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- Thread starter helofrind
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When we use a BJT as a switch (saturation / cut-off) we use forced beta Ic/IB = 10.

Or Ib = Ic/(βmin * K ) where

K = 3...10 - overdrive factor

But if we designs a amplifier we use a minimum beta value from the data sheet as well.

And we use voltage divider to bias the base and Re resistor. And we choose voltage divider current much larger (ten times larger) than the base current.

Or Ib = Ic/(βmin * K ) where

K = 3...10 - overdrive factor

But if we designs a amplifier we use a minimum beta value from the data sheet as well.

And we use voltage divider to bias the base and Re resistor. And we choose voltage divider current much larger (ten times larger) than the base current.

Last edited:

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thanks for the reply. i built a discrete switch with hysteresis that has an upper and lower threshold of 2-2.5V, which is what i wanted. i set the parameters by trial and error. when the voltages at the base of Q1 raise to 2.5 volts the the output of Q2 is high. the output of Q2 does not go low until the voltage at the base of Q1 decreases to 2 volts. I know RE is one of the factors that determines the threshold. i understand Q1 is driving Q2 into saturation and cutoff, and i never heard of overdrive factor or forced beta, that helps a little when i try to research how this circuit functions thanks. could you help me better understand this circuit by analysis? the two transistors are 2n3904's

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But let as first assume that Q1 is Cut-off and Q2 is in saturation region.

So to open Q1 the voltage at Q1 base must be Vbe larger then emitter voltage.

So the first think we need to do is to find a voltage across RE resistor (Ve) when Q1 is OFF and Q2 in saturation region.

We know that emitter current is always (even in saturation) equal to

So we have

Ib = (Vcc - Vbe - Ve))/R2

Ic = (Vcc - Vd - Vce(sat) - Ve)/Rc

Where

Vcc = 5V; Vd = 2V --->LED forward voltage drop

Vbe = 0.7V

Vce(sat) - Vce saturation voltage ---> 0.1V.

And now if we solve this for VE we have this

[itex]\Large Ve = (\frac{Vcc - Vbe}{R2} + \frac{Vcc - Vd - Vce_{sat}}{RC}) * R2||RC||RE[/itex]

[itex]\Large Ve = 1.42V[/itex]

So Q1 will start to conduct if voltage at Q1 base is larger than

Q1 will also starts to steal some R2 current from Q2 base. As the input voltage rising Q1 switches from Cutt-off to saturation and Q2 switch from saturation to cutt-off.

Q2 will start to comes out form saturation for the base current smaller than.

Ic2_sat = (Vcc - Vd - Vce_sat - Ve)/Rc = (5V - 2V - 0.1V - 1.42V)/250Ω ≈ 1.5V/250Ω = 6mA

IB2 = 6mA/200 = 30μA

So the IC1 is now equal to

Ans this Ic current caused by IB1 current equal to:

IB1 = Ic/Hfe = 3.3mA/200 = 16.5μA. And this base current will flow through RB resistor and

causes a voltage drop.

VRB = IB*RB ≈ 0.38V

So now we have all information needed to find upper threshold voltage.

Now let as try to find a lower threshold voltage.

We have Q1 in saturation and Q2 in cut-off. So to open Q2 Vce1 voltage mus be larger than Vbe2.

Ic = Ie = (Vcc - Vbe)/(R2+RE) = 4.4mA and Ve = 0.660V.

If the voltage at Q1 base is smaller than Ve + Vbe1 = 0.660V + 0.7V = 1.36V Q1 starts comes out form saturation region. This will happen for the base current equal to

IB1 = Ic/Hfe = 4.4mA/200 = 22μA. This current will give as a voltage drop across RB equal to:

VRB = 0.5V.

And the lower threshold voltage is equal to:

Designing this circuit also is ton easy task. And this is why we almost always use a op amp as a Schmitt trigger or we add more resistors.

http://www.johnhearfield.com/Eng/Schmitt.htm

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