how to use transistor to increase resistance

Thread Starter

saeidbb

Joined Jan 21, 2022
25
hi.
I have a variable resistor for calculating temperature that works in the range of 0.3 to 10 kilo ohms and I have a temperature indicator that works in the range of 0.8 to 20 kilo ohms. How can I change this resistor to the desired range of 3 to 7.5 using a transistor?

I edited this post and replaced the values close to reality
 
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Alec_t

Joined Sep 17, 2013
12,744
Welcome to AAC!
It probably will need more than just a transistor.
We don't have sufficient info. Can you post a schematic of your temperature sensing circuit and your temperature indicator?
 

Thread Starter

saeidbb

Joined Jan 21, 2022
25
Welcome to AAC!
It probably will need more than just a transistor.
We don't have sufficient info. Can you post a schematic of your temperature sensing circuit and your temperature indicator?
In fact, I used a digital thermometer, the picture of which is in the link below.

https://dkstatics-public.digikala.c...=image/resize,m_lfit,h_600,w_600/quality,q_80

And instead of its own resistive temperature sensor, I used the car's temperature sensor

The resistance range of the digital thermometer sensor itself is about 3 to 7.5 kOhm. And the resistance range of the car temperature sensor is about 1 to 3.5 kOhm
 

MrChips

Joined Oct 2, 2009
25,930
hi.
I have a variable resistor for calculating temperature that works in the range of 1 to 3.5 ohms and I have a temperature indicator that works in the range of 3 to 7.5 ohms. How can I change this resistor to the desired range of 3 to 7.5 using a transistor?
Did you omit the k as in kΩ?
 

MisterBill2

Joined Jan 23, 2018
11,623
The simple way to produce the scaling change you want is not simple. I am guessing that one side of your temperature sensor resistor is tied to common (ground) and not changeable. AND I am guessing that one side of your temperature display device input is pulled up toward some higher voltage. What has not been mentioned is if both anticipate a positive temperature coefficient, or a negative one, or perhaps one of each. An op-amp circuit with adjustable gain and offset can drive a transistor as a variable current sink and make it work. But calibration will be rather tedious at best. And there will probably be some linearity issues as well. Thermisters are often not linear over a wide range.
 

MisterBill2

Joined Jan 23, 2018
11,623
OK, the first step will be to use a variable resistor with the digital temperature display and record the resistance versus temperature display. Doing that for every ten degrees in the range that you are interested in. That data will allow us half the information that we need.
Then record the resistance of the temperature sensor at a few points in the range of interest.
If the change in resistance between temperatures is the same for both devices then a series resistor is all that you need. But in any case, that information will allow somebody to come up with the required circuit.
 

Alec_t

Joined Sep 17, 2013
12,744
You'll be hard pushed to operate any resistance-simulator off of just a 1.5V supply. If this is for use in a vehicle, do you also have access to a 12V (or regulated 5V) supply in the vehicle?
 

Thread Starter

saeidbb

Joined Jan 21, 2022
25
You'll be hard pushed to operate any resistance-simulator off of just a 1.5V supply. If this is for use in a vehicle, do you also have access to a 12V (or regulated 5V) supply in the vehicle?
I have access to 12 volts power supply but the digital thermometer I use works with 1.5 volts
 

Thread Starter

saeidbb

Joined Jan 21, 2022
25
OK, the first step will be to use a variable resistor with the digital temperature display and record the resistance versus temperature display. Doing that for every ten degrees in the range that you are interested in. That data will allow us half the information that we need.
Then record the resistance of the temperature sensor at a few points in the range of interest.
If the change in resistance between temperatures is the same for both devices then a series resistor is all that you need. But in any case, that information will allow somebody to come up with the required circuit.
I have done this before for a temperature range but the change in resistance was not the same as the change in temperature in the two resistance sensors and was done with a factor.
So I used a transistor with the code LB 13003C, which it solved the problem to some extent, but the problem was that the temperature shown was different with the cooling and heating of this transistor, and with the heating of the transistor at about 10 degrees, the temperature shown Was changing
 

Alec_t

Joined Sep 17, 2013
12,744
Providing the old temperature sense resistor has one terminal connected to battery negative, and providing the gauge works by passing current (constant or not) through that resistor, the following circuit should simulate a 3k-7k5 resistance. R1,R2,D1,D2,C1,C2 are to suppress transient voltages (especially in an automotive environment). D3 protects against supply polarity reversal. The MOSFET and D3 choices are not critical.
The top trace shows the simulated resistance in kΩ, represented by kV, as the new sensor resistance sweeps through the 1k to 3k5 range in 1 sec. The traces for v(gauge) and v(ref) coincide almost exactly. V(gauge) is generated at the input to the circuit. For comparison, V(ref) is synthesised using a modelled resistance sweeping through the 3k to 7k5 range.
SensorRaisedR.jpg

Edit: Rx in the sim should be R3+R4=180k
 

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Thread Starter

saeidbb

Joined Jan 21, 2022
25
Providing the old temperature sense resistor has one terminal connected to battery negative, and providing the gauge works by passing current (constant or not) through that resistor, the following circuit should simulate a 3k-7k5 resistance. R1,R2,D1,D2,C1,C2 are to suppress transient voltages (especially in an automotive environment). D3 protects against supply polarity reversal. The MOSFET and D3 choices are not critical.
The top trace shows the simulated resistance in kΩ, represented by kV, as the new sensor resistance sweeps through the 1k to 3k5 range in 1 sec. The traces for v(gauge) and v(ref) coincide almost exactly. V(gauge) is generated at the input to the circuit. For comparison, V(ref) is synthesised using a modelled resistance sweeping through the 3k to 7k5 range.
View attachment 258988

Edit: Rx in the sim should be R3+R4=180k
First of all, thank you for your help
I was finally able to build this circuit. But there is a problem. After placing this circuit, the indicated temperature also increases with increasing resistance, but with the digital thermometer sensor and the car sensor, the resistance decreases with increasing temperature.
Because the digital thermometer I was using used a 1.5 volt battery and I was using it in the car, I got another digital thermometer (W1209) that works with 12 volts. the thermometer sensor itself is still the same as the previous thermometer sensor and has the same operating temperature range and performance controls.
In order to be more precise, I have prepared a table of temperatures and resistors that I want to convert, which is given below.

car sensor :

themp (C)resistans (kilo Ω)
107.85
472.07
541.56
700.97
810.64
860.55
900.48
930.45
960.41


W1209 Digital thermometer :

temperature (C)resistance (kilo Ω)
358.81
308.24
398.01
454.88
47.54.48
50.54.15
543.68
702.24
811.65
861.44
90.21.28
95.11.13
96.21.11
1000.98
1020.92
1060.84

i want use car sensor in W1209 Digital thermometer .
 

MisterBill2

Joined Jan 23, 2018
11,623
Graphing that data on one chart will certainly help us see how close the curves are. Just looking at the tabulated data shows us that for both sensors the resistance drops as the temperature rises. That is good for us because it means that there does not need to be an inversion section. All that is required is adjustable gain and offset. Now if the data is ploted we will be able to see if there is anything like a LINEAR change in resistance with temperature. That would be handy and certainly make accuracy a lot easier. THAT is why I asked about plotting the data points. It quickly shows us how much gain and offset will be needed.
 
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