I'm trying to build a simple temperature controller by using a thermistor and a potentiometer attached to a Schmidt trigger to control a relay.
I realize that I have some serious design flaws in this circuit:
- The power dissipated through the thermistor is to large (discovered after nicking my finger).
- The op-amp's output hits almost no resistance, leading to a short.
Iv'e tried solving this by adding more resistors, but I'm not sure how to proceed while keeping the op-amp's inputs away from the (one-sided) supply's limits.
Any help apreciated. 
Answer
Main problem is 20r is far far far too low for thermistor value. Use of a 20k when cold thermistor is about right. Why use a 20r? (eg may be a special part etc).
For temperature measurement thermistor self heating must be small - say 1 mW in a small package. You would need Vthermistor << 1V if thermistor = 20r.
Use a "single supply" opamp whose input common mode includes ground. Vout going to ~= ground is also useful. Cheap and available LM324 (quad) of LM358 (dual) is OK for this circuit.
You could make a Vsupply/2 reference point to allow input to not be ground referenced but this is not needed if Rtherm higher and single supply opamp used.
Use a rail to rail op amp OR at least one that ncludes ground in its cmmon mode range. The very available common and cheap LM358 (dual) and LM324 (quas) include ground in input common mode range and can be run as "single supply" amplifiers on a 9V (or 5V or ...) single supply system.
You do not say what IC is used for the amplifier or why.
If this is going to work as a temperature controller then self heating of the thermistor MUST be < to << the heating from heat sources. So a design target of say 1 mW self heating to start gives SOME idea.
Power thermistor = V^2/R = 0.001 Watt.
So V therm = sqrt(Ptherm x Ttherm) = sqrt (0.001 x 20) ~~~= 0.04 Volt.
The problem is that the thermistor is vastly too low in resistance for 9V operation - how did you choose it?
Is it really 20 ohms?
If it is 20k then Vtherm = sqrt(20k x 0.001) =~ 4.5V = much better.
If you MUST use a 20 Ohm thermistor ad if you MUST use a non single supply IC then you can make an eg half supply reference point.
You can take eg 1k+1k in series from 9V to ground and use centre point as 4.5V dummy-grounjd.
Return thermistor chain and opamp + input chain to there.
BUT the dummy ground divider must be "stiff" so it is not altered by changes in inputs from thermistor and opamp output. Opamp drive is via 100k so that is OK with a 2 x 1k divider.
BUT Op amp + in ref chain of 220R + 500R pot + 312R? resistor is very low. Why?
You would need a
Increase thermistor drive to a level that meets thermistor max dissipation at very very worst. BUT you really need a stiffer reference point and a regulator (LDO 3 terminal or TL431 or ...) can be used for ref. BUT a single supply op amp makes more sense.
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