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Author Topic: Understanding Transistor (TIP120) Datasheet  (Read 1353 times)
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For reference I'm using the following datasheet: http://www.fairchildsemi.com/ds/TI/TIP120.pdf

I am going to use the TIP120 transistor as a switch to turn off and on a group of LEDs; however I'm getting confused on the research I'm doing and the datasheet.  When looking at the table on page 2 ,it states the hFE is a min of 1000.  However in Figure 2 (page 3) there is a graph of Base-Emitter Saturation Voltage and Collector-Emitter Saturation Voltage where it states Ic = 250*Ib (upper right corner of graph); this leads me to believe that the hFE is 250.   Which value do I use to calculate Ib, 1000 or 250 (Ib = Ic/hFE)?

Secondly, on Figure 2 (page 3) can I use that graph to determine the Vbe at a particular Ic?  The reason for asking is because I want to determine the resistance at the base I need (Rb = (Vin - Vbe)/Ib) and during my research it said that most Vbe is around 0.7V.  Looking at that graph the Vbe can be anywhere between 1.25V to 2.5V for the TIP120, so should I use that value to determine the Rb?

Any help is appreciated.
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Vbe for a single transistor is typically one diode drop's worth of voltage, which is where the 0.7V comes from.
TIP120 however is 2 transistors in  a Darlington configuration with one transistor driving a 2nd transistor so that the source can supply less current and still achieve higher gain.
Since you are intending the TIP120 to be used as a switch, you will drive it hard enough to ensure it goes fully into saturation when turned on.
From Figure 2, the lowest Vbe(sat) is 1.25V.
So base resistor of (5v  - 1.25V)/20mA = 187.5 ohm, so use a 180 ohm resistor.
Actual collector current flow will then be determined by the current limit resistor used with the LEDs, typically seleceted for a max continuous current of 20mA max.
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Thank you for the reply and information.  I didn't realize it was a Darlington configuration (even though it is clearly stated) and that is the reason the Vbe would not follow convention.
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