We assume that the LED will light as soon as the transistor comes out of the cut-off region, therefore the BJT is on and operates in the active region. I know that that Vbe = 0.7V, Ib>0, ic=BIb and Vce>=0.7. Is this inequality for Vi expected for the BJT in the active region? If so, what value should we expect fot saturation mode?

NPN BJT circuit


  • \$\begingroup\$ It is important to remember hFE drops to 10% of the max value when fully saturated or =10 typ. at rated value of Vce=Vce(sat) \$\endgroup\$ Commented Oct 27, 2019 at 3:14

1 Answer 1


You are very close. You unfortunately dropped the minus sign on your third last line. After correcting that, you will find Vi <= 1.66 V


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