Rate my voltage divider

:man: : On this episode of Rate my Voltage Divider! we have a returning contestant: Pix.
What you got for us today Pix?

:man_beard: : Well Gary I have this unstable garbage, it’s supposed to provide an analogue voltage meter values between 50mv and 900 mv:

:man: : Oh Pix, you meek fool! That won’t work! The resistors values are too low and the inherit impedance of the voltage meter will significantly affect it’s own input voltage!

:man_beard: : What do I win?

:speaking_head: crowd boo-ing noises

I think my voltage meter wants values between 0v and 1v.
Maybe I should consider 0.05v and 0.7v just to be on the safe side and not burn out the meter.

The challenge is that I would love to use this 10k Digital Potentiometer whilst still achieving a nice range of voltages out of my divider. Are there any tricks?

2 Likes

Hi Pix

Use the whole 10k and forget about the others.
The other resistors are used where you want to limit the range.
This type of thing is only predictable where the load is much higher than the lower resistive element (10X???).
If the load is too low the resistive (or reactive) value has to be considered in parallel with the lower resistive part of the divider.
On the other hand if the source resistance (or impedance) is too high it has to be considered in series with the upper resistive part of the divider.

So you see it is not just a matter of connecting a couple of resistors.

If it is a modern DMM it should be auto ranging. If it is an older manual ranging device or an analog meter it is normal (I stress “normal”) practice to start on the highest range if the voltage is unknown. An analog meter should be left on the highest range possible when not in use . Or that is what I have always been taught anyway.
Cheers Bob

Just had another look at your sketch.
What sort of a volt meter is only 50Ω.
A DMM is 10MΩ

1 Like

Using the values in the circuit the ratio across the meter varies from 0.0033 to 0.0654.
If the supply was 5V this would give a range from 0.0165v to 0.437v.
This is not considering the current needed to trigger the meter.
(50Ω is very very low, analog meters are normally in the k ohms range)

You say the meter has a range from 0 to 1V at 50Ω, so it looks like you are trying to limit the voltage across the meter to that range. What is the supply voltage ??

Without knowing the rest of the circuit it is has to say how well it would work or not.
Sorry, but I don’t really see what you are trying to achieve.

Cheers
Jim

PS amusing intro !!!

1 Like

Supply

:man_facepalming: I should have put it on the diagram. It’s 5V. :slight_smile:

Meter impedance.

I’m basing that off my multi-meter :slight_smile:
I took a photo to show my working. i figure this thing must be decommissioned from a power plant (based on the kilovolt range)

Clarifying the math.

That’s precisely my problem.
Just to fully illustrate this let me walk through the math so we’re all on the same page.

I make it

// Where VoltDiv Output is  (Vin*R2) / (R1 + R2)
Lower Bound Vout = (5 * 110) / (10500 + 110) = 0.0518v
Upper Bound Vout = (5 * 110) / (500 + 110) = 0.902v

However, that’s in a vacuum, when you add the voltage meter in parallel with the 110 ohm resistor it changes everything.

// Where Impedance of Parallel Resistors is  (Vin*R2) / (R1 + R2)
R2 = (110 * 51) / (110 + 51) = ~35
Lower Bound Vout = (5 * R2) / (10500 + R2) = 0.0166
Upper Bound Vout = (5 * R2) / (500 + R2) = 0.327

And that’s assuming that the meter’s impedance doesn’t change as current is applied.
That’s what I meant when I said the circuit is “unstable”.

Limiting the volts

I think I do want to limit the range. As I understand it from the person I bought it off if I throw 5v into this meter it might “burn out”. He recommended nothing over a volt.

What am I trying to achieve? :man_shrugging:

I guess the point of my post was to learn how to build a better voltage divider (that’s stable when I draw from it’s output). I wanted a voltage divider because I thought it would be a good way to map scale 5v to lower voltages between 0 and 1. It may be that a voltage divider is not the best solution and I’m all ears for an alternative.

The point of my voltage divider was to play with this cool multi-meter I bought.
I’d like to use the digital 10k potentiometer because then I could control it with a micro controller and I thought that would be fun. I didn’t really have a goal beyond that.

Thanks for your replies @Robert93820 & @James46717
Pix :heavy_heart_exclamation:

Hi pix
Firstly what does it say on the meter scale. KILOVOLTS. The last time I looked 1kV was 1000V that is the meter is designed to fit into a system that measures up to 150000V. This would not be the meter itself but would have quite a few high voltage resistors in series as multipliers.

All analog meters are basically current meters and have multiplier resistors externally to operate as Volt meters
As it says on the scale (the print is not that small to be missed if you had looked) FSD (Full Scale Deflection) 1.2mA and also "use with Ext Res.
image

Almost time you had a read up on some basics methinks.
Cheers Bob

Hi Pix
I missed this one

Not possible if the current you “draw” is not constant.
As described above the minute you take any current from a “voltage divider” it means the load resistance is low enough th require that current so all calculations will go out of the proverbial window and you can start again and the same thing will apply every time the load and thus the current changes.

Th put it as simply as I can. This arrangement of resistors is a VOLTAGE divider NOT a CURRENT divider.
Cheers Bob

Thanks for the info, I now understand what is going on.
The circuit with those values will damage the meter.
The range of current through the meter will be 0.327/51 = 6.4mA to 0.0166/51 = 0.3mA.

You want most of the current to flow through the parallel resistor so the 110 ohm needs to be smaller which will bring the voltage down.
If they were both 51 ohms.

 // Where Impedance of Parallel Resistors is  (Vin*R2) / (R1 + R2)
R2 = (51 * 51) / (51 + 51) = ~26
Lower Bound Vout = (5 * R2) / (10500 + R2) = 0.0123v, 0.0123/51 =  0.24mA
Upper Bound Vout = (5 * R2) / (500 + R2) = 0.247v, 0.247/51 = 4.8mA

Another way.
51 ohms, max current 1.2mA, therefore max voltage will be 0.0612v.
If you did not have the parallel resistor, meter in series, the top resistor would need to be higher.
5 - 0.0612 = 4.9388v, current 1.2mA, therefore resistor = 4k2 ohms.
This is the situation without the 10k pot, with it the current would be 0.35mA. It may be enough that the meter does not fully go to zero, but that was always going to be the case.
So varying the resistor varies the current through the meter from 0.35mA to 1.2mA.
As long as the current does not exceed 1.2mA the meter would be ok.
Changing the value of the 10K may be necessary to get a better swing of the needle.

Pretty good exercise in theory now to test it.

Regards
Jim

1 Like

Hi All
Why not do what should have been done in the first place…
Analog meters are rated in Ohms per Volt. Thus a movement of 1mA with a series resistor of 1kΩ (total including the meter resistance) would be said to have a sensitivity of 1kΩ/V.Then if you want a FSD of 20V the series R would be 20kΩ.

Basic DC metering theory

The best bet would be to make this meter a Volt meter with a FSD of 1.5V which fits into the scale markings good. So the TOTAL resistance would have to be 1.25kΩ. The meter is 50Ω so if you insert a 1.2kΩ resistor in series with it and treat the whole thing (meter and resistor) as a volt meter with a FSD of 1.5V.

Problem solved. This is still quite low to be measuring a voltage divider output but the low value resistor in the bottom R in this case would allow this. The meter resistance falls into the 10X range.
Cheers Bob

1 Like

All good mate.

Yep. That’s a nice way to phrase my question :slight_smile: .

FS = 1.2ma.

Ah man I was wondering what that print on the front meant. Good to know.

Ah yeah I follow.
Let’s make it 5k ohms for a buffer.

// Lower Bound
5v/15kΩ = 0.34ma
// Upper Bound
5v/5kΩ = 1ma

Just to clarify we’re saying that the needle on this high voltage meter is current controlled?

1 Like

Hi Pix

Nearly time you realised this
ALL analog meters of the moving pointer (except electrostatic meters) type are current meters. THAT METER MOVEMENT on its own IS NOT A HIGH VOLTAGE METER. It is only the scale markings that indicate this. These scale markings could be anything.
Cheers Bob

2 Likes

Yeah cool. :slight_smile:
How did people measure voltages before digital multi-meters?
When my DMM measures voltage, is it actually measuring current and then doing some ohms law math to interpret it as volts?

Success and thank yous


How good is math! These values are super tight to the calculations (I used a 4.7k resistor in series with a 10k pot). Thanks @James46717 and @Robert93820.
I would never have figured out the needles was current controlled not voltage controlled without your experience. :slight_smile:

3 Likes

Hi Pix

With 1mA movements with multipliers to produce 1kΩ/V volt meters
Then along came more sensitive movements (50µA) to allow 20kΩ/V meters.
Then Vacuum tube voltmeters (VTVM) with a 10MΩ input resistance across all ranges.
A bit of a stopgap which was FET input sort of VTVM. Not sure exactly how these worked.
DDMs have been around in some form for a long time now.

Bit of a side issue. Analog meters are still extensively used. Try adjusting a tuned circuit in a transmitter or receiver with a DMM. They are so slow this is almost impossible so all is not lost with “old” technology… Another use is very high voltages. Most are analog like the one you are fiddling with.
History lesson over.
Cheers Bob

2 Likes

Ohhhh, so because the meter’s internal resistance in a known quantity it can run ohms law to calculate the voltage in real time?
Thanks for the history lesson. :slight_smile:

Running off amps was way easier than a volt divider. Good system. :slight_smile:

2 Likes

Hi Pix

Not quite sure what you mean here but if you are happy I am happy. Will leave it at that.
Cheers Bob

1 Like