Basic NPN Transistor Amp

YouTube TUTORIAL https://www.youtube.com/watch?v=-itA0p_Uikk The output voltage has a gain of 8.75. The supply voltage is 12Vdc. (The original input signal from the AC supply source is 1.3 volts peak to peak. This simply means positive (+)650mv peak and negative (-)650mv peak, giving you a total of 1.3v peak to peak. The input amplitude is 650mv volts.) The input signal from the capacitor to the transistor base is 1.2 volts peak to peak. The voltage divider provides the transistor base with a 1.25vdc signal. The 1.25vdc positions the collector vdc signal and the emitter vdc signal just right so at maximum amplitude the collector ac signal will not top out, emitter ac signal will not bottom out, and both collector and emitter ac signals will not meet in the middle. In other words no clipping. The transistor stays on because the 1.2vdc is more than enough to keep it on even with the capacitor signal at Neg Vp. (1.25v (+) - 650mv = .6v) Only .7volt is required to turn the transistor on. (((Transistors are know to start turning on below .7vdc which means you can still avoid clipping at the negative peak))). AC Source signal 0vdc center: .650mv peak: -.650mv peak: (From .-650mv to +.650mv is a total of 1.3 volts peak to peak.) Transistor base input signal 1.25vdc Center: 1.85v peak: .650mv peak: (Peak volts =1.25v + 0.6 volts =1.85v) (Negative peak volts = 1.25v - .6v = 650 mv.) 0.6volts plus 0.6volts = 1.2 volts peak to peak. The voltage divider places the sine wave coming out of the capacitor on a 1.25vdc signal as if the 1.25vdc was virtual ground to the sine wave. This is done to make accommodations for the 700mv needed to turn the transistor on. Collector max output signal: 6.81vdc center signal: Vp = 11.3 vdc (+) -1.3 vdc = 10.5vpp / 2 = 5.25v peak. Emitter Signal .53vdc Center: Vp = 1.09v (+) -24mv = 1.066vpp / 2 = .533vp Voltage gain is Vpp out / Vpp in = 10.5Vpp / 1.2Vpp = 8.75 Changing the supply ac input voltage to the base changes the collector output voltage amplitude. Changing the Potentiometer percentage changes the gain, because it will reduce the gain from the 10 to 1 ratio that it is currently at. (4.7k ohms to 470 ohms) In avoiding clipping, leaving a little play, and the forward bias .7 diode voltage takes up some of the Ratio. I placed open dc sources to reference peak voltage levels. Opening the switch to remove the ac signal shows how the voltage divider keeps the transistor on and how the collector dc level is centered for maximum gain.

published 7 years ago