@3:02 I believe you mean "increase the voltage at the emitters", as it cannot be decreasing it at that point: the current at the emitter current from VT2 is increasing, hence increasing emitter voltage, which in turn, shuts off VT1
how do you calculate the lower threshold of the schmitt trigger? is it perhaps the voltage accross Re? when calculating it it sure seems like it but I am not entirely sure.
Excellent video. Thank you. Question: What exactly are T1 &T2? I don't believe the cost of this approach can be lower than the jelly bean Schmitt trigger chips. T1 & T2 will have to be very high speed switching transistors and cost each as much as an 8 gate chip.
Signal conditioning and processing of all sorts! If you have an analog signal which you need to process you have to convert it to digital first. This can be easily accomplished using this circuit. Processing the digital signal might be to retrieve the frequency, pulse width etc of the original analog signal
I'm sorry to disappoint you but the hysteresis will be the same if you add a diode(or 2, or 3....). Adding diodes at the input(T1), only increases the threshold but the hysteresis remains the same. Hysteresis only depends on voltage drop on Re.
Both Ra and Rb are useless in modern silicon transistor triggers. It's a legacy from old germanium transistors when saturation condition for T1 was not met in order to switch off T2[Vbe was comparable to Vce_sat (≈0.15V)]. With silicon transistors it is no longer the case. I suggest you to get rid of them. Another thing... R1 is too small. Usually R1 is 10 times R2.
@Afterbytes--->>> Frankly, I don't get it... Could you be more specific about the threshold? And the LED stays ON all the time no matter what input voltage is, or what???
The lower boundary of the input voltage trigger point. Going past ~1.9V turns the LED on....though moving back down below ~1.3V does not turn it off using those values you suggest.
@Afterbytes--->>>It has to be two thresholds. Get this: upper th. 5V×[0.33k/(0.33k+0.9k)]+0.6V=1.941V and lower th. 5V×[0.33k/(0.33k+10k)]+0.6V=0.759V where 0.6V=Vbe(T1), 0.33k is Re, 10k is R1, 0.9k is R2//R1(when T2 is saturated) (Ra and Rb are omitted) ...and the hysteresis is: 1.941-0.759=1.182V
@3:02 I believe you mean "increase the voltage at the emitters", as it cannot be decreasing it at that point: the current at the emitter current from VT2 is increasing, hence increasing emitter voltage, which in turn, shuts off VT1
Hmm this might be a mistake in the vid 😊
Great !I liked your explanation. I could not find good explanation of it last two days. Thanks.
great explainaatinon!
Thanks
How do we set the upper/lower threshold value?
You adjust the resistors R1 Ra Rb for correct voltage divide.
how do you calculate the lower threshold of the schmitt trigger? is it perhaps the voltage accross Re? when calculating it it sure seems like it but I am not entirely sure.
Thank you very much! I understand how it works now.
Sjajno objašnjeno! Hvala puno!
Very good, congratulations !
Thanks project is successfully
Done
before showing this video is really great
can we use 5Mohm resistor for lowest energy need
Tq sir it's more informative keep going
good job
Excellent video. Thank you. Question:
What exactly are T1 &T2? I don't believe the cost of this approach can be lower than the jelly bean Schmitt trigger chips.
T1 & T2 will have to be very high speed switching transistors and cost each as much as an 8 gate chip.
You only need 2n2222 which is a great transistor
how make low power(nA) transistor schemitt triger ?
Thanks.
what's the use of this circuit ?
Signal conditioning and processing of all sorts! If you have an analog signal which you need to process you have to convert it to digital first. This can be easily accomplished using this circuit. Processing the digital signal might be to retrieve the frequency, pulse width etc of the original analog signal
how to increase the difference voltage between upper and lower threshold aka as hysteresis voltage ? thanks
Increase Re value.
the logic of the circuit is the opposite of the one in the video before it right ?
If adding a diode at base of Tr1 we can have a bigger hysteresis gap of 0.7 Volts 2 Diods will rase the gap with 1.4Volts etc.....
I'm sorry to disappoint you but the hysteresis will be the same if you add a diode(or 2, or 3....). Adding diodes at the input(T1), only increases the threshold but the hysteresis remains the same.
Hysteresis only depends on voltage drop on Re.
Make an inverting one too please
Just add a transistor at the T2 output. Done
Спасибо
What is the value of the Vref?
1:00 5 volts
I need something with far greater precision... this would be perfect for a boiling water sensor or something
LM311
Simple
Both Ra and Rb are useless in modern silicon transistor triggers. It's a legacy from old germanium transistors when saturation condition for T1 was not met
in order to switch off T2[Vbe was comparable to Vce_sat (≈0.15V)]. With silicon transistors it is no longer the case. I suggest you to get rid of them.
Another thing...
R1 is too small. Usually R1 is 10 times R2.
Very good comment- I saw both circuits (with and without the 2 resistors) and could not explain why the difference.
Just tried this circuit with R1 = 10K and R2 = 1K as you suggest. There no longer seems to be a lower threshold and the LED stays on...
@Afterbytes--->>> Frankly, I don't get it... Could you be more specific about the threshold? And the LED stays ON all the time no matter what input voltage is, or what???
The lower boundary of the input voltage trigger point. Going past ~1.9V turns the LED on....though moving back down below ~1.3V does not turn it off using those values you suggest.
@Afterbytes--->>>It has to be two thresholds. Get this:
upper th. 5V×[0.33k/(0.33k+0.9k)]+0.6V=1.941V
and lower th. 5V×[0.33k/(0.33k+10k)]+0.6V=0.759V
where 0.6V=Vbe(T1), 0.33k is Re, 10k is R1, 0.9k is R2//R1(when T2 is saturated)
(Ra and Rb are omitted)
...and the hysteresis is: 1.941-0.759=1.182V
Explanation 100
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Ёр аксент из вери рекогназибл.
Найс рофлишь, бро!))
Rb is useless and can make the circuit not working if not chosen precisely, just get rid of it.