# [SOLVED]1ma constant current source for PT100 sensor (Circuit problem

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#### Ranbeer Singh

##### Full Member level 5
Hello

I am making 1ma current source for constant

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Sorry for mistake

Continue......I am making 1ma constant current source for PT100 sensor using OP07 op-amp. I am using a voltage divider for negative pin and a series resistance for positive input pins. At the negative input pin i am generating 4.57V and same on positive pin by a 501E resistance it will drop approximately 0.5V. Same volt at both inputs... and transistor should be generate 1ma constant current.
It's not working according my calculation. When i connect 100E resistance it makes 115mv.

V=I*R
0.001*100 = 0.1v

Circuit

http://obrazki.elektroda.pl/9844962300_1455512660.png

#### Warpspeed

Collector current may be 1mA but the emitter current will be 1mA plus the base current, whatever that happens to be.

Replace the bipolar transistor with a Jfet.
The source current will then be exactly the same as drain current, because there is always zero gate current.

#### Ranbeer Singh

##### Full Member level 5
You are right but i am worry with extra voltage.

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I have decided to use JFET for my circuit. What will be suitable JFET for my application.

#### c_mitra

Use the multimeter and measure exact values of the resistors and calculate the current by hand and the transistor specs.

#### KlausST

##### Super Moderator
Staff member
Hi,

I assume VCC is 5V. But OP07 is not specified to operate at 5V.
Additionally there may be an issue wit opamp common mode input voltage range.
Additionally consider a capacitive feedback from opamp output to inverting input and an additional feedback resistor to avoid oscillation.

--> Look for a suitable Opamp. RR input, RR output, unity gain stable.
And as suggested a fet instead of the bjt.

According your voltage reading your supply voltage is slightely higher than 5.00V. Your current is proportional to VCC, so a higher VCC causes higher current. Use a zener or a voltage_reference to improve in this.

Klaus

#### Ranbeer Singh

##### Full Member level 5
Hi Klaus

Datasheet says Wide input voltage range: ±14 V typical, Wide supply voltage range: ±3 V to ±18 V but you says...

I assume VCC is 5V. But OP07 is not specified to operate at 5V.

#### KlausST

##### Super Moderator
Staff member
Hi,

Supply: +/-3 V gives a total of 6V minimal, but you use 5V only.

Input voltage range depends on supply voltage. The given input voltage range is for a supply voltage of +/-15V.
(So about 1V from supply voltage)

Sadly in my (Analog devices) datashheet there is only "input voltage range" specified.
This usually is the differential input voltage range. (With your circuit about 0V)

Good datasheets show:
* differential input voltage range, and
* common mode input voltage range (with your circuit about VCC - 0.5V)

Klaus

Ranbeer Singh

### Ranbeer Singh

points: 2

#### Warpspeed

Thinking about it a bit more, a small n channel mosfet such as a 2N7000 might be better, as it would go straight into your existing circuit without making any other changes.

https://www.fairchildsemi.com/datasheets/2N/2N7000.pdf

Gate threshold for 1mA is 2v typical.
That would fit in quite well with your op amp and +5v supply.

#### Ranbeer Singh

##### Full Member level 5
Oh

I did not think of it. I have tested my all resistance values by a high quality multimeter. The exact valve was given in my circuit but my theory and practical was not matching to each other.
I will check it with 12v supply.

I am using a 0.1uf capacitor on transistor output...Is is write or wrong?

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what type of BJET will be suitable for it?

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JFET

N-channel or P-channel?

#### Warpspeed

The 2N7000 is a small plastic N channel mosfet, and I gave a link to the data sheet.

I never thought to look at the op amp. Duh !

If you have +12v available, that would be much better.

Ranbeer Singh

### Ranbeer Singh

points: 2

#### c_mitra

Can you please explain why you are sending the feedback into the non-inverting input? How it is going to work in this case?

#### Warpspeed

Can you please explain why you are sending the feedback into the non-inverting input? How it is going to work in this case?
The circuit layout is correct.

If current tries to increase, the voltage drop across the 510 ohm resistor increases.
This causes the voltage on pin 3 (non inverting) to fall.
Op amp output falls, current through NPN falls.
So overall feedback is negative.

Ranbeer Singh

### Ranbeer Singh

points: 2

#### c_mitra

I am permanently confused between NPN and PNP! Thanks for the explanation.

#### FvM

##### Super Moderator
Staff member
I presume you are referring to the circuit linked in post #1

I would suggest some modifications. If no FET is available, you should at least use a Darlington (e.g. BC517 for the NPN configuration) to keep the base current error negligible.

Secondly, the output transistor involves a voltage gain for Rsens < 500 ohm which can make the OP loop unstable, because it's not compensated for feedback factors > 1. The problem can be overcome by using a PNP transistor (or PFET) and an inverting OP configuration, otherwise compensation (e.g. a capacitor from OP output to inverting input) is required.

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Below an implementation from a real instrument. Vref should be also used for the ADC to make a ratiometric measurement.

Ranbeer Singh

### Ranbeer Singh

points: 2

#### KlausST

##### Super Moderator
Staff member
Hi,

Can you please explain why you are sending the feedback into the non-inverting input? How it is going to work in this case?
My mistake. I had another circuit in mind with the resisor on the low side.

But a capacitor from output to inverting input is correct...it will stabilize the whole circuit.

Klaus

#### c_mitra

I also have another comment: it is not a good practice to run a complete circuit with a reference voltage because there may be unwanted loading on the Vref. There are reliable and cost-effective solutions for a voltage reference and they should have been used at the appropriate stage rather than using the power supply voltage.

#### FvM

##### Super Moderator
Staff member
I also have another comment: it is not a good practice to run a complete circuit with a reference voltage because there may be unwanted loading on the Vref. There are reliable and cost-effective solutions for a voltage reference and they should have been used at the appropriate stage rather than using the power supply voltage.
Generally correct, but the presented circuit depends at on Vref sourcing the 1 mA current. Using the (analog) power supply as reference can be reasonable if a ratiometric measurement is implemented

#### c_mitra

Using the (analog) power supply as reference can be reasonable if a ratiometric measurement is implemented
If the VCC changes by 20% (say) then the current will also change in a similar manner. This current is being used to measure the resistance of a Pt-resistance thermometer by converting the resistance to a voltage. That purpose will be defeated.

#### FvM

##### Super Moderator
Staff member
We are expecting a regulated analog supply, e.g. +/- 5% tolerance, +/-1 % temperature drift. By using ratiometric measurement (same reference for current source and ADC), the error can be completely eliminated.

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