# Operational amplifier - comparator with LED 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?

• 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). Nov 6 '17 at 5:34
• @Thanks Peter Bennett! So, V+ would be 7.5? V+ = 15 x R2/(R1+R2) Nov 6 '17 at 5:34
• Yes thats correct
– user94729
Nov 6 '17 at 5:39
• @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. Nov 6 '17 at 5:51
• 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. Nov 6 '17 at 14:19