![]() ![]() 2N3904 has a gain value of 300 this value determines the amplification capacity of the transistor. And 0.3 V (which is half of the nominal 0.6 V bias of the BE junction) is a reasonable rule of thumb.Īs the above simulation shows, the actual values you get will vary a bit from transistor to transistor - the 2N3904 is considered a "switching" transistor with low Vce, but other transistors may have higher values. Brief Description on 2N3904: 2N3904 is a NPN transistor hence the collector and emitter will be left open (Reverse biased) when the base pin is held at ground and will be closed (Forward biased) when a signal is provided to base pin. For this reason, the collector can be pulled some way below the base, but not too much. Now when the collector is pulled towards the emitter, the BC junction starts to be forward biased: once that happens, it will "steal" current from the base, and less current is available to flow in the BE junction. Consider an n-p-n transistor with its base - emitter junction forward biased and collector base junction reverse biased. The total current gain of a transistor is the ratio of output current to the input current. 1 Current gain of transistor Q1 2 Current gain of transistor Q2 The total current gain of the Darlington Transistor is D. The relationship is exponential - so the change in voltage needed to double the current is quite small. Total base-emitter voltage is a summation of the base-emitter voltage of both transistors. The transistor starts to conduct when there is significant current flowing in the BE junction - because that is a (si) junction, this happens at around 0.6 V. If you are testing PNP transistor, you should see OL (Over Limit). ![]() For an good NPN transistor, the meter should show a voltage drop between 0.45V and 0.9V. Hook the negative meter lead to the EMITTER (E) of the transistor. Note, you will likely need a resistor between the collector of the first transistor (NPN) and the base of the second (PNP). The base forms two circuits, the input circuit with the emitter and. Base The middle section of the transistor is known as the base. The collector section of the transistor is moderately doped, but larger in size so that it can collect most of the charge carrier supplied by the emitter. When I run the DC simulator, I get the following values:Īs you can see, the collector voltage is about 134 mV - well below the base voltage, which is 702 mV (and less than your "rule of thumb" of 0.3 V). Step 1: (Base to Emitter) Hook the positive lead from the multimeter to the to the BASE (B) of the transistor. The collector-base junction is always reverse biased. Simulate this circuit – Schematic created using CircuitLab Here is a very simple transistor circuit: ![]()
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