Battery powered Wheatstone? - Page 1
Forum section: electronics / beginners

author: logictom
author_rank: Unknown
timestamp: July 29, 2011, 01:45:29 pm
content: I have a small changing resistance I wish to measure - ~8.2-8.6ohm but need to reduce the current consumption, using three 8.4ohm and the sensor will meaning drawing ~400mA @3.3v which I cannot afford to do.If I change the two lower resistors to 10k to reduce the current draw I go from a swing of ~30mV down to ~30µV, I'm going to feed this into an instrumentation amp and see if this is still usable.Is this the best way to measure a small resistance whilst limiting the current draw or are there better/alternative methods?Thanks as always :)

author: bilko
author_rank: Frequent Contributor
timestamp: July 29, 2011, 01:54:15 pm
content: You could try supplying a constant current and measuring the voltage across the resistor

author: Rufus
author_rank: Super Contributor
timestamp: July 29, 2011, 06:00:44 pm
content: Only the ratio of resistors in each side of a bridge need to match you don't have to match current in each side. If you have a high input impedance amplifier you don't need to waste much current in the reference side.There is no way around the amount of current flowing in the measured resistance determining how much voltage you are going to get to work with. Consider the current only needs to flow while you are actually measuring.

author: IanB
author_rank: Super Contributor
timestamp: July 29, 2011, 06:13:14 pm
content: 8 ohms is loads of resistance to work with. If you put 1 mA through it you will generate 8 mV, which is easily measurable. To what precision do you need to measure the the resistance? This is probably going to determine your measurement technique.

author: logictom
author_rank: Unknown
timestamp: July 29, 2011, 06:30:17 pm
content: Thanks for the feedback, the current source means less current than the original wheatstone but now Rufus pointed out it's the ratio (don't know how that escaped me before) I'll check what I can do with that.Just to clarify Ian, the resistance varies roughly 8.2 to 8.6 ohms, I'm measuring a change of 0.4 ohms.What I'm trying to do is create a basic respiratory sensor with some conductive fabric which changes resistance when stretched - breathing in and out, I'd like to be able to distinguish between shallow and deep breaths but depending on how it goes detecting the breathing would suffice.Something I've just thought about, if fabric is in contact with the skin with a voltage applied what would happen if the person were to touch a grounded object, it will be battery powered so would this be a concern in terms of either safety or affecting the signal?

author: IanB
author_rank: Super Contributor
timestamp: July 29, 2011, 11:05:02 pm
content: To start with your last question, any electrical devices attached to the skin in the area of the chest are going to cause considerable safety concerns. As a practical matter I don't think operating voltages of less than, say 5 V, are going to pose any hazard, especially if the 5 V potential is not directly in contact with the skin, but there may well be some institutional or regulatory paranoia in this area. You would do well to seek expert advice.As to technical considerations, a Wheatstone bridge is designed to measure voltages without loading the device under test. In your case the no load consideration does not apply, so a Wheatstone bridge is an unnecessary part of the design.You want to measure a change of about 0.4 ohms in 8 ohms, so about 5%. That is quite a large change and should be easily detectable.Suppose you use a voltage regulator to generate a fixed 3.3 V supply (just for instance, could be any low voltage). Now place this in series with a 1k resistor and your 8 ohm detector. You will have what is nearly a constant current 3 mA supply across the detector. When the resistance is 8.2 ohms the current will be 3.3 V / (1000 + 8.2) ohms = 3.27 mA and the voltage across the detector will be 3.27 x 8.2 = 26.8 mV. When the resistance is 8.6 ohms the current will be 3.3 / (1000 + 8.6) = 3.27 mA and the voltage across the detector will be 3.27 x 8.6 = 28.1 mV. You have a nearly linear variation in voltage of 1.3 mV in 28 mV or so. You could feed this voltage through a differentiating op amp circuit to detect and amplify the variation.I don't know what kind of cost considerations you have? If you are able to include a microcontroller in the design the possibilities are limitless. If you want something much simpler and you are limited to discrete components then you will need to be a bit more creative of course.

author: Rufus
author_rank: Super Contributor
timestamp: July 30, 2011, 12:16:04 am
content: Quote from: IanB on July 29, 2011, 11:05:02 pmAnd when you hang two resistors across the 3.3v supply to get 28mV to differentiate from with your differentiating op amp you just created the other half of that 'unnecessary' Wheatstone bridge.

author: IanB
author_rank: Super Contributor
timestamp: July 30, 2011, 01:12:39 am
content: Quote from: Rufus on July 30, 2011, 12:16:04 amNo, I don't think so. Differentiation (d/dt) is not the same as differencing (x-y).You would use the differentiating op amp to compute the time derivative of the input voltage, giving positive and negative swings in voltage that correspond to breathing in and out. You would measure the magnitude and frequency of these pulses to indicate the frequency and magnitude of each breath. There is no need to create a 28 mV (or any mV) reference for this circuit. The beauty of differentiation is that all constant offsets are zeroed out. (The downside of differentiation is that noise signals are amplified, but all good things come with a price.)Yes I know that an op amp measures the difference between two signals, but technically a Wheatstone bridge has a particular definition in engineering terminology. Let's not be loose with our technical terms...

author: logictom
author_rank: Unknown
timestamp: August 01, 2011, 11:37:54 pm
content: Thanks for the replies. I've been hoping between this and a few other bits over the last few days.This is forming part of a larger project and will be connected to an ADC for processing.I've got a circuit built after some testing and simulating but still considering alternatives.I tried using a basic inverting opamp and also a differentiating setup, the problem I had was with the ac coupling cap, because the signal is slow changing the cap had to be oversized to pass the signal. I tried to ac couple so I could remove the dc offset which eliminates the need for the second set of resistors to form the same ratio (instead using a Vdd/2 ref for multiple opamps).Is there anyway to ac couple the signal without large caps or am I stuck with using the resistor divider as a ref?Thanks.

author: Zad
author_rank: Super Contributor
timestamp: August 01, 2011, 11:45:07 pm
content: Maybe I have the wrong end of this (its possible, as it is still 26C in here at 1am, and horrible humidity) but you seem to be using a capacitor to remove a DC offset from a wanted signal that is itself very nearly DC?

author: logictom
author_rank: Unknown
timestamp: August 01, 2011, 11:58:50 pm
content: Pretty much, so I can offset it to around Vdd/2.... is there another way?Ditto to the conditions, I've had fans blowing all day and it's still too warm.

author: IanB
author_rank: Super Contributor
timestamp: August 02, 2011, 12:18:56 am
content: Jeri Ellsworth is doing something rather similar here with a motion detector (filtering out fast and slow changing signals and keeping just the ones of interest):Maybe some clues there for your own project?(The time constant of an RC network depends on both R and C, so maybe you can increase R instead of having a bigger C?)