Electrical – the turn-on and turn-off condition of the npn-transistor


Following is the circuit I have been analyzing. My intention is to calculate the turn on and off condition of the transistor Q1. R1 and R2 (10 kΩ each) are internal to the transistor.


simulate this circuit – Schematic created using CircuitLab

Following is my understanding, whenever there is voltage at the input V1, it will equally divide across R1 and R2. When it reaches 0.7 at Vbe, Q1 will be turned on. If I want to increase the turn on voltage of the transistor, introducing voltage divider at the input (before resistor R1) will be sufficient.

I want to on and off the transistor above to a fixed voltage, say 5 V. Please advise my understanding.

Best Answer

This makes a lousy comparator R1/R2 0.8min 1.0typ 1.2max

Thus +/-20% of 0.6V becomes 20% of 5V is best case and Leakage current affects R3 Ambient temperature variation adds another 10%.

General design not for this Q1

  • Assuming hFE=100 (bad assumption) and Vce=Vce(sat) = 0.2V at 10% of hFE or Ic:Ib=10:1
  • Ic=15.8V/0.1MΩ = 158uA, Ib=16uA

    • At high temp Leakage current can be 20uA edge of graph not spec'd

    - so 100k load is a poor choice for this Q1 and hFE can be as low as 30

  • if Vin =5V and Vbe=0.6 ( not 0.7 at this low current)
  • Rb=4.4V/16uA=275k

  • Next the threshold for turn-on needs to be 5V input.

    • thus 0.6V/5V= 12%

    • thus adding external R1 to internal R1= 75k-10k= 48k or **external R1' ~65k

Threshold at room temp may be 4.4 to 5.7V for 80% output swing. Simulated with triangle wave. enter image description here


simulate this circuit – Schematic created using CircuitLab


Define Cost limit, Ambient range, V threshold tolerance and go back to drawing board