I have a MMBT3904L SMT transistor (datasheet http://www.onsemi.com/pub_link/Collateral/MMBT3904LT1-D.PDF) and I am getting wierd behavior accross the emitter and collecter terminals. The base is held high at 3.3v measured with a multimeter but the resistance between the emitter and collector is about 1.4M ohms compared to the 1.6M ohms when the base is held low.

I am very confused as to why the circuit is not working as there should be enough current to saturate the transistor but the resistance only dropped slightly.

I have tested this on multiple transistors and they all exhibit this behavior.

I would appreciate if someone tells me what is wrong with my setup.

  • \$\begingroup\$ Have you tried loading the collector? And held high how, directly to 3.3V? Or with an appropriate resistor? \$\endgroup\$ – Passerby Mar 12 '16 at 2:16
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    \$\begingroup\$ Please show your complete schematic. It does no good to know the voltage at the base if we don't know the voltage at the emitter or collector. \$\endgroup\$ – The Photon Mar 12 '16 at 2:19

I am afraid there is a misunderstanding here.

  • you have killed the transistor by putting its base to 3.3V
  • you're measuring the resistance of the E-C which is pointless (and you may likely measure it with a multimeter with wrong polarity

This is the standard way of driving an NPN transistor: schematic
(source: elinux.org)

Note that:

  • you always need a series resistor for the base. A transistor base will be and stay at 0,7V over the emitter (so in this circuit, at 0.7V) when the transistor conducts. If you force it to anything higher, then you killed the transistor. The other side of the series resistor is what you can drive with 3V or 5V. The resistance of the series resistor is not critical, anything around 1k-10k will do

  • the transistor will conduct in one direction (therefore using a multimeter and measure it in the wrong direction can give misleading results). That is, it is switching the GND to the load as seen on the schematic. So instead of measuring resistance, you can measure voltage between 5V and the collector pin.

In the real life, the transistor "amplifies" current. I.e. when you put 3.3V to the resistor, there is a small current flowing from that to the base, and from the base to the emitter. The transistor "amplifies" this by its amplification factor (some datasheets shows as beta) which is around 200..500.

The amplification is in fact a trick, it is not generating bigger current for real. Rather, it can change the current flowing via the load, i.e. the current between collector and emitter. If you're using the transistor as a digital switch, a very small base current (say, 1mA) is enough to make C-E conduct and let up to 200-500mA (see above the apmlification factor) flow between the load (and C-E).

Transistor amplifiers use this principle to drive a power hungry speaker from a very small signal (however transistor power amplifiers are a more complicated topic).

| improve this answer | |
  • \$\begingroup\$ Nice observations... Still, the simplest explanation of the "amplification trick" is to think of the transistor as of a "controlled resistor". By varying its "resistance" from zero to infinity, it changes the current through the load with resistance RL from Vcc/RL to zero. \$\endgroup\$ – Circuit fantasist Sep 12 at 9:37

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