# Stk412 750 Datasheet 16.pdf

Stk412 750 Datasheet 16.pdf

Here I have written my problem to the best of my knowledge, can some one please tell me where am I wrong. Thank you for your time A: Your understanding is close, but incorrect. If the open-circuit voltage of D1 is Vout, you are observing the correct thing. If D1 doesn’t turn on (because it only conducts when it’s configured for high-side switching), R2+R1 will conduct up to a maximum of Vout = 2 x Iout + (Vout – Vin). If Vin = 5 V, then the maximum Vout that R2+R1 will conduct is 5 V, and the most positive current that R1+R2 can conduct is 2 x 0.5 = 1 ampere. That leaves the possibility of \$I_R = 2 \frac{Vin-Vout}{R1+R2} \$ amperes flowing through D2, so D2 conducts up to at least (2 / 1) = 2 V. The real issue is that the D1+D2 junction cannot conduct enough current to light up the LED when you have a 2 V input. That’s because the D1+D2 junction will always conduct the maximum amount of current needed to drive the LED. The LED will be conducting the rest. You can get \$V_{LED}=I_C \times R_C\$ where \$I_C\$ is the collector current of the transistor. So, with \$I_C = I_{D1+D2} \approx 3.0 – 3.5 mA\$ and a 0.01 V LED, you’ll be getting 0.5 V or less at the LED, which is barely enough for it to barely glow. Furthermore, while the LED can turn on with \$I_{D1+D2} < I_{R2} \le I_{D2}\$ (so \$V_{LED} = I_{D1+D2} \times R_C \le \frac{V_{LED}}{R_C} \le \frac{V_{LED}}{2R_2} = \frac{1}{2} \frac{V_{LED}}{R_2} < I_{R2} \times R_C\$), it can't turn on