Electronic – Do Fixed Bias BJT ‘s have a cut in voltage

The most important thing to recognize in this circuit, is that the current through R1 and R2 is not equal and that everything can be solved algebraically. Let's just write down everything we know and see where we land. In this notation, \$I_x\$ means the current through x

1. \$(I_{R_2}+I_B)R_1 + I_{R_2}R_2 = 20 V\$
2. \$(I_B+I_C)R_E + V_{CE} + I_CR_C = 20 V\$
3. \$I_C = \beta * I_B\$
4. \$(I_B + I_C)Re + V_{BE} + (I_{R_2} + I_B)R1 = 20 V\$

This system of equations has too many unknowns: \$I_{R_2}, I_B, I_C, V_{BE}\$ and \$V_{CE}\$. We need to either find a fourth independent equation or guess one of them. I assume that you are being expected to assume that \$V_{BE}\$ is constant, as per your statement that it is 0.7V. This allows you to solve this system of equations:

• \$I_B = (20 - 0.7 - {R_1* 0.7\over R_2})/(R_1 + R_E*(1+\beta) + {Re*R_1*(1 + \beta)\over R_2}) = 28.75904 µA\$
• \$I_C = 50*I_B = 1.43795245 mA\$
• \$I_E = I_B + I_E = 1.46671149 mA\$
• \$V_{CE} = 20 - I_ER_E - I_CR_C = 20 - 1.613382639 - 8.91530519 = 9.471312171 V \$

You are using the wrong topology for what you are trying to achieve. Here is a better circuit:

First, we figure out R2. This is what sets the LED current. You want 700 mA. D1 drops 1.4 V, and let's say the transistor saturates to 300 mV. That leaves 3.3 V across the resistor. (3.3 V)/(700 mA) = 4.7 Ω. Note that this resistor will dissipate (3.3 V)(700 mA) = 2.3 W.

Now for the base resistor. You need 700 mA collector current, so you need at least that divided by the transistor gain as base current. Let's say the minimum guaranteed gain of the transistor at 700 mA is 50. (700 mA)/50 = 14 mA minimum required base current. Figure the B-E drop will be 750 mV, so that leaves 4.25 V across R1. (4.25 V)/(14 mA) = 304 Ω. So the largest common value that guarantees everything within spec is 270 Ω.