0
\$\begingroup\$

I was trying to analyse and understand the following circuit. enter image description here

My intention was to calculate the hysteresis of the comparator.

Circuit working: For both compartor input, there will be voltage inputs ranging from 40 mV to 250 mV. This will vary as per the connected circuits and will trigger the output of the comparator either as HIGH or LOW. I have worked on comparators with one input pin as fixed-reference voltage and other pin as the varying-input voltage.

Comparator Part Number: TS332 comparator from ST.

For me this seems bit new and I was not able to understand the working of the comparator completely. Basically I am looking for a guide to calculate the upper and lower threshold and finally the hysteresis.

Could someone help me with this.

\$\endgroup\$
1
  • \$\begingroup\$ Your Vout of the comparator will be constrained to 0.0 volt and 0.7 volts. Perhaps insert 100K ohm resistor in the base lead. \$\endgroup\$
    – analogsystemsrf
    Commented Sep 3, 2018 at 22:02

1 Answer 1

Reset to default
2
\$\begingroup\$

In this circuit the transistor acts as a switch which shorts R4 or it doesn't.

The approach to finding the levels at which the output switches is to distinguish the two cases:

1) when the transistor is off

2) when the transistor is on

When the transistor is on (case 2) the right side of R3 is connected to ground so you can ignore R4 and R5.

After simplifying the circuit it is simply a matter of finding the situation where the + and - inputs of the comparator have the same input voltage.

Sidenote:

The way the output of the comparator, R6 and the NPN transistor isn't what it should be. If the comparator had an "open collector" or "open drain" output (only pulling the output low) then this circuit would be OK. But the TS332 has a rail-to-rail output. That makes R6 unneeded. Also when the output is high will try to pull the base of the NPN to the supply. This will fail as the TS322's output cannot deliver that much current. There should really be a base resistor in series with the base of the NPN. You can just use R6 so move it so that it's in series with the base of the NPN.

\$\endgroup\$
3
  • \$\begingroup\$ understood it. See the threshold voltage calculated. 3.3 V * ( R5 / (R5 +R6 ) ) = 3.3 V ( 2.7 kΩ / ( 2.7 kΩ + 97 kΩ ) = 0.089 V. Can I call this value as lower threshold value, fundamentally will there be absolute lower and upper threshold values for this circuit? Or will it depend on the input voltage as well ? \$\endgroup\$
    – vt673
    Commented Sep 3, 2018 at 13:19
  • \$\begingroup\$ You can call it whatever you like, just calculate both and the lower is the lower. Do the same calculation with and without connecting a 1 V voltage source on the left side of R2. Do the threshold voltages change? \$\endgroup\$
    – Bimpelrekkie
    Commented Sep 3, 2018 at 13:23
  • \$\begingroup\$ When comparator output is off, Threshold will be = 0.089 V. This will be default voltage at the -ve input of the comparator. When the +ve input of the compartor just cross this voltage ( = 0.089 V), output will be high. This means, transistor is turned on. From that moment, -ve terminal voltage should be ( = 0.089 V * (R3/ (R2+ R3)) = 0.089 V / (97 kΩ/(97 kΩ + 10 kΩ)) = 0.0075 V ) more than 0.0075 V to get low output at the comparator output. \$\endgroup\$
    – vt673
    Commented Sep 3, 2018 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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