I connected the collector to positive (through a 330 Ohm resister) , base to positive (through a 10k ohm resistor) and emitter to an LED which I connected to ground . I measure with a multimeter the voltage drop across LED ; by first pulling the LED from the ground {I'm going into pointless detail to leave out no detail because what happened is strange) and connecting the loose end of the LED to the one of the multimeter leads and one multimeter lead to the ground . I got a voltage drop of 1.2 (as expected). Now I reconnected the LED to ground and touched the multimeter leads to the emitter and ground , here is the kicker , I still got a voltage drop of 1.2 . How is there a drop before the LED ? What happeend ?

I'm sorry that I have no picture for proof , I'm not sure how I can do it but I'll try .

EDIT : It is also important to note that when I removed the LED and connected the emitter directly to ground there was no voltage drop

enter image description here

  • 2
    \$\begingroup\$ It would be much easier to understand your circuit if you just included a schematic instead of describing it with words. \$\endgroup\$ – The Photon Aug 6 '15 at 15:19
  • \$\begingroup\$ @Spehro Pefhany the positive voltage is 6 so that the base is saturate. I am aware of the fact that there is a. 6 voltage drop but I'm getting a voltage drop off 1.2 \$\endgroup\$ – Vrisk Aug 6 '15 at 15:21
  • \$\begingroup\$ @The Photon added, I hope this helps. I measured voltage across D1 and from Q1 emitter to ground. Both have me a voltage drop off 1.2 \$\endgroup\$ – Vrisk Aug 6 '15 at 15:38
  • \$\begingroup\$ @AnirudhGanesh its hard to understand what exactly you are asking or describing. But if I understand correctly, you are asking why are you getting 1.2V between the emitter and ground with the LED connected ? \$\endgroup\$ – efox29 Aug 6 '15 at 15:45

Okay, I think somewhere in here one of these matches your verbal description. I ran the simulation with the built-in circuit simulator and the voltages are correct for the type of LED used.


simulate this circuit – Schematic created using CircuitLab

enter image description here

R7 just simulates the input resistance of your meter, so the emitter is open.

In cases 1 and 3 the transistor is saturated and has < 0.1V across it (collector to emitter).

In case 2 the transistor has about 0.5V across it (less than the usually quoted 0.6 or 0.7V because the meter draws almost no current).

I don't see anything that resembles 1.2V there. If it's an IR LED then you could get 1.2V and then V1 would be 1.2V but the other voltages should be the same as shown above.

  • \$\begingroup\$ Yes ! I think I remember now , I got 2 Volts (Which is V1) at emitter not a voltage drop of 1.2 . The simulation seems fairly consistent with what I got . My transistor was saturated . Okay, why does this happen ? I thought the voltage drop is only across the LED ? \$\endgroup\$ – Vrisk Aug 6 '15 at 16:22
  • \$\begingroup\$ When the LED is there, the voltage drops across the resistor too. The transistor drops almost no voltage (< 0.1V) so it's similar to if you had the resistor in series with the LED- 2V across the LED and 4V across the resistor, approximately. Otherwise the LED and transistor would burn out. \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 16:26
  • \$\begingroup\$ Why would there be a voltage drop across the resistor ? Has it got something to with the non ohmic properties of LEDS ? \$\endgroup\$ – Vrisk Aug 6 '15 at 16:33
  • \$\begingroup\$ @AnirudhGanesh There's current going through it (about 12mA), so it drops about 4V. \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 16:49
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    \$\begingroup\$ @AnirudhGanesh The battery has 6V, the LED has about 2V almost regardless of current- the non-ohmic property you mention, and that leaves 4V for the resistor, so the current will be ~12mA. I = E/R. If the current was 1uA or 1A it would not be true, but for reasonable currents you can say the LED acts almost like a fixed voltage. \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 17:53

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