Influence of switch resistance in Hamon Dividers

[e61_phil]

Hi,

I tried to investigate the importance of the switch resistance for the parallel to series switch (calibration to divide) in a 1:10 Hamon divider.

For a first investigation I assumed everything else than the switch is perfect. In a perfect trim (0V bridge voltage) the resistors of the upper and lower leg are exactly the same.

I started calculating the resistance of the upper arm in calibration mode (parallel) (see net1.png, net2.png btw. is it possible to embedd pictures without an external source?).

I wrote a small function to calculate the resistance as a function of R and the switch resistance Rs. I kept all resistors to have room for playing in the end. (rleg_function.png)

The result with 40k output resistance and 10mR switch resistance (as described in the Fluke 752A manual) leads to 40.000009k (rleg_calc.png). That is already a deviation of 0.2ppm and not only 0.025ppm as described in the Fluke manual.

To verify my calculation I tried the same circuit in LTSpice (spice1.png and spice2.png). That gave the same results.

Therefore, I have to ask: Where is my fallacy?

Best regards
Philipp

[Gyro]

I did the same (well, back of an envelope) calculation and ended up using solid screw terminal barrier strip for my '2:1 switch' on my primitive 10k divider. I've no idea how Fluke manage it [EDIT: for their stated spec].

https://www.eevblog.com/forum/metrology/anyone-else-built-a-hamon-divider/


P.S. It's not just the (variable) switch contact resistance, it's the interconnect wiring too.

[Conrad Hoffman]

I go with the shorting bar. Remember that you can parallel wafer switch wafers, so if you start with a good switch you can probably do better than you'd think.

[e61_phil]

Thanks for your feedback!

My biggest concern is, that I got 0.2ppm error and not 0.025ppm as described by Fluke. The Fluke 752A is very old and therefore, I think it is verified over many years and my calculation must be wrong anywhere.

If the switch resistance is really that critical, I wonder why it isn't verified periodically. Fluke wrote the Fluke 752A is a ratio standard and don't need any traceable calibration. It seems it is possible to Null the 752A and it might be off anyway, if the contact resistance has increased.

[Vgkid]

A really good quality rotary switch, especially one with paralleled connections will be very repeatable.
Some test here:
https://www.eevblog.com/forum/metrology/4-wire-switch-selectable-resistance-standard/msg1192644/#msg1192644

[e61_phil]

A really good quality rotary switch, especially one with paralleled connections will be very repeatable.
Some test here:
https://www.eevblog.com/forum/metrology/4-wire-switch-selectable-resistance-standard/msg1192644/#msg1192644

Thanks, I had my various Pt100 simulators in mind. All of them need a restoration of the switches to get back into specification.

[Vgkid]

A really good quality rotary switch, especially one with paralleled connections will be very repeatable.
Some test here:
https://www.eevblog.com/forum/metrology/4-wire-switch-selectable-resistance-standard/msg1192644/#msg1192644

Thanks, I had my various Pt100 simulators in mind. All of them need a restoration of the switches to get back into specification.

The good thing is that for a rtd simulator, a super low switch contact resistance is not needed.

[beanflying]

About to get around to making one with a long awaited collection of goodies from Edwin. Given the cost of quality rotary switches I was going to use plated shorting bars and good terminals instead for mine.

Much as it will slow down usage I am only building it for my own use so speed isn't really an issue in my case at least.

[Kleinstein]

The difference between the Fluke manual and the simple calculation could be the base value for the PPMs. An error of 0.2 ppm for the 40 K resistors would result in an error of about 0.02 ppm if the input voltage.

Companies tend to provide the better looking numbers if they can. 

[e61_phil]

The difference between the Fluke manual and the simple calculation could be the base value for the PPMs. An error of 0.2 ppm for the 40 K resistors would result in an error of about 0.02 ppm if the input voltage.

Companies tend to provide the better looking numbers if they can. 

The idea is good, but they wrote 1:10 0.2ppm of ratio. And they also wrote 0.2ppm of output.

My concern is still, why the switch alone already causes 0.2ppm. The calculation is no shortcut nor something difficult. I can not imagine that the Fluke 752A is worse than specified and it wasn't noticed until today.

In the manual, they have the following formular to calculate the influence of the switches:

Vo/Vi = 1 / ( N + (dR/R) )

where:
N = ratio -> 10 for 10:1
R = output resistance (40k)
dR = switch contact resistance (10mR)

That gives -0.025ppm as described in the manual. But where does the formula come from?

[Dr. Frank]

You need to calculate the ratio, that is
Ratio = R/(10R + R_switch),
with R =40k, and R_switch is the failure from the 1:1 balancing of the Hammon bridge arrangement.
The error of the ratio is then
(Ratio - 0.1)/0.1 with 0.1 being the default ratio.
This error then calculates into
-R_switch/(10R+R_switch) which is about R_switch/10R, which gives 0.025ppm with  R_switch =10mOhm.
The only thing I could not find out, how the switch resistance only counts once, not 2 or more times.
It could be important to adjust one of the outer chains instead of the middle one.
Frank

[e61_phil]

You need to calculate the ratio, that is
Ratio = R/(10R + R_switch)
With R =40k and R_switch will be the failure from the 1:1 balancing of the Hammon bridge arrangement.
The error of the ratio is then
(Ratio - 0.1)/0.1 with 0.1 being the default ratio.
This error then calculates into
-R_switch/(10R+R_switch) which is about R_switch/10R, which gives 0.025ppm with  R_switch =10mOhm.
The only thing I could not find out, how the switch resistance only counts once, not 2 or more times.
It could be important to adjust one of the outer chains instead of the middle one.
Frank

I calculated the ratio and that gives also 0.2ppm with 10mR switches.


They wrote one have to add the 10mR in series to the resistance. That is roughly correct (8.8mR is more precise for the two switches). But you have to add that to 40k and not to 10x 40k.

They idea of the formula might be for the 1:10

gain = Rout / (Rout + 9R + Rswitch)

And as long Rout = R it is the same as

gain = R / (10R + Rswitch)

but that seems to be incorrect.


Can anyone show that the formula given by Fluke is correct?


PS: I tried the "nulling" with every of the four resistors. All experiments results in the same 0.2ppm

[Kleinstein]

The Fluke formula (3-1)  in the manual seems to apply to a different switch somehow in series with the divider , not the series / parallel switches.
In this case the fluke formula makes sense - though I don't see the switch involved.

Under 3-30 they handle the series / parallel switches  (RS1, RS2) separately - though without many details.

So the formula in question is for the wrong problem.

[e61_phil]

The Fluke formula (3-1)  in the manual seems to apply to a different switch somehow in series with the divider , not the series / parallel switches.
In this case the fluke formula makes sense - though I don't see the switch involved.

Under 3-30 they handle the series / parallel switches  (RS1, RS2) separately - though without many details.

So the formula in question is for the wrong problem.

THANK YOU!!

That explains everything

It seems there is some black magic involved in the 752A to achieve the impressive specifications. They wrote "The design of the instrument is such that most of the effects of this errors are reduced by adjusting the interconnections resistances." (F752a.png)

I already saw during playing around that some detunning of the Rout (intentionally setting Rout higher or lower than the parallel resistance of R1, R2 and R3) can compensate for the switch resistance (see dirty.png). Maybe that is what they are doing. The curve will change with the switch resistance of course.

PS: The graph is achieved by setting Rout to Rout + detunning and than nulling the bridge by variation of R2.

[e61_phil]

A more coloful graph

If the switches really have 10mR for the parallel to series switch and they do it by detunning (or different selection) of Rout, the switch should still stay within ~2mR to keep the 0.042ppm.

It would be also possible to increase the switch resistance intentionally and compensate later for that. That might give more stable behaviour over time. It would also be interesting how guildline solved this issue in their automatic divider. With relay contacts the problem might be even bigger.

[Kleinstein]

One could compensate for the known fixed part of the switch resistance (e.g. the wires), but there is still the possibly variable part, e.g. due to oxide and the exact switch position than is not easy to compensate. So it needs a construction the is very repeatable.
Before critical measurements checking the switches (for low resistance) is definitely a good idea.
A way for compensating for some fixed switch resistance would be some intentional asymmetry in the 120 K resistors. This would lower the resistance of the parallel combination a little.

[Lesolee]

Therefore, I have to ask: Where is my fallacy?

I would say that at least part of the problem is that you are showing a 2-wire connection, when a Hamon build-up resistor requires a 4-wire connection to function correctly.

This image is from Hamon's 1954 paper.



The idea of adding resistors into the potential leads dates back to 1912 (according to Hamon's references).

[e61_phil]

Therefore, I have to ask: Where is my fallacy?

I would say that at least part of the problem is that you are showing a 2-wire connection, when a Hamon build-up resistor requires a 4-wire connection to function correctly.

This image is from Hamon's 1954 paper.

(Attachment Link)

The idea of adding resistors into the potential leads dates back to 1912 (according to Hamon's references).

You're talking about about a complete different thing.

The fallacy was already revealed by Kleinstein.

[Kleinstein]

The circuit from the old Hamon article could still be used to a certain degree to compensate some of the switch resistance.  I don't know if this works all the way, but it could compensate at least some and would be active during the 10:1 divider adjustment step.
The main idea is using 4 way junctions and separate switches for voltage and current.  The voltage sensing switches would only carry a small current cause by errors in the approximate current sources through extra resistors. It needs about 2-3 times the number of switches and extra, less critical resistors, and could compensate something like 99% (depends on the accuracy of the extra resistors and adjustment range) of the switch resistance.

I don't know if the 752 uses this technique, but it is plausible.

[e61_phil]

The circuit from the old Hamon article could still be used to a certain degree to compensate some of the switch resistance.  I don't know if this works all the way, but it could compensate at least some and would be active during the 10:1 divider adjustment step.
The main idea is using 4 way junctions and separate switches for voltage and current.  The voltage sensing switches would only carry a small current cause by errors in the approximate current sources through extra resistors. It needs about 2-3 times the number of switches and extra, less critical resistors, and could compensate something like 99% (depends on the accuracy of the extra resistors and adjustment range) of the switch resistance.

I don't know if the 752 uses this technique, but it is plausible.

How do you bring the current and voltage path together to form the bridge point?

[Lesolee]

You're talking about about a complete different thing.

I agree. I downloaded the 752A manual and the simplified circuit is exactly as you have it.

I have not seen “Hamon” mentioned in the manual at all, and the circuit seems entirely unrelated to a Hamon transfer standard like the Guildline 9350 for example.

Frustratingly the last few pages showing the actual schematic diagrams are missing from the manual (downloaded from the Fluke site). 

I am suspicious of their simplified circuit, since it seems likely that you would want to run each equal resistance at the same current in order to minimise self-heating and voltage coefficient errors.

[Kleinstein]

Bringing the voltage and current paths together can work a little similar to the kelvin bridge. It is a little tricky and limited in how much of the resistance can actually be compensated. At the bridge side the current sharing resistors have to be quite accurate and small (e.g. 1 Ohms range).
When used as the 10:1 divider some of the current sharing resistors would be shorted out by an additional switch. This could explain why there is a switch in series with the full bridge at all.

attached is a LTspice picture and sim file, that used JFETs for the switches.  Due to the high resistance the sharing resistors are larger.

[Lesolee]

I have written it out fully, just for completeness.

[e61_phil]

I have written it out fully, just for completeness.

(Attachment Link)

Thanks!

Just for the clarification: You're describing the series resistance from the switch to the input of the divider.

I thought they are describing the switch resistance for the series- to parallel switch. But for this switch (and the simple hamon circuit) my calculations from the first posts are correct. Therefore, if you build a hamon divider with the textbook design and you use 40k for Rout and 10mR switches, you will end up with 0.2ppm (not 0.025ppm) uncertainty just from the switches alone (everything else will add up).

The Fluke 752A has a 1:10 uncertainty of 0.2ppm including all effects. Therefore, they must use some black magic like Kleinstein describes (or simillar).

[Dr. Frank]

Philipp,
no, in Lesolees calculation (it's exactly the same I have done) that dR is NOT describing the series switch resistance.

This dR comes from the Hamon adjustment .. the series/parallel switch resistance is compensated by the adjusted leg of the three 120kOhm resistor chains.
That means, this leg is mathematically reduced by the switch resistances.
Either you simply assume that all three 120k (plus 2 * 10mOhm) are in parallel, or you elaboratley calculate the star/ triangle transformation, that might give 3 *10mOhm, maybe, but the Hamon adjustment condition says, that this paralleled circuit has to match exactly the 40k of the lower leg.

This Hamon adjustment condition you always miss in your calculations!

Therefore there is no black magic behind, but simple math.

Btw.: there is also no difference between the described switches in the assembly, because all switches / wipers are sitting on the same switch and are made from the same material. 

The schematics are also available.
Frank