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Transducer digital output to analogue


johndeverill

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Hi John,
It's a pressure transducer with a positive DC differential analog output that is powered from a positive supply. Digital has nothing to do with it.

1) Its wheatstone bridge schematic is shown in your Image 3-D.
2) Its 4 terminals (2 for power and 2 for differential output) are shown in Image 4-D.
3) Seven sensors are listed in Image 6. Do you have the one for 30psi?
4) Its spec's in Image 7 lists a 5K Ohm input resistance for its load.

Since the 30psi sensor gives a full-scale differential output of 330mV DC, if you want only a 330mV ground-referenced output, you need an opamp that is wired as a differential amplifier with a gain of one.

What is the resistance of whatever you are connecting to the output of the opamp?
What is the minimum voltage that you need its output to be?

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Johndervill,
You already have an analog signal. You just need to amplify it and scale it to read correctly.
We build quite a few variations of baro sensors.
Are you going to display it on a LED display, LCD, or just use a voltage out like 0-1 or 0-5? Will you be feeding the input to a micro? Just curious of where you are going from here.

MP

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Hi Guys,

Many thanks for the info. I am very new to electronics so what I may be stating above may seem like a load of gobbledeegook, apologies. What i am trying to build is a pressure display/switch circuit. I have a LDR switch circuit currently and somehow was trying to replace the ldr with the pressure transducer so that it would switch dependent upon pressure rather than light. I think the circuit is using a transistor to switch as opposed to a op-amp. I also have a panel meter, which has differential inputs, so I assume would connect without a problem. The confusion I am having is getting my head around differential signals as I have seen transducers that output via one wire as opposed to +/-.

Does this sound a feasable concept?

Thanks

John

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The LDR is just changing resistance dependent upon the amount of light. If you can measure the resistance on this device out of circuit or calculate it while it is powered, then you will only need to make the pressure switch output cause the same resistance and/or voltage changes.

MP

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Hi MP,

That's great. At least I feel I am going in the right direction now. A transistor will have one conection with a varying voltage that acts to switch the circuit won't it?

If so, would i then need to pass the differential output of the tranducer to an op-amp and connect that single output to the transistor? The circuit uses a pot. to vary at which point it switches. If the voltage ranges produced by the LDR and Transducer vary Iassume it would then be a case of varying the pot/associated resistor to get the switching range correct.

Sorry without a circuit diag it is hard to really explain but if the logic is right I think I could get it to work with a bit of tinkering.

Many Thanks

John

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Hi John,
If you need any accuracy of the pressure switch-point, use a comparator IC instead of a transistor. Also use a 10:1 resistor ratio in the differential amplifier opamp to boost the 330mV of the transducer up to 3.3V for the comparator. The reference voltage for the comparator will be variable from a voltage-divider pot for you to accurately adjust the pressure switch-point.  ;D 

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Hi John,
To get any kind of range out of the unit, you would normally use two different op amps. This will allow you to have an offset adjustment and a zero adjustment if need. Not sure what kind of accuracy you need or which pressure sensor you have. I have attached a schematic below. This is how you would get a 0 to 5 volt output that is TTL compatable and can feed directly into a A/D converter if needed. Your pressure sensor probably has a different output range, but this should give you an idea. Hope it is helpful.

MP

post-555-14279142177163_thumb.jpg

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