enter image description here

I am trying to find the differential input voltage(Vd) for the circuit above. Since this circuit is in open loop, Us is going to be either +Vsat or -Vsat.

Rt is a thermistance that varies with temperature

Rh is a variable resistor.

I am supposing that R,R,Rt,Rh are in series.

In this case,

The inverting input(V-) would be

V- = 15 * Rh/(2R+Rh+Rt).

The non-inverting input(V-) would be

V+ = 15 * R/(2R+Rh+Rt)

Thus, Vd = (V+) - (V-) = 15*(R - Rh)/(2R+Rh+Rt).

I am not exactly sure if I am doing this right... I am having difficult time with analyzing circuits... Could you help me?

  • \$\begingroup\$ The two leftmost Rs (it would help if you called them R1 and R2) are in series, and Rt and Rh are also in series - those four resistors make two separate voltage dividers. V- is 15 x Rh/(Rt+Rh) (the first two Rs are not involve in this calculation at all). \$\endgroup\$ Nov 6 '17 at 5:34
  • \$\begingroup\$ @Thanks Peter Bennett! So, V+ would be 7.5? V+ = 15 x R2/(R1+R2) \$\endgroup\$
    – user167987
    Nov 6 '17 at 5:34
  • \$\begingroup\$ Yes thats correct \$\endgroup\$
    – user94729
    Nov 6 '17 at 5:39
  • \$\begingroup\$ @Mikey Thanks Vd = 7.5 - 15xRh/(Rt+Rh) In order to light up D1, Vd should be greater than 0, which makes Rt > Rh In order to light up D2, Vd should be smaller than 0, which makes Rt < Rh. Right? Thanks a lot guys. \$\endgroup\$
    – user167987
    Nov 6 '17 at 5:51
  • \$\begingroup\$ Do be careful to power the op-amp from a plus-and-minus DC supply. If it is powered between +15V and ground, LED D1 will light up but LED D2 will not. \$\endgroup\$
    – glen_geek
    Nov 6 '17 at 14:19

Assuming that the input differential voltage is not restricted internally with protection diodes, the voltage at the inverting input is the potential divider formed by RT and RH acting on the 15 volt supply. The non-inverting terminal has 7.5 volts because the two resistors are equal (R=R).

Not considered in this simple explanation are: -

  1. Input offset voltage (maybe up to 10 mV)
  2. Input bias currents (pA to uA) that make a small error in the analysis.

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