0
\$\begingroup\$

When I simulate the circuit on (OrCad) I got an output voltage between (0 to 1) V changing with temperature . but when I connect the circuit I got the output voltage = -1.85 in minus and different from simulation value , so where is the problem .

and I am sorry about my weak language the circuit

the Vo orcade simulation values

\$\endgroup\$
  • 1
    \$\begingroup\$ but when I connect it - connect what? \$\endgroup\$ – Eugene Sh. Dec 20 '18 at 21:17
  • \$\begingroup\$ when I connect the circuit (do it on real life) <<<<sorry about my weak language \$\endgroup\$ – Ali Ali Dec 20 '18 at 21:27
  • 2
    \$\begingroup\$ Well, we don't know how you do it in real life. Probably connecting something in a wrong way. Check carefully, step by step. \$\endgroup\$ – Eugene Sh. Dec 20 '18 at 21:28
  • 1
    \$\begingroup\$ How did you check? Disconnect the second stage, check the first one only. Does it work as expected? \$\endgroup\$ – Eugene Sh. Dec 20 '18 at 21:32
  • 2
    \$\begingroup\$ Where did you find 680.79 and 602.663 ohm resistors to build the real circuit with? \$\endgroup\$ – The Photon Dec 20 '18 at 21:45
0
\$\begingroup\$

The analysis of this circuit is fairly straightforward.

  • The first stage (U1, Q1, V2 and R2) put a constant current through Q1's B-E junction, creating a voltage that's more-or-less proportional to Q1's temperature — around 650 mV at room temperature. The current is V2/R2 = 1.667 mA.

  • The output voltage is fed through a divider (R3 and R1), creating a Thévenin source (relative to V2) of 0.87192 × Vout and a Thévenin resistance of 87.192 Ω. If Vout is 0.65 V, then the Thévenin source voltage is -73.65 mV.

  • This source voltage is then multiplied by \$-\frac{602.663}{87.192}\$ by U2 and R4. If Vout is 0.65 V, then the output of U2 is 0.509 V.

If Q1's voltage ever rises as far as 735 mV, the output voltage will go negative. If you're getting -1.85 V, then the output of the first stage must be sitting at around +1.04 V.

This all assumes that the opamps are ideal. Of course, friends don't let friends use the 741 in the first place.

\$\endgroup\$
  • \$\begingroup\$ my task is to find the resistors values to give an output of 10 mV per degree Centigrade , so the output voltage increase as temperature increase , so this my calculation ..... I assumed the Vo =0 at T=0 ,and I found Vbe=734.436mV at T=0 ,so I(r1)=I(r3) so I assumed r3=100 so I got r1=680.79 ..... At T=100 , Vbe=568.508mV , Vo=1000mV...... so I did a KCL ..... Vo/R4 + Vbe/R3 = 5/R1 .....so I got R4=602.663 ......my calculations is right or I have a problem ?? \$\endgroup\$ – Ali Ali Dec 20 '18 at 23:52
  • \$\begingroup\$ You definitely have a problem, because the output voltage goes down with increasing temperature -- the second stage is an inverting amplifier. Also, why do you think that R1 and R3 carry the same current? \$\endgroup\$ – Dave Tweed Dec 20 '18 at 23:56
  • \$\begingroup\$ but the question required to gives an output of 10 mV per degree , so how I can increase the Vo when the temperature increase? \$\endgroup\$ – Ali Ali Dec 20 '18 at 23:59
  • \$\begingroup\$ because I assumed at T=0 that Vo=0 so the V(r4)=0 ,because of that R1 and R3 carry the same current at T=0 \$\endgroup\$ – Ali Ali Dec 21 '18 at 0:03
0
\$\begingroup\$

There is basically nothing wrong with your circuit (0.00~1.00V out for 0~100°C), conceptually, except the variations between the models and reality (and the variation between samples of a given BJT) will make a complete mockery of your 5 or 6 digits of precision calculation of the resistors.

There does appear to be something wrong with your implementation, but ignoring that for the moment--

I would suggest replacing R1 with a ~500 ohm resistor in series with a 500 ohm trimpot. Replace R4 with a ~400 ohm resistor in series with a 500 ohm trimpot.

Center the two pots. Put the sensor into an ice-water slurry and adjust R1 to get 0.00V out. Put the sensor into boiling water and adjust R4 to get 1.00V out. Because 741s have a lot of Vos you might need to iterate one time or so to get the maximum accuracy.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.