Finding threshold voltage of BJT based schmitt trigger

Thread Starter

omar-rodriguez

Joined Jun 24, 2015
67
Hi, I'm trying to find the high (Vth) and low(Vtl) threshold voltages on this schmitt trigger circuit:

- The first thing I did was to asume that Q1(off) - Q2(on) so I wrote this ecuation to find Ve2 (emiter voltage of Q2):



I wrote those equations to find the emiter voltage:



Once I found the emiter voltage of Q2, then I found the condition to turn on the transistor Q1, Q1 turns on when it's base emiter junction is forward biased, so with the inequality above, I found the high threshold voltage.

- Now I asume that Q1(On) - Q2(off)

I wrote those equations to find the emiter voltage:



As you can see the emiter voltage of Q1 depends of the imput voltage, I have tried to find the lower threshold voltage of the schmitt trigger, finding the input voltage to cut off Q1 and turn on Q2 again, but I have not succeed, , What should I do, I don't understand how to find Vtl?

I did a simulation, Vin is the triagle wave and the blue one is the output,


Vth is about 6.32V and Vtl is about 4.77V, but I have to find Vtl analytically, and without using approximations as say that the collector and emitter currents are the same, the only approximations I have used is to fix the colector emiter saturation voltage to 0.3V and the base emiter voltage as 0.7V
 

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Jony130

Joined Feb 17, 2009
5,241
In case one the Q2 is not in the saturation region. So, repeat your calculation for this case.

We can use a trick to found VTL.
First assume that the Q1 is in saturation and solve for VR4 voltage. This is the Q2 base voltage So, to open the Q2 the emitter voltage must drop below this value. So, the VTL ≈ VR4 when Q1 is in saturation.
 
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MrAl

Joined Jun 17, 2014
8,390
Hi, I'm trying to find the high (Vth) and low(Vtl) threshold voltages on this schmitt trigger circuit:

- The first thing I did was to asume that Q1(off) - Q2(on) so I wrote this ecuation to find Ve2 (emiter voltage of Q2):



I wrote those equations to find the emiter voltage:



Once I found the emiter voltage of Q2, then I found the condition to turn on the transistor Q1, Q1 turns on when it's base emiter junction is forward biased, so with the inequality above, I found the high threshold voltage.

- Now I asume that Q1(On) - Q2(off)

I wrote those equations to find the emiter voltage:



As you can see the emiter voltage of Q1 depends of the imput voltage, I have tried to find the lower threshold voltage of the schmitt trigger, finding the input voltage to cut off Q1 and turn on Q2 again, but I have not succeed, , What should I do, I don't understand how to find Vtl?

I did a simulation, Vin is the triagle wave and the blue one is the output,


Vth is about 6.32V and Vtl is about 4.77V, but I have to find Vtl analytically, and without using approximations as say that the collector and emitter currents are the same, the only approximations I have used is to fix the colector emiter saturation voltage to 0.3V and the base emiter voltage as 0.7V
Hello there,

Can you solve this knowing only the assumed *constant* voltages Vbe and Vce sat?
I ask because i just took a quick look and it looks like you may be able to get away with Q2 as a switch but it seems that you have to consider Q1 as being in the linear mode during the time when the base current starts to decrease. That's because the collector voltage will change gradually as the emitter voltage changes and thus both the emitter and collector voltages will be responsible for the determination of where it 'snaps' back into the Q2 'on' mode. This means you may have to assume some value for Beta too and incorporate that into the equations.

If you dont figure this out, i'll look into more carefully. To start, look at the Q1 collector voltage in your simulation and perhaps post that wave here too.
 

Jony130

Joined Feb 17, 2009
5,241
The first thing I did was to asume that Q1(off) - Q2(on) so I wrote this ecuation to find Ve2 (emiter voltage of Q2):



I wrote those equations to find the emiter voltage:

Why didn't you include R4 resistor in your calculations?
Also you have assumed that the Q2 is in saturation region but you didn't check whether this assumption is true.

If Ve = 6.05V we have:

Ie = 6.05V/7.5kΩ = 0.8066mA

Ic = (15V - 6.35V)/10kΩ = 0.865mA

As you can see we have Ic > Ie so, there is something wrong (in saturation Ie = Ib + Ic is true).

If we assume Q2 in active region and β = ∞ ;

we have

Vb2 = 15V *(9.1k)/(8.2k Ω+ 3.9kΩ + 9.1kΩ) = 6.439V

And

Ie2 = (6.439V - 0.7V)/7.5k ≈ 0.765mA

Ic2 = 0.765mA

Vc = 15V - 10kΩ*0.765mA = 7.35V

Vce = (7.35V - 5.74V) = 1.61V

As you can see the Q2 is indeed in the active region even for ideal BJT.

Therfore the Vth = 5.74V + 0.6V = 6.34V

Vtl voltage is not so easy to find.

If I assume Q1; β1 = 150; and Vbe1 = 0.68V and Vbe2_on = 0.55V we can solve for Vin.
But first we need two equation. The first equation for VR4 voltage and the secend one for Ve1 voltage.

VR4 = Vc1 * (R4)/(R2 + R4)

Vc1 = ( (Vcc - Ic1*Rc1)*(R2 + R4) )/(R2 + R4 + Rc1)

Ve1 = Ie * RE = (Vin - Vbe1)/( RE + R1/(β+1) ) * RE

Ie ≈ Ic ≈ (Vin - Vbe1)/( RE + R1/(β+1) )


And all you need to do is to solve for Vin from Vbe2_on = VR4 - Ve1

And I get Vtl = 4.751V

But as I said earlier there is a simpler way. Have you try it?


 

Thread Starter

omar-rodriguez

Joined Jun 24, 2015
67
go intoYeah, I did include R4 in my calculations but unlike your calculations didn't assume that base current of Q2 is zero, did you remember I said that I don't want to use approximations.. at least in this exercise.

Anyway, I realized that Q2 is not in saturation region because the base current is negative, so I repeated the calculations as you advised to me.

The first thing I did was to find de colector current of Q2 in terms of the current gain (hfe=beta), then I looked into the 2N3904 datasheet to find a value of hfe that is consistent with the equation and datasheet.





Finally:


Now If you look at the 2N3904, you'll find this plot


Through trial and error between this plot and the ecuation of Ic2 I finally found that,



For Q1 I realized that as Vin starts to decrease, Q1 gets into saturation until Vin=7.35V, then gets into active mode until Vin=4.61V and finally it cut off, and Q2 turns on again. so VTL=4.61V
 

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MrAl

Joined Jun 17, 2014
8,390
Hi,

Here's a simulation plot showing the almost pure switching actions of Q2 and the pseudo but significant linear action of Q1. Vc1 is the collector voltage of Q1 and Vc2/5 is the collector voltage of Q2 divided by 5 for clarity in the plot.

From this plot looking at Vc1 we can see that when the input voltage is rising the transistor Q1 acts almost like a pure switch, but when the input is falling the transistor linear operation is apparent and significant. This is why i suggested to do a plot of the collector voltage of Q1.

This plot also suggests that this isnt a very good Schmitt Trigger circuit because the hysteresis is so small. Maybe they wanted that though for some reason.
 

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