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My transistor works when its collector is disconnected. That is: The LED in the following circuit glows, albeit somewhat dimmer than when I connect the collector. (but it still glows considerably good)

schematic

simulate this circuit – Schematic created using CircuitLab

Is this normal? If yes, could you explain why is this happening?

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    \$\begingroup\$ The junction between the base and emitter is essentially a diode, so, yes, as long as your base resistor is not super large, this is exactly what I'd expect. Consider putting the diode (with some series resistor) between the the source and the collector... \$\endgroup\$ – MikeP Jan 5 '16 at 19:07
  • \$\begingroup\$ How does it work? As a diode. \$\endgroup\$ – Brian Drummond Jan 5 '16 at 19:36
  • \$\begingroup\$ A transistor can actually be used to generate white noise this way. \$\endgroup\$ – 3871968 Jan 6 '16 at 6:22
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Yep, it works.

The base to emitter connection with collector floating is in fact just a diode. You're downgrading a NPN transistor to a PN junction in your case. You won't get the current amplification feature of the transistor because you've not connected the collector.

In this configuration you have a silicon PN junction and it does all what a PN junction does: rectification, about 0.7v voltage drop and so on. It's not much different to what a vanilla diode like the common 1N4184 diode would do. On the other hand diodes are optimized to do the diode job while a base-emitter junction is not, so you'll likely get better performance from a dedicated diode.

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  • \$\begingroup\$ Then what is the function of the collector? For which purposes do we use the collector? \$\endgroup\$ – Utku Jan 5 '16 at 19:28
  • \$\begingroup\$ @Utku - Now connect the collector to 12 volts. Do you see any difference? Might that give you an idea? \$\endgroup\$ – WhatRoughBeast Jan 5 '16 at 19:49
  • \$\begingroup\$ @WhatRoughBeast Yes I do. I just meant that what causes this difference? What is the magic behind the scenes? But I guess this would be another question itself. \$\endgroup\$ – Utku Jan 5 '16 at 20:59
  • \$\begingroup\$ @Utku there is no magic, this behavior is reason for the ubiquity of semiconductors in electronics. I would suggest wikipedia, book, or a tutorial on how transistors work since the question you are suggesting is beyond the scope of ee.se \$\endgroup\$ – crasic Jan 5 '16 at 21:04
  • \$\begingroup\$ the collector is where the extra current comes from that makes the LED light brightly, if there's no collecter connection the transistor can't amplify the signal that enters at the base. try a higher value for Rbase, eg: 330K this effect will be even more pronounced. \$\endgroup\$ – Jasen Jan 5 '16 at 22:36
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The emitter current (which is your LED current) is the sum of the base and collector currents. With the collector disconnected, you effectively have a voltage+resistance+diode+LED loop, resulting in a base current of about Ib =(5-0.5-1.5)/10680 = 0.3mA, which results in a dimly glowing LED.

Because of that 0.3mA base current, the transistor will be pretty strongly "on", meaning the collector will conduct (a lot more) current than the base. If you then connect a voltage source between the collector and the bottom of your existing 5V source, then additional current will flow through the collector and the LED.

For the chosen transistor in this configuration and a sane collector voltage (say 12V), the transistor will probably be saturated, which means the Vce (collector-emitter) voltage will be at a minimum, and the current flowing through it will be determined primarily by the rest of the circuit. In this case, say you have Vc = 12V, that gives Ic = (12-0.7-1.5)/680 = 14.4mA.

Now the emitter current Ie = Ic + Ib = 14.7mA and you have a nicely-glowing LED.

where the random-looking numbers come from:

Vbe = 0.5V (transistor base/emitter acting as diode)

Vcesat = 0.7V (saturated collector/emitter voltage)

Vf = 1.5V (LED forward voltage)

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