How to calculate Q point in a BJT biased circuit?

Start with the base voltage divider, assuming base current is low compared to divider current. If we later find out it is not, we can correct.

I = 9 / (24k+96k) = 75µA
Vb = 24kx75µA = 1.8 volts.

subtract 0.6 volts to get VE, 1.2 volts.

1.2 volts across 1k gives us 1.2 mA IE which is also IC

1.2 mA causes voltage drop across Rc of 1.2 x 5.25 = 6.3v, so Vc = 9?6.3 = 2.7 volts.

So we have Vc = 2.7 v, VCE = 2.7?1.2 = 1.5 volts.

Checking IB, = 1200mA/80 = 15µA

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15µA is significant, so we have to start over with that correction. With some math, however.
 I1 = (9?Vb)/96k
 I2 = Vb/24k
I2 = I1?15µA
Vb/24k = (9?Vb)/96k ? 15µA
4Vb = (9?Vb) ? 0.36
5Vb = 8.44
Vb = 1.7

subtract 0.6 volts to get VE, 1.1 volts.

1.1 volts across 1k gives us 1.1 mA IE which is also IC

1.1 mA causes voltage drop across Rc of 1.1 x 5.25 = 5.8v, so Vc = 9?5.8 = 3.2 volts.

So we have Vc = 3.2 v, VCE = 3.2?1.1 = 2.1 volts.

Checking IB, = 1100mA/80 = 14µA

close enough

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The Q, or quiescent point, is the DC level, or where the circuit produces no noise.
The first thing to do would be to Thevenize the left side of the circuit; Vth = 24K/(24K+96K) * Vcc, Rth = 96K in parallel with 24K.

Once you solve those, you will have a circuit with an a voltage source on the left (Vth) and a resistor, Rth, between Vth and Vbe.

To solve for the current i(b), use i(b) = (Vth - Vbe)/Rth

Once you have i(b), you should be able to find i(c) and Vce, and then verify that the circuit is in active mode.