There is transistor tutorial on sparkfun.com (https://learn.sparkfun.com/tutorials/transistors/all). I am wondering why there is 1.3 V drop depicted across the LED1 on the picture, since when the base is low (open switch) the collector's current is 0 (almost, 50nA or similar), and it is correctly depicted (0mA). Where does the 1.3V drop come from? Shouldn't it be 0 and 3.7V drop be 5V?

enter image description here

  • \$\begingroup\$ I gather you are asking why there might be a voltage drop across an LED with very little current (leakage?) Are you aware of the Shockley diode equation? It's all there. \$\endgroup\$
    – jonk
    Commented Dec 27, 2018 at 0:49
  • \$\begingroup\$ No, I am just not sure what is the source of voltage drop in this picture because they do not explain voltage across LED at all. But I guess I have answered my own question: the 1.3V drop comes from that tiny current through collector when transistor is in off state. How do I find it in datasheet, would that be collector's cut-off current, I_CB0? \$\endgroup\$
    – 4pie0
    Commented Dec 27, 2018 at 0:55
  • 1
    \$\begingroup\$ In the On Semi datasheet for 2N3904 there's a line for "collector cutoff current", given as 50 nA. \$\endgroup\$
    – The Photon
    Commented Dec 27, 2018 at 1:14
  • \$\begingroup\$ The approximate equation will be something like \$V_\text{LED}=\eta\frac{kT}{q}\operatorname{ln}\frac{I_\text{leakage}}{I_\text{SAT}}\$. Assuming \$I_\text{leakage}\approx 50\:\text{nA}\$, \$I_\text{SAT}\approx 1\:\text{pA}\$, and \$\eta\approx 4.5\$, you might see about that voltage. \$\endgroup\$
    – jonk
    Commented Dec 27, 2018 at 1:28
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    \$\begingroup\$ It could also be that the folks at Spark Fun measured the voltage at the collector of the transistor and then made some assumptions, neglecting the resistance of the multimeter (probably 10M\$\Omega\$, if it was a digital VOM). \$\endgroup\$
    – TimWescott
    Commented Dec 27, 2018 at 1:34


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