analog logic of shmidt trigger bjt circuit

[john23]

Hello, the following circuit turns slow pulse into fast one.
how the BJT makes it happen?
Thanks.

[Andy Watson]

Positive feedback.

The emitters of the transistors are coupled together. When one of the transistors begins to conduct it robs some drive from the other transistor ... which causes the original transistor to conduct more, etc., until the circuit has flipped state.

[magic]

With Q4 turned off, R7 turns on Q5 and the output is driven low. R7 + R8 currents flow through R9, dropping almost 200mV.
With Q4 turned on, Q4 saturates and turns off Q5. Only R7 current flows through R9, dropping less than 100mV.

Reminder: bipolar transistor behaves very much like a diode, in that its current increases sharply when base-emitter voltage crosses some 0.6~0.7V, depending on transistor type, temperature and magnitude of current being switched.

Hence Q4 switches when its base is about 650mV above R9 voltage.
The low to high threshold is 200mV + 650mV or 0.85V.
The high to low threshold is 100mV + 650mV or 0.75V.
This explains the hysteresis.

During transition, R9 voltage begins to change in the direction which accelerates the transition.
This gives the fast output edges.

[john23]

Hello,how did you get  100mv 200mv values?

"Hence Q4 switches when its base is about 650mV above R9 voltage.
The low to high threshold is 200mV + 650mV or 0.85V.
The high to low threshold is 100mV + 650mV or 0.75V."

[magic]

It's the voltage at the top of R9, so Ohm's law

In low state, there is over 4V across R7 and almost 5V across R8, and both of those currents (almost 10mA) flow into R9.
In high state, there is almost 5V across R7 and this current flows into R9.

[john23]

Hello,how this feedback mechanism works in this circuit?

[john23]

Hello Andy, yes both currents go into the same R9.Why why Q4 has more current then Q5 then it will cause it to take even more current?

"The emitters of the transistors are coupled together. When one of the transistors begins to conduct it robs some drive from the other transistor ... which causes the original transistor to conduct more, etc., until the circuit has flipped state."

[Andy Watson]

Why Q4 has more current then Q5 then it will cause it to take even more current?

As Q4 turns on, the voltage on its collector will start to drop. When there is less than approx 0.6V across Q4, the drive for Q5 is reduced. Less current through Q5 causes more current in Q4 because it is trying to maintain the voltage on the common emmiter resistor. Go back to the start of this paragraph and repeat until the circuit has flipped state.

[john23]

Hello ,I tried to simulate this circuit in LTspice attached.Q1 is input Q2 is output. When you described the dynamics you said that we have Q1 open gradually on the expence of Q2 so the less Q2 has current threw it the more Q1 Vbe increases and opening even more.I did a ramp input as shown in the attached photo in the simulation and photo of the link in the zip.
However as you can see the cuurent of Q1 is proportional to Vin, but the current in Q2 is not inverse proportional to the current of Q1.
most of the time the current at Q2 is much much lower then Q1
Why is that?

When I am trying to write mathematcal equations for example for Vin=5V.
I need to make assumptions and to  prove or desprove them.
I can assume for each transistor that its either in saturation active or cuttoff, so there are 9 options for this circuit.
what is the assumpion strategy you suggest me to analize this circuit by equations?
Thanks.

[mawyatt]

Hello ,I tried to simulate this circuit in LTspice attached.Q1 is input Q2 is output. When you described the dynamics you said that we have Q1 open gradually on the expence of Q2 so the less Q2 has current threw it the more Q1 Vbe increases and opening even more.I did a ramp input as shown in the attached photo in the simulation and photo of the link in the zip.
However as you can see the cuurent of Q1 is proportional to Vin, but the current in Q2 is not inverse proportional to the current of Q1.
most of the time the current at Q2 is much much lower then Q1
Why is that?

When I am trying to write mathematcal equations for example for Vin=5V.
I need to make assumptions and to  prove or desprove them.
I can assume for each transistor that its either in saturation active or cuttoff, so there are 9 options for this circuit.
what is the assumpion strategy you suggest me to analize this circuit by equations?
Thanks.

You need to add a Vdd supply of 5V in your schematic.

Best

[john23]

Hello mayatt , yes i forgot the vdd.
When I am trying to write mathematcal equations for example for Vin=5V.
I need to make assumptions and to  prove or desprove them.
I can assume for each transistor that its either in saturation active or cuttoff, so there are 9 options for this circuit.
what is the assumpion strategy you suggest me to analize this circuit by equations?
I was told to calculate the current of the base and compare it to full-on collector current divided by the minimum Beta spec for the transistor.
given the plot below how do I make this analysis based on math and datasheet.

https://www.onsemi.com/pdf/datasheet/p2n2222a-d.pdf

[mawyatt]

Hello mayatt , yes i forgot the vdd.
When I am trying to write mathematcal equations for example for Vin=5V.
I need to make assumptions and to  prove or desprove them.
I can assume for each transistor that its either in saturation active or cuttoff, so there are 9 options for this circuit.
what is the assumpion strategy you suggest me to analize this circuit by equations?
I was told to calculate the current of the base and compare it to full-on collector current divided by the minimum Beta spec for the transistor.
given the plot below how do I make this analysis based on math and datasheet.

https://www.onsemi.com/pdf/datasheet/p2n2222a-d.pdf

Don't worry too much about Beta, look at the Vbe change in each transistor Q1 and Q2 when the output is High and Low. For a 10X change in collector current then ∆Vbe with be ~60mV. Use classic bipolar relationship between Ic and Vbe, and define what Vsat is for Q2 which relates to what Vout low is.

You'll need to define what is High and Low on the output to solve, and you can solve for Vin in each state, High and Low output. The difference in Vin for these two states is the Schmidt Hysteresis which will be the ∆Vbe1 + ∆Ve where ∆Ve is the difference voltage across the emitter resistor.

The solution will likely be transcendental and require more assumptions and multiple irritations to home in on the solutions (use Netwon's method or simple geometrical convergence to help with numerical solutions.). Compare your analysis results with LTspice simulations.

The operation (Vout low) is when the input is ~ below Vbe1 + Ve, then Q2 is saturated since it's base current is ~(Vdd-Vbe-Ve)/R7, or ~4ma which flow thru emitter resistor R9. Q2's emitter current is ~ (Vdd-Vsat-Ve)/R8 or ~4.6ma which also flow thru R9 and thus Ve is ~ (4ma + 4.6ma)*20Ω or ~172mV. Iterate thru this multiple times until the results converge for Ve and solve for Vin.

For (Vout high) when Vin is above Vbe1 + Ve, then Q2 is off as it's base is pulled down to Vsat + Ve, where Ve is now just the emitter current from Q1 times R9 or {(Vdd -Vsat- Ve)/R7 }R9 or ~ 94mV. So ∆Ve is ~172-94mV or 78mV. Iterate thru this state a few times and now ∆Vin can be estimated as ∆Vbe1 + ∆Ve.

Here we've added a LTspice simulation showing the two states with ∆Ve and ∆Vin.

Anyway, hope this helps understand this circuit and how to approach an approximate analysis.

Best

[PGPG]

Hello,how this feedback mechanism works in this circuit?

You have to understand positive feedback mechanizm and it seems you don't.
Imagine: Vin=0, Q4 Ic=0, Q5 is switched on by R7 current so it is saturated. Both R7 and R8 currents have to go out of Q5 through its emitter making some voltage drop at R9.
Now you increase Vin. At some voltage Q4 begins to conduct. Part of R7 current begins to go to R9 not through Q5 but through Q4 but total R9 current don't change. R7 current is about 100 times higher than Q5 needs to saturate so when Q4 stoles some of R7 current nothing changes for Q5. But as Q4 takes more and more R7 current at some point Q5 will reduce its collector current. Think of it as at the beginning a very small reduction (not switching yet). But when Q5 Ic current reduces (even a little) voltage drop at R9 reduces. As we don't change Vin at that moment Q4 Vbe gets little bigger. But making Vbe bigger makes Q4 Ic bigger what makes Q5 Ic smaller what makes R9 voltage drop smaller what makes Q4 Vbe bigger so its Ic bigger. The process is self-reinforcing until Q5 Ic drops down to 0. As it can't be reduced farther everything stops here.