I'm just working on a basic Schmitt trigger circuit. But I have a question about the rise time of Q3. As shown in figure the collector fall is very sharp, but on the other hand when the transistor goes to the off state, the collector voltage rises slowly (around 15.82 us).

How can I change the rise time of the collector voltage (when Q3 turns off)?

enter image description here enter image description here enter image description here enter image description here

  • \$\begingroup\$ Q3's base appears to be getting a lot of voltage and certainly more than your diagram suggests. Look at Q2's collector for clues about the slow rises. You should also look into why people give their help for free and recognize that for prolonged and professional advice you need to upvote useful answers and formally accept the best answers to questions you have raised previously. \$\endgroup\$
    – Andy aka
    Jan 3, 2022 at 15:33
  • \$\begingroup\$ Anytime you see what amounts to an RC curve, you need to realize that this means a passive pull. To improve that, look to create an active pull. End of story. \$\endgroup\$
    – jonk
    Jan 3, 2022 at 15:42

2 Answers 2


You need to add an active pull-up stage to the output, rather then just relying on the collector resistor to pull the collector voltage up. Look at what's used in a typical TTL output stage, what's called a totem-pole output structure.

Look at this question: TTL, Totem Pole vs. Open Collector Output

  • \$\begingroup\$ hi SteveSh, thank you for you comment.I added Q6 and Q5, but I think nothing change. As you can see in photo. \$\endgroup\$ Jan 4, 2022 at 19:14
  • \$\begingroup\$ What you added is basically an emitter follower, push-pull output stage, where the output follows the input. This is why the blue trace and the green traces are similar. You need to figure out why Q3 is slow turning off. \$\endgroup\$
    – SteveSh
    Jan 4, 2022 at 21:52

The answer from SteveSh is one reason for a slow rise time. Another is that saturation in Q3 will cause it to be slow to turn-off.

One improvement that can be done is to add a small capacitor across R7, maybe 100pF. This will cause a negative base current and more quickly turn-off Q3.

Also a resistor from Q4 collector to ground, maybe 1k, will help speed up that stage. Currently the charge from C1 has to discharge through R7 and the base of Q3 slowing down the rise time of the output.

  • \$\begingroup\$ Agree. C1 is discharging through 22 K, a time constant of 220 us. The gain of Q3, along with the small voltage change in Vbe that transitions the transistor from saturation to cutoff, work to quicken this. Still, I think C1 is slowing things more than a speed-up capacitor across R7 can compensate. \$\endgroup\$
    – AnalogKid
    Jan 3, 2022 at 17:14
  • \$\begingroup\$ @AnalogKid - Agreed. That's why a resistor from Q4 collector to ground will help. It will provide a better discharge path avoiding the base of Q3. \$\endgroup\$ Jan 4, 2022 at 0:02
  • \$\begingroup\$ AnalogKid, actually I could not understand your comment. How C1 is discharging through 22k? If it is pıossible can you draw, current path? \$\endgroup\$ Jan 4, 2022 at 6:25
  • \$\begingroup\$ @ivanmateo C1 capacitor is charged via Q4 and R6 resistor. But the main discharge path is C1_upper _plate --->R6---->R7--->R8---->GND--->C1_bottom plate. Do you see it now? – \$\endgroup\$
    – G36
    Jan 4, 2022 at 13:31
  • \$\begingroup\$ AnalogKid, sorry about that I understood that you mean R3 because it is also 22k :). Thank you for your information. \$\endgroup\$ Jan 4, 2022 at 16:21

Your Answer

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

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