centre point, the 'end' conversions fall within the claimed accuarcy.

This method is not uncommon.
If you visualise a 'curved' response over the full range of an instrument then the centre calibration will give the best accuracy near zero and fsd.

EDIT:
Visualise a 'tangent' line to the curve.

__________________
Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
PIC tutorials: Nigel's www.winpicprog.co.uk/ Bill's: www.blueroomelectronics.com/

 

The ICL7106 datasheet gives typical of only +/-0.2 count for full non-linearity over the entire 0-200mV range so it is indeed very linear. As such it is still better to calibrate at top of range instead of mid range. Not true though for other types of conversion which fits your description and so is best to calibrate at mid-range.

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L.Chung

 

I see the linearity stated for the 7106 as +/-0.2 over the full range, but as I pointed out it claims this 'accuracy' based on a V/2 calibration.

If you calibrated at full range ie: close to Vmax, the bottom end linearity would be less than claimed, a rough guess would either +/- 0.4 [ not knowing the sense of the non linearity.

QED.

__________________
Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
PIC tutorials: Nigel's www.winpicprog.co.uk/ Bill's: www.blueroomelectronics.com/

 

Eric, you're comment is exactly true. Sorry I've missed that.

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L.Chung

 

Any suggestions for how I could easily go about obtaining a 199.85V supply for calibration? Or a 100V? Or any voltage over the maximum output of the LM317 which is currently the only adjustable voltage source I have at my disposal.

Also, there are 2 ICs on the meter. A 7107 and a CD4069.

 

You don't need a high voltage to set your display.

Since your application is to display the output voltage of the LM317 power supply, you just set the LM317 output to its highest and connect it to another DVM to get a reading. Then set the PM-129B display to match. That's all you'll need to do.

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L.Chung

 

OK, I can do that. Thanks.