I would like to use a wheatstone bridge for a sensor circuit and was thinking to use a 3V coin cell (50 mAH) due to limited space. The resistors will be close to the 2.8 ohms mark (slightly varying around this figure). The output will feed into an ATtiny10. How would I work out how long the battery would last? My calculation turned out to be less than an hour of use so I think that I have gone wrong.

Many thanks for your time and help.

 

If your Wheatstone bridge consists of four 2.8Ω resistors the resistance across the drive terminals will be 2.8Ω. The current drawn by the bridge will be (according to Ohm's law, anyway) 3.0V / 2.8Ω = 1.07 amps, or 1070 mA. If your coin cell has a capacity of 50 mAH, its lifetime driving the bridge will be 50 mAH / 1070 mA = 0.047 hours, or 2.8 minutes.

 

Why such a low resistance?
You need much high bridge resistor values if you want to operate for any reasonable amount of time from a coin cell.

 

Thanks. Could I just add 4 extra resistors to increase the overall resistance on each leg? Say for example 1k, so each leg would then be 1002.8 ohms? If not, is there any other way I could increase the battery life?

Thanks again

 

You could but the output from the bridge would only be about 1/300 of the output for the same change of resistance of one of the 2.8 ohm resistors. If you explain exactly what you are trying to achieve we could probably come up with some suggestions.

Les.

 

hi m,
Download the PDF from this link.
http://r.search.yahoo.com/_ylt=A7x9...s_en.pdf/RK=1/RS=adb_0MVLYh3BwKe.rsjIBzKmQPQ-
E

 

The 2.8 ohms come some sensors that I have made. I got this figure by just connecting the multimeter up to it at rest position. The resistance will only vary very slightly during operation. Am I correct in thinking the the resistors that come from manufactures vary from what the colour bands represent? Seen as I am dealing with small tolerances would this mean that my idea of adding extra resistors would not work? If I added an OpAmp at the output would I be able to get it back up to something that the ATtiny10 could use?

Thanks again for your time and help

 

If the 1K resistors you plan to put in series with your 2.8 ohm resistors are 1% tolerance then there value can be betwee 990 and1010 ohms so if you are unlucky there could be 20 ohms difference between samples. which is about 6 times the value of the resistors that you are trying to measure. Is this a strain gauge you are making or are you not even going to give us that much information ?

Les.

 

Could I just add 4 extra resistors to increase the overall resistance on each leg? Say for example 1k, so each leg would then be 1002.8 ohms? If not, is there any other way I could increase the battery life?

Have you calculated the battery life you would get if you added a 1000 ohm resistor to each leg of your Wheatstone bridge? I showed you how to do that in post #2 above. What is the resulting battery life? Is that adequate?

Do the calculations yourself, but here's the answer: 16.7 hours.

As others have pointed out, there's another serious problem with adding those resistors: they drastically reduce the sensitivity of your bridge, by a factor of 1000/2.8 = 357 times. If, as you say, your 2.8 ohm sensor resistors change only slightly while sensing whatever it is you're trying to sense, that resistance change will be only 1/357th as significant if you add fixed 1000 ohm resistors in series with them.

You'll have a difficult enough time getting usable results from this Wheatstone bridge even without adding those 1000 ohm resistors; adding them will make the job impossible.

The 2.8 ohms come some sensors that I have made. I got this figure by just connecting the multimeter up to it at rest position. The resistance will only vary very slightly during operation. Am I correct in thinking the the resistors that come from manufactures vary from what the colour bands represent? Seen as I am dealing with small tolerances would this mean that my idea of adding extra resistors would not work?

Yes, that's one more reason why adding extra resistors is a poor idea: their tolerance, whether it be ±1%, ±0.1%, or even ±0.01%, would completely overshadow any resistance change in your 2.8 ohm sensor resistors.

Another factor, closely related, is that all resistors are somewhat sensitive to temperature. Unless you want to spend a LOT of money on those 1000 ohm resistors, their temperature sensitivity will most likely be rated at 100 ppm (0.01%) per °C. That means that for every °C your ambient temperature changes, your 1000 ohm resistors could change by as much as 0.1 ohms, rendering your measurements meaningless.

The bottom line regarding powering a device containing a 2.8Ω Wheatstone bridge from a 50 mAH button cell is "you can't get there from here." You need to re-think this whole thing from scratch.

 

Thanks. Could I just add 4 extra resistors to increase the overall resistance on each leg? Say for example 1k, so each leg would then be 1002.8 ohms? If not, is there any other way I could increase the battery life?

Thanks again

You do not need four additional resistors. All you need is one resistor in series with the battery.
Of course you will have to amplify the signal from the bridge using a differential instrumentation amplifier.

 

Thanks for your help and advice. I did get the same numbers in terms of time, it seemed quite low so just wanted to check that it was correct- it was not what I intuitively thought. Much appreciated! Yes they are basically strain gauges. As you recommend I am rethinking the idea. In order to get around micro amps of usage and mitigate the tolerance issue I am thinking that means the gauges themselves would have to be much larger resistance values. To increase the 2.8 ohm resistances I could experiment with different geometries and choose a less conductive material. With resistances around 10K I think this gives me the micro amps of usage but the output of the wheatstone bridge is very small. Could a normal Op Amp work with this or would a differential instrumentation amplifier be needed? Never used Op Amps before so any advice on this would be most gratefully received.

 

Check out the ebook pages on AAC about opamps and instrumentation amplifiers:

https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/introduction-operational-amplifiers/

 

With resistances around 10K I think this gives me the micro amps of usage but the output of the wheatstone bridge is very small. Could a normal Op Amp work with this or would a differential instrumentation amplifier be needed?

In simplest terms, an instrumentation amplifier is just one or more op amps integrated with all the supporting components needed to make it function as an IA, all on one chip. One thing you need to consider is that your Wheatstone bridge is not the only thing that will be drawing current from your button cell: the op amp or IA will require supply current, too, and you must consider that in figuring your button cell operating life.

Never used Op Amps before so any advice on this would be most gratefully received.

If you use an operational amplifier, you will end up having to configure it as an IA anyway, to deal with a strain gage Wheatstone bridge; so you might as well use an IA instead.

One that I've used successfully in several low-power strain amplifier designs is the LT1789. It will work on supply voltages as low as 2.5 volts, and has a maximum supply current of only 95 μA. I'm sure there are others, from other manufacturers, that would probably work but I'm not familiar with them.

 

Thanks for the info. Does anyone know of anything that is cheaper that could work? One that I have found is called AD620. Do you think that would be worth a try?

Data: http://www.analog.com/en/products/amplifiers/instrumentation-amplifiers/ad620.html#product-overview

Cheap on eBay: http://www.ebay.co.uk/itm/5Pcs-Sop-...228741&hash=item1a29c29b46:g:66YAAOSwo4pYIOTh

 

hi,
I use the AD623 , its capable of working from a single supply voltage.
What is the application.?

E

 

Thanks for the info. Does anyone know of anything that is cheaper that could work? One that I have found is called AD620. Do you think that would be worth a try?

Data: http://www.analog.com/en/products/amplifiers/instrumentation-amplifiers/ad620.html#product-overview

Cheap on eBay: http://www.ebay.co.uk/itm/5Pcs-Sop-...228741&hash=item1a29c29b46:g:66YAAOSwo4pYIOTh

If you look into the details on the AD620 data sheet, you can easily see that it would not be suitable at all: it requires a supply voltage of at least 4.6 volts (±2.3 volts); it draws up to 1.3 mA supply current (there goes your battery life); and its output cannot swing close enough to either of the supply rails to give a useful output.

Yes, the LT1789 is expensive; but you've got some pretty demanding requirements and I don't know that you're going to find anything much cheaper. As always, performance costs.

 

It would seem that some design compromise is needed in order to try and keep the cost down. Assuming that it is possible from an ergonomics point of view, I am now considering the case of adding a second 3V coin cell to give 6V.

When it says 1.3mA max supply current, is that related to the gain? I.e 1.3mA at the max gain of 10000? Is the current just worked out from using Rg, (+Vs) - (-Vs) and ohms law?

Assuming that I know the output of the Wheatstone bridge and the required gain (by looking at the input requirement of the ATtiny). Is my thinking correct regarding how to go about optimising the design:

1. Rg = 49.4K / (G - 1) <- obtained from the data sheet. I also know that G = Vout / Vin

2. I = ( (+Vs) - (-Vs) ) / Rg

3. Use total current in circuit (including bridges and ATtiny) to work out the battery life

4. If not acceptable, change the supply voltage, bridge resistors or both and then try again

Thanks again for your time, support and constructive criticism, it is really helpful and I am learning a lot! Much appreciated!

 

When it says 1.3mA max supply current, is that related to the gain? I.e 1.3mA at the max gain of 10000? Is the current just worked out from using Rg, (+Vs) - (-Vs) and ohms law?

No. There is absolutely no relation between the gain and the supply current.

1. Rg = 49.4K / (G - 1) <- obtained from the data sheet. I also know that G = Vout / Vin

Correct.

2. I = ( (+Vs) - (-Vs) ) / Rg

No. Rg only determines the gain. It has no bearing on the supply current.

3. Use total current in circuit (including bridges and ATtiny) to work out the battery life

Correct.

4. If not acceptable, change the supply voltage, bridge resistors or both and then try again

Correct.

 

2. What is it that determines the current?

 

2. What is it that determines the current?

The power supply current draw of an amplifier is determined by its design.
Typically there's little you can do to change it.