Sensor Monitor

Hello,

I have got a sensor that outputs a 0 to 1mA DC signal and would like to convert that to 2 to 12V signal, with safe guard limitations to avoid going beyond 0V or above 14V. What sort of circuit/components should I use?

Thanks

andy_king said:

Hello,

I have got a sensor that outputs a 0 to 1mA DC signal and would like to convert that to 2 to 12V signal, with safe guard limitations to avoid going beyond 0V or above 14V. What sort of circuit/components should I use?

Thanks

Click to expand...

hi,
What load resistance will the 0/1mA drive.?

hi Eric

thanks for your reply.

The load circuit is to be designed from scratch, it is basicly a block with that 0-1mA signal coming in and 2-12V coming out. It is the detailed circuit for that block that I am struggling with to design.

Cheers

andy_king said:

hi Eric

thanks for your reply.

The load circuit is to be designed from scratch, it is basicly a block with that 0-1mA signal coming in and 2-12V coming out. It is the detailed circuit for that block that I am struggling with to design.

Cheers

Click to expand...

hi andy,
Reason for asking about the load resistance is there enough voltage overhead on the 0/1mA loop so that we could develop a voltage across a resistor.

EDIT: do you follow what I am asking.?

andy_king said:

sensor that outputs a 0 to 1mA DC signal

so your source is a 1 mA current source.

to convert that to 2 to 12V signal

Click to expand...

"so that we could develop a voltage across a resistor"

(12-2)/1 mA = a 10 kΩ resistor. Shunting the current source with this gives 0 to 10 V out.

Then add an independently powered 2 V level shifter so now we have a 10 kΩ resistor in series with a 2 V source (use one of those LMxxx adjustable bandgap references) to ground, so now we have 2 - 12 V out with 0 to 1 mA in.

Add any diode in series with the 0-1 mA current source so that the 2 V source can't push current back into the 0-1 mA current source.

So far no power gain or impedance level changing is necessary.

Drawing 10 µA by loading this circuit with a 1.2 MΩ resistor will give you a 1% output voltage error.

If the load resistor is smaller than 1.2 MΩ you need a current amplifier (emitter follower or opamp voltage follower).

Willbe said:

"so that we could develop a voltage across a resistor"

(12-2)/1 mA = a 10 kΩ resistor. Shunting the current source with this gives 0 to 10 V out.

Then add an independently powered 2 V level shifter so now we have a 10 kΩ resistor in series with a 2 V source (use one of those LMxxx adjustable bandgap references) to ground, so now we have 2 - 12 V out with 0 to 1 mA in.

Add any diode in series with the 0-1 mA current source so that the 2 V source can't push current back into the 0-1 mA current source.

So far no power gain or impedance level changing is necessary.

Drawing 10 µA by loading this circuit with a 1.2 MΩ resistor will give you a 1% output voltage error.

If the load resistor is smaller than 1.2 MΩ you need a current amplifier (emitter follower or opamp voltage follower).

Click to expand...

hi willbe,
This will only work, if the sensor suppy voltage to the sensor is at least [10V+ Voltage required by the sensor]
and the output impedance of the sensor is low.

I hoping the OP will give us more information on the sensor and its power supply.

Hi, Mr. G!

ericgibbs said:

hi willbe,
This will only work, if the sensor suppy voltage to the sensor is at least [10V+ Voltage required by the sensor]
and the output impedance of the sensor is low.

Click to expand...

If the sensor really is an ideal current source it has infinite output impedance and an infinite open-circuit voltage. Real world current sources bring up the issue of compliance. This link is kind of weird but it does explain the basic idea. http://www.keithley.com/data?asset=6465

In the worst case, probably he should use an opamp as a current-to-voltage convertor and stick in a level shifter somewhere in the feedback loop. He might then need dual supplies. It gets messy quickly.
With no constaints yet on accuracy I'd kind of like to squeeze as much as I can out of the source, first. If the load is a DMM, we're almost done.

Willbe said:

Hi, Mr. G!




If the sensor really is an ideal current source it has infinite output impedance and an infinite open-circuit voltage. Real world current sources bring up the issue of compliance. This link is kind of weird but it does explain the basic idea. http://www.keithley.com/data?asset=6465

In the worst case, probably he should use an opamp as a current-to-voltage convertor and stick in a level shifter somewhere in the feedback loop. He might then need dual supplies. It gets messy quickly.
With no constaints yet on accuracy I'd kind of like to squeeze as much as I can out of the source, first. If the load is a DMM, we're almost done.

Click to expand...

hi,
Perhaps, hopefully Andy will give us more to think about.

EDIT: thanks for link Willbe, but I'm upto speed on Compliance.

hi Eric and Willbe

Thank you very much for your help and I would like to apologise for not being able to get back to you earlier than this.

The circuit is part of a design exercise, it just says "0 to 1mA DC output signal sensor, that needs to be converted to 2 to 12V signal, with safe guard limitations to avoid going beyond 0V or above 14V". So I am afraid there aren't any information about the sensor itself or the load. The 2 to 12V will probably go to ADC-capable MCU for signal analysis and feedback, which is not a problem for me. It is the matter of designing the right circuit to do the 0 to 1mA signal to 2 to 12V signal conversion.

Eric, I didn't understand what you meant by your second post, about the voltage overhead, can you elaborate more please?

Willbe, which of your 2 solutions do you recommend for the 'ideal' design purpose and which in case I ever get around and actually build it?

Many thanks

andy_king said:

Willbe, which of your 2 solutions do you recommend for the 'ideal' design purpose and which in case I ever get around and actually build it?

Click to expand...

I'd go with the first, simpler one. It literally meets your specs.
The under/over voltage prot. may just need two 1N400X diodes and a +1 V source and a +13 V source, or some such thing. This network might get messy.
Do you need a schematic?

A schematic will be brilliant!

Thank you so much!

andy_king said:

Eric, I didn't understand what you meant by your second post, about the voltage overhead, can you elaborate more please?

Click to expand...

hi,
While Willbe is doing his schematic I'll try to explain the overhead voltage bit.

Assume your 1mA source was derived from a voltage source of 10volts.

If we connected in the 1mA loop say a 1000R resistor we would measure 1Volt across the resistor
when a current of 1mA was flowing in the loop.
This could be amplfied to the output voltage you required.

A problem could arise at the sensor source due to the added 1000R resistor reducing the voltage
across the sensor to only 9volts.

If however the sensor required only 5volts to operate correctly, it would still have a 4volt 'overhead' to drive the 1mA loop.

This is the way that a 4mA to 20mA loops works.

The sensor will work OK with a supply voltage of say 12V and drive 4 to 20mA thru the loop.
If we power the sensor with say 24V, that leaves a 12V overhead so that we can add resistors in series with loop in order to generate an output voltage that we can use.

Hope you follow OK.

andy_king said:

A schematic will be brilliant!

Thank you so much!

Click to expand...

I put it in my album called Tech. See if you can retrieve & post it. If so it was my mistake; I had the photo as a folder and not as a pic.

Thank you Willbe for the schematic, I was able to retrieve it, however when I tried to simulate it, the result seemed a bit strange, it is mostly in the voltage shifter. I have uploaded my design and simulation results in an album called "i2v", will be you be kind enough to have a look and tell me where I went wrong?

Eric, I do apologise, but I am still a bit struggling to follow you up with the overhead voltage. I got lost after "If however the sensor required only 5volts to operate correctly, it would still have a 4volt 'overhead' to drive the 1mA loop." If 1V is measured already across the 1000R resistor (which is what I am after) and the sensor got its 5V to operate, what will be the problem with the overhead 4V?

Cheers

andy_king said:

Thank you Willbe for the schematic, I was able to retrieve it, however when I tried to simulate it, the result seemed a bit strange, it is mostly in the voltage shifter. I have uploaded my design and simulation results in an album called "i2v", will be you be kind enough to have a look and tell me where I went wrong?

Eric, I do apologise, but I am still a bit struggling to follow you up with the overhead voltage. I got lost after "If however the sensor required only 5volts to operate correctly, it would still have a 4volt 'overhead' to drive the 1mA loop." If 1V is measured already across the 1000R resistor (which is what I am after) and the sensor got its 5V to operate, what will be the problem with the overhead 4V?

Cheers

Click to expand...

hi Andy,
Perhaps this dwg will help.

The Voltage connected to the two wires in this example is 12Vdc.
The Sensor Unit has an inbuilt 5Vreg.
The reg reduces the voltage to the sensor head electronics to +5Vdc

The 1mA flowing in the loop develops 1Volt across the 1000R.

These leaves only 11V to the input of the sensor unit 5V reg, but as the reg will work down to say 7V
[that is the 5V reg requires at least 2V more in than out, to work correctly]

If I we now fitted a 10,000R in place of the 1000R, we would loose 10V across the 10,000R and the sensor unit would be 'starved' of voltage and would not work correctly.
So this means in practice we are only allowed to drop 5V across the current sense resistor ie: so a 5000R would be the maximum allowed.

We could of course use a 24Vdc PSU and the 10,000R would be acceptable.

Is this any better.?

andy_king said:

Thank you Willbe for the schematic, I was able to retrieve it, however when I tried to simulate it, the result seemed a bit strange, it is mostly in the voltage shifter. I have uploaded my design and simulation results in an album called "i2v", will be you be kind enough to have a look and tell me where I went wrong?

Eric, I do apologise, but I am still a bit struggling to follow you up with the overhead voltage. I got lost after "If however the sensor required only 5volts to operate correctly, it would still have a 4volt 'overhead' to drive the 1mA loop." If 1V is measured already across the 1000R resistor (which is what I am after) and the sensor got its 5V to operate, what will be the problem with the overhead 4V?

Cheers

Click to expand...

How do I expand your album pic (t's too small to read)?
That circuit doesn't look much like what I wrote.

**broken link removed**

The bottom of the current source should be grounded.

Secondly, the two diodes in series pointing down are supposed to be a 2 V reference voltage.
If you're going to build this thing I'd substitute a "programmable shunt regulator" that goes down to 2 V . Some versions of the LM431 go that low.
The LM10 is another way to do this.

I need to get a thicker pencil.

Eric, that does clear it out and explains it nicely, it does make sense, I think I was confused earlier with the different numbers and couldn't work out which goes with what. Thank you.

Willbe, please find attached circuit schematic as taken from the simulator, I left the +13V and +1V sources not connected with the voltmeter for now. I think I got confused which is the entry point of the 2V shift to the circuit. Simulating the circuit with only the 1mA source, diode, 10K resistor and 2 diodes in series then ground, gets the voltage across the resistor to be the required 10V, however, as I understood from the 2 upper schematics in your diagram, that these are the voltage shifter and would fit in with the voltage across the 10K resistor to produce the required 2 to 12V, please correct me if I am wrong. Not sure what do you mean by the diodes being 2V ref voltage. I do apologise for being pain in the backside, I am still fairly new at this and trying to get my head round many things. Thanks

andy_king said:

the 2 upper schematics in your diagram, that these are the voltage shifter and would fit in with the voltage across the 10K resistor to produce the required 2 to 12V, please correct me if I am wrong.

Not sure what do you mean by the diodes being 2V ref voltage.

Click to expand...

The two upper schematics are voltage sources that can sink current for the protection diodes so that the 2-12 V doesn't ever go outside that range.
The upper 10 K output pin can't go above 13 + 1 = 14 V and the lower one can't go below 1 - 1 = 0 V.
Some voltage sources can only source current; driving current into them makes their output voltage higher. These two sources can sink and source current.

A forward biased conventional diode (like a 1N4148 or 1N4001) can act like a ~1 V Zener diode when you push rated current through them; that is, the V-I curve is almost horizontal [with V on the Y-axis and I on the X-axis] at around 1 V, but diodes used like this are not precise and are temperature sensitive.
So, two diodes in series at rated current approximates a 2 V Zener. This is what shifts the 0-10 V output into 2-12 V out.

You can see by your meter that silicon diode D7 is barely turned on, because ~14 V/15 kΩ = ~1 mA, and that's not enough current to turn it on fully. You can reduce the 15 KΩ resistor to get more current and a higher Vdrop across that diode.

D2 and D3 are also not fully turned on and that's why you got only 11.3 V instead of 12 V. You could increase their current to 10 mA by running a 1300 Ω resistor from the anode of D2 up to +15 V. I considered putting in this "keep alive" resistor but I left it out for the sake of simplicity.

Note that the Vdrop across a diode is due to the Vdrop across the PN junction plus a small drop across the diode "bulk resistance", this resistance resulting from the material they used to make the diode.

I should have labelled all these diodes D1, D2, etc.
The one on top of the current source (your computer calls this D1) prevents the 2 V diode drop from pushing current back into the source through the 10 KΩ resistor. 2 V/10 kΩ is 0.2 mA so this would introduce a (100%)*(0.2/1) = 20% error in the current source output.

This should explain the purpose of all the components.

Revision A: Change the 13 V Zener to 12 V; I just noticed this minute that when the output is at 12 V the protection diode D4 is turned on. Since it's not connected in your circuit, you didn't notice this effect.

I must say this thread has been fun, and I didn't expect SW would be breadboarding this back-of-an-envelope circuit.




How do you do the attached thumbnail that can be expanded? I hate posting mostly white paper.

hi willbe.

How do you do the attached thumbnail that can be expanded? I hate posting mostly white paper.

Click to expand...

Open the Reply as normal.
Scroll down until you see the 'Manage Attachments' button.
Left click and it opens an upload window,
Use the 'Browse' to find the files on your PC
Click 'Upload',,, when done close window.

Hit the 'Submit Reply' button when your'e ready to post.