Standard values of commercially available resistors

     In designs of your electronic circuits, calculated values of resistors may not be commercially available. This table lists standard value of resistors available at electronic component stores worldwide. Choose the resistance values from the table that are nearest to your calculated results. 

Ohms Kilo-Ohms Mega-Ohms 0.10 1.0 10 100 1000 10 100 1.0 10.0 0.11 1.1 11 110 1100 11 110 1.1 11.0 0.12 1.2 12 120 1200 12 120 1.2 12.0 0.13 1.3 13 130 1300 13 130 1.3 13.0 0.15 1.5 15 150 1500 15 150 1.5 15.0 0.16 1.6 16 160 1600 16 160 1.6 16.0 0.18 1.8 18 180 1800 18 180 1.8 18.0 0.20 2.0 20 200 2000 20 200 2.0 20.0 0.22 2.2 22 220 2200 22 220 2.2 22.0 0.24 2.4 24 240 2400 24 240 2.4 0.27 2.7 27 270 2700 27 270 2.7 0.30 3.0 30 300 3000 30 300 3.0 0.33 3.3 33 330 3300 33 330 3.3 0.36 3.6 36 360 3600 36 360 3.6 0.39 3.9 39 390 3900 39 390 3.9 0.43 4.3 43 430 4300 43 430 4.3 0.47 4.7 47 470 4700 47 470 4.7 0.51 5.1 51 510 5100 51 510 5.1 0.56 5.6 56 560 5600 56 560 5.6

0.62 6.2 62 620 6200 62 620 6.2 0.68 6.8 68 680 6800 68 680 6.8 0.75 7.5 75 750 7500 75 750 7.5 0.82 8.2 82 820 8200 82 820 8.2 0.91 9.1 91 910 9100 91 910 9.1

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2010-09-22 Bipolar Junction Transistor / Thevenin equivalent circuit

Find the voltages at Base, Emitter, Collector terminals of the bipolar junction transistor using Thevenin equivalent circuit theory. Is this BJT in saturation state?

Solution

Transform the circuit on the left side of BJT into an equivalent circuit containing a voltage source V_th in series with a resistor R_th. 

Cut the circuit on the left side from that on the right at the Base terminal and calculate the V_th at the cutting point.

V_15kΩ = [15 V - (-15 V)][15kΩ/(15kΩ+15kΩ)] = (30)(1/2) = 15 V

V_th = -15V + V_15kΩ = -15V + 15V = 0 V

To find R_th, make all DC power sources to zero (15 V to 0 V and -15 V to 0 V). At this point, both 15 kΩ resistors are in parallel.

R_th = 15kΩ||15kΩ = (15kΩ)(15kΩ)/(15kΩ+15kΩ) = 7.5 kΩ

Therefore the equivalent circuit for the left side of BJT is simply a ground in series with a 7.5 kΩ resistor.

Now, we assume that the BJT is in active mode (we will check whether this is true or not later). Therefore V_BE = 0.7 V. With β = 100, I_C = (100)I_B and I_E = (101)I_B, we set up a KVL equation from -5 V to 0 V.

-5 V + (2kΩ)(101)I_B + 0.7 V +(7.5kΩ)I_B = 0 V

I_B = 0.0205 mA, I_C = 2.05 mA and I_E = 2.07 mA

Finally, use KVL equations to find V_B, V_E, and V_C as following.

V_B + (7.5kΩ)(0.0205 mA) = 0 V;     V_B = - 0.15 V     Ans

-5 V + (2kΩ)(2.07 mA) = V_E;     V_E = - 0.86 V        Ans

V_C + (3kΩ)(2.05 mA) = 10 V;     V_C = 3.85 V     Ans

At saturation, 

V_CE = 0 V, I_C(sat) = [10 V- (- 5 V)] / [2kΩ + 3kΩ] = 3.0 mA

Both I_C and I_E are less than I_C(sat). The BJT is therefore not saturated.     Ans

Check the results;

V_BE = (- 0.15 V) - (- 0.86 V) =  0.71 V ==> OK

V_CE = (3.85 V) - (- 0.86 V) =  4.71 V

0 V < V_CE < 15 V  ==> BJT is not saturated (in active mode as we assumed at the beginning).