Question
I have to determine the operating region for thefollowing: npn, Vbe = .8V, Vce = .4V npn, Vcb = 1.4V, Vce = 2.1V pnp, Vcb = .9V, Vce = .4V npn, Vbe = -1.2V, Vcb = .6V My Prof was rather unclear on how to go about this types ofproblems. Thank you for your help ahead of time. I have to determine the operating region for thefollowing: npn, Vbe = .8V, Vce = .4V npn, Vcb = 1.4V, Vce = 2.1V npn, Vcb = 1.4V, Vce = 2.1V pnp, Vcb = .9V, Vce = .4V npn, Vbe = -1.2V, Vcb = .6V My Prof was rather unclear on how to go about this types ofproblems. Thank you for your help ahead of time. pnp, Vcb = .9V, Vce = .4V npn, Vbe = -1.2V, Vcb = .6V My Prof was rather unclear on how to go about this types ofproblems. Thank you for your help ahead of time. npn, Vbe = -1.2V, Vcb = .6V My Prof was rather unclear on how to go about this types ofproblems. Thank you for your help ahead of time.
Explanation / Answer
Assuming the transistor is made of silicon For a transistor Vbe + Vec + Vcb = 0 For a transistor Vbe + Vec + Vcb = 0 1) For npn transistor if Vbe = 0.8V and Vce=0.4V Vbe = 0.8V forward biases the emitter junction Vcb = - Vbe + Vce = -0.8+0.4 = -0.4V forward biases thecollector junction The transistor is in SATURATIONregion 2) For an npn transistor if Vcb = 1.4V andVce=2.1V Vbe = -Vcb - Vec Vbe = -Vcb + Vce Vbe = - 1.4 + 2.1 = 0.7V The emmiter junction is forwardbiased Vcb = 1.4V reverse biases the collector junction Vcb = 1.4V reverse biases the collector junction Hence the transistor is conducting and is in active region 3) For an pnp transistor if Vcb = 0.9V and Vce=0.4V Vbe = -Vcb - Vec Vbe = -Vcb + Vce Vbe = - 0.9 + 0.4 = - 0.5V just forward biasesthe emitter junction Vcb = 0.9V , forward biases the collector junction Hence the transistor is just conducting and is in near to saturation region 4) For npn transistor with Vbe=-1.2V and Vcb=0.6V Vbe = -1.2V reverse biases the emitter junction Vcb = 0.6V reverse biases the collector junction The transistor is in cutoffregion Vbe = -Vcb - Vec Vbe = -Vcb + Vce Vbe = - 0.9 + 0.4 = - 0.5V just forward biasesthe emitter junction Vcb = 0.9V , forward biases the collector junction Hence the transistor is just conducting and is in near to saturation region 4) For npn transistor with Vbe=-1.2V and Vcb=0.6V Vbe = -1.2V reverse biases the emitter junction Vcb = 0.6V reverse biases the collector junction The transistor is in cutoffregion