I read somewhere on this site that a transistor in saturation mode dissipates less energy than a transistor in active mode. I would like to know why. Please help as soon as possible.
y = x*( m*x + b) = m*x^2 + b*x
for m < 0, b > 0
where m*x + b is the equation of the load line in a CE configuration.
You must also remember that if voltage drops are positive then a votage rise is negative. So the base is 0.5 volts above the collector and that by symmetry Vbc is the additive inverse or 0.5VThanks guys - Alexk,Papabravo, and hgmjr. Comments were really helpful. Out of curiousity, i would like to know: if Vbe = 0.7 and Vce = 0.2, would it be right to say
Vcb = Vce - Vbe = -0.5 volts?
by Luke James
by Jake Hertz
by Luke James
by Jake Hertz