| Electronics > Beginners |
| Question on NPN Logic Inverter From Textbook |
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| rstofer:
Assuming an hFE of 40, I did a little MATLAB thing to figure out what the base resistor should be. I am also going to assume VceSat = 0.2V and Vb = 0.7V. In any event, I come up with around 20k Ohms for the base resistor to get the device into saturation. If I were to actually build the circuit, I would use 10k just to make absolutely sure I got the transistor into saturation. --- Code: ---[font=courier] format shortEng format compact Vcc = 5.0 % given Vsignal = 1.0 % given VceSat = 0.2 % assumed Vb = 0.7 % assumed R2 = 8200 % given VR2 = Vcc - VceSat IR2 = VR2 / R2 Ib = IR2 / 40 % assumed hFE = 40 VR1 = Vsignal - Vb R1 = VR1 / Ib [/font] --- End code --- The results look like: --- Code: ---[font=courier] Vcc = 5.0000e+000 Vsignal = 1.0000e+000 VceSat = 200.0000e-003 Vb = 700.0000e-003 R2 = 8.2000e+003 VR2 = 4.8000e+000 IR2 = 585.3659e-006 Ib = 14.6341e-006 VR1 = 300.0000e-003 R1 = 20.5000e+003 [/font] --- End code --- This little arithmetic problem will also run in the free Octave (I believe based on other tests). Biasing to use the transistor as a common emitter amplifier is an entirely different matter. Here the issue is just to create an inverter. |
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