# Transisor beta to design transistor circuit

i was wanting to know if anyone knows how to choose the value of beta when trying to design a simple transistor circuit? lets say we're using a 2n3904 NPN 12Vcc with with three resistors RC, RB, and RE. how do i determine what value of beta i should use? i want to determine beta so i can add that into my formulas (β=IC/IB) when trying to find what base current is needed for saturartion or cutoff. im trying to better understand the beta aspect of a bjt. i know how beta is changed by temp and where it is on the data sheets but i dont exactly fully understand how i can use it in my calculations.

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.

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• 1 person
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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is what im saying is i would like to know how the math works in this circuit so i can set the upper and lower threshold at certain voltages

Full analysis of this type of a circuit is not so easy.

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

Ie = IB + IC

So we have

Ie = Ve/RE

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.

Ve/RE = (Vcc - Vbe - Ve))/R2 + (Vcc - Vd - Vce(sat) - Ve)/Rc

And now if we solve this for VE we have this

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

$\Large Ve = 1.42V$

So Q1 will start to conduct if voltage at Q1 base is larger than Ve + Vbe ≈ 2.12V.
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.

IB2 < Ic2_sat/Hfe

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 Ic1 = IR2 - IB2 ≈ IR2

IR2 = (5V - 0.7V - 1.42V)/850Ω ≈ 3.3mA

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.

Vt2 = Ve + Vbe1 + VRB ≈ 2.5V

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:

Vt1 = Ve + Vbe + VRB = 0.66V + 0.7V + 0.5V = 1.86V

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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wow i guess i got in a little over my head lol. but then again i now know what it takes. thanks this really helps out with my experiments. op amps are pretty easy and i got them down. transistors are what im trying to get better at. i only knew the basic theory coming out of school and not too much about designing the circuits