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am a beginner trying to understand transistors. I am using a npn transistor.

I have a 6V battery.

The base and collector are connected to 6V Emitter has the load (LED + 100ohm resistor) to the ground. Current passes and the LED lights up. I understand, it is in saturation mode. With or without a base resistor.

Case 2: The base is connected to 6V. The collector has the load (LED + 100ohm resistor) connected to 6V as well. Emitter is connected to the ground. The LED does not light up when powered.

Case 3: Further to Case 2: I use a 1K resistance with the base and the LED lights up.

Why does not the LED light up in case 2?

This seems similar to Emitter Follower LED Circuit - LED in Collector, Resistor in Emitter, but I did not get the concept.

Thanks.

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  • \$\begingroup\$ You need to take a step back and understand that an NPN transistor can act like two diodes back to back (sometimes!). In case 2, you're forward-biasing the B-E diode, and it would burn out quickly if the power supply was beefy enough. \$\endgroup\$ – Daniel Oct 19 '16 at 19:49
  • \$\begingroup\$ read more how good designs work as switches using Ohm's law and V=I*R drop + diode Vf \$\endgroup\$ – Sunnyskyguy EE75 Oct 19 '16 at 22:10
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    \$\begingroup\$ This would be a much better question with actual diagrams of the circuits in your different cases. There is a schematic capture capability when you edit the question. Look for the icon above the text entry field that looks like a schematic being drawn (to the right of the one that looks like a picture). \$\endgroup\$ – Makyen Oct 19 '16 at 22:22
  • \$\begingroup\$ Which transistor are you using as a model? \$\endgroup\$ – EM Fields Oct 19 '16 at 22:23
  • \$\begingroup\$ @Daniel: Thanks. I get am forward biasing the BE diode. But so is the case, in case 1 too, right? What sort of biasing is there in the CB diode? \$\endgroup\$ – electronicsdummy Oct 20 '16 at 6:54
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The LED does not light up because the B-E junction of the transistor is passing enough current to pull the 6V source down to <1V, which is too low to light the LED. It's amazing that the transistor doesn't burn out in this situation, but perhaps it's a rather beefy one, and your 6V supply has a relatively high impedance.

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  • \$\begingroup\$ To which instance are you referring? \$\endgroup\$ – EM Fields Oct 19 '16 at 22:41
  • \$\begingroup\$ @EMFields: Case 2, the one the OP explicitly asked about: "Why does not the LED light up in case 2?" \$\endgroup\$ – Dave Tweed Oct 20 '16 at 3:34
  • \$\begingroup\$ @DaveTweed: Is it not the same case when the LED load with 100ohm resistor is on the emitter side? Since there is no 1K resistor connected to the base, in Case 1 too, the B-E diode will pull the source to < 1V? It is not burning because probably I have a switch between my 2, 3V supplies and I switch it on for just a second or two to test my circuit. My motive is in understanding, what is the difference with the load on emitter or collector. The load on emitter does not need a 1K resistor and the load when on collector, needs it. \$\endgroup\$ – electronicsdummy Oct 20 '16 at 7:04
  • \$\begingroup\$ No, it is not the same. When you have the LED in series with the emitter, the B-E drop across the transistor is still just 0.7 V or so, but this is not directly across the 6 V power supply -- you end up with about 5.3 V across the LED and its resistor in that configuration. \$\endgroup\$ – Dave Tweed Oct 20 '16 at 10:41
  • \$\begingroup\$ @Dave Tweed Thanks. Just to reiterate, you mean: Load (LED + 100ohm resistor) on the collector, the voltage across it is 5.3 V \$\endgroup\$ – electronicsdummy Oct 21 '16 at 14:08
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Why does not the LED light up in case 2?

There are two possible reasons assuming you wired it as you state: -

  • The 6V power source was "clamped" to about 1 volt because you forced a raw base-emitter region across it. It's a diode after all.
  • You blew the transistor base-emitter region up because you didn't use a current limiting resistor.
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  • \$\begingroup\$ Thanks. Please see my extra queries on my comments to Dave Tweed \$\endgroup\$ – electronicsdummy Oct 20 '16 at 7:06
  • \$\begingroup\$ When you have an emitter resistor, this is equivalent to having a base resistor that is hFE times bigger. \$\endgroup\$ – Andy aka Oct 20 '16 at 7:35
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Case 1: with the LED and its series ballast connected between the emitter and ground and the collector and base connected to Vcc, the transistor can never go into saturation because the voltage drop across the LED and its ballast resistor will never allow the emitter to see ground.

Case 2. Please post a schematic and post the part number of the transistor in question.

Case 3. Please see case 2

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  • \$\begingroup\$ EMFields: Thanks. I did not get what you say in Case 1. I assumed, it is in saturation and my transistor is acting like a switch here and hence the LED lights up. \$\endgroup\$ – electronicsdummy Oct 20 '16 at 6:15
  • \$\begingroup\$ I am using a S8050-D331 : NPN transistor in the SNAP circuits kit. And a switch between the two 3V batteries. And in Case 1: Saturat \$\endgroup\$ – electronicsdummy Oct 20 '16 at 6:17
  • \$\begingroup\$ Sorry, pressed enter errorneously. \$\endgroup\$ – electronicsdummy Oct 20 '16 at 6:18
  • \$\begingroup\$ Case1: If it is not in saturation, why does the LED light up? Transistor is acting like a switch here, was my assumption. Do you mean, it is an amplifier here?**Question2:** What is the effect of having the load on collector and emitter and the practical consequences. \$\endgroup\$ – electronicsdummy Oct 20 '16 at 6:32
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@Daniel: Thanks. Your comment, got me thinking and I read articles on biasing again and in Case 2: without a resistor the FF biasing mode is happening and hence goes into saturation. Putting a resistor on base makes it in FR with the load on collector. http://www.vsagar.org/how-npn-transistor-works-tutorials-of-forward-reverse-biasing/

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