3
\$\begingroup\$

As part of our lab' design project, we have to calculate suitable resistance values for this circuit below and for others like it with different sensor resistances. However, in lectures we have never seen an op-amp being used as it is here, presumably as some sort of switch.

I have no idea how to start tackling this circuit as all the information we've been given is the typical resistances for the sensors. In this case, the thermistor will be 1 kOhms at 80 C and 4 kOhms at 20 C. This circuit must turn on the LED on the right when the temperature is high.

Another question, rather than asking how the circuit is works, is: how should I connect components for simulation in LTspice? Should the trailing end of the LED be connected to ground? And, since LTspice doesn't allow voltage rails, should the voltage source's negative terminal be connected to the ground at the bottom?

Temperature Sensor Circuit

Thanks for any help and sorry for any dumb questions, I'm just first year. The online pre-recorded lectures aren't the ideal learning environment and our tutors won't respond to any emails.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ A very warm welcome to the site and thanks for a well-written question. It's an op-amp used as a voltage comparator. Google 'voltage comparator op-amp' or suchlike and you should find plenty on the subject that will answer this for you. Again, welcome. \$\endgroup\$ – TonyM Nov 12 '20 at 17:23
  • 3
    \$\begingroup\$ You might be interested in this question about op amps vs. comparators. Note that, as I explained in my answer, it's usually not optimal to use an op amp as a comparator -- use an actual comparator for it. Obviously you have to use an op amp for this lab but it's something to think about when you finish school. \$\endgroup\$ – Null Nov 12 '20 at 17:42
1
\$\begingroup\$

In your lectures, you should have been introduced to the basic equation for an ideal op amp: Vout = G (V+ - V-). Where G is very large, like on the order of a million or so for modern op amps.

So, when V+ is one millivolt less than V-, the output will attempt to be a -1000 volts (.001 times 1,000,000), but will be at zero since that is the negative supply.

When V+ is one millivolt more than V-, the output will attempt to go to +1000 volts. Since the op amp cannot produce an output greater than its positive supply voltage, that will be what it does instead.

Note that this assumes that the op amp is what's called a rail-to-rail output amp. You should be aware that many op amps, especially older models like 741s, 356s, TL081/82/84, etc, will not be able to drive closer to the power supply lines than about 3 volts, which may affect a given circuit.

\$\endgroup\$
1
\$\begingroup\$

A starting point would be to decide at what value of RTemp you want it to switch (maybe 2K). Then set R1, R2 and R3 all to that value.

R4 is the current limiting resistor for the LED and should be set to allow a few milliamps through. Yes, the negative terminal of the LED will need to be grounded for anything to happen.

\$\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.