I am currently learning about transistors from allaboutcircuits.com.

He presented the following schematic:

transistor circuit

He claims that when the transistor is in its cutoff state, V.output will read the full voltage of the battery, which he stated earlier to be 15V. Why is this? Shouldn't the voltage reading be smaller than 15V since of the voltage needs to be dropped across the resistor at the top of the circuit? Is he neglecting the resistance of the resistor for teaching purposes, or am I missing something?

  • \$\begingroup\$ Byron, could you edit your post and add a link to the page on allaboutcircuits so we have more context? \$\endgroup\$ – Nick Alexeev Apr 16 '15 at 19:58
  • \$\begingroup\$ Ohms law "says" V = I x R. If I=0 then V = 0 | When transistor is fully off (no light) I = 0 so IR=0 so there is no V drop in resistor. When transistor is on current flows and there is voltage drop. \$\endgroup\$ – Russell McMahon Apr 17 '15 at 0:55

Yes - You are correct in thinking of the transistor as another resistor in series - but what is the off resistance of a transistor ?
Or alternately what is the current flowing through R when the transistor is off?

  • \$\begingroup\$ It might be more useful for future readers if this answer was complete instead of simply guiding the asker in the right direction. \$\endgroup\$ – user 5061 Apr 16 '15 at 19:38
  • \$\begingroup\$ I'm assuming the resistance of the transistor when it's unsaturated is considered "infinite". If so, would that be why it reads the full voltage drop? \$\endgroup\$ – Sam D20 Apr 16 '15 at 19:42
  • 1
    \$\begingroup\$ exactly right. When the transistor is in cutoff, it does not allow current to pass through the transistor, thus making it like an open circuit (infinite resistance path). No current through a resistor means no voltage drop across a resistor. \$\endgroup\$ – TylerH Apr 16 '15 at 19:56

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