Author Topic: Bipolar Junction Transistors "BJT"  (Read 2493 times)

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Offline Lazarus

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Bipolar Junction Transistors "BJT"
« on: January 02, 2015, 09:15:05 PM »
This is a work in progress, and as such is subject to change, so i will be editing this post as i think of more detail and better ways of presenting it, but if i'm gonna do this then i've got to start somewhere, so for now here goes,

How it Works:

as illustrated in the figure below of an NPN transistor, if we apply a small amount of base current it will cause a much larger amount of collector current to flow, this is how a transistor is able to amplify a signal, and the ratio of base to collector current is called the current gain, symbolized by the greek letter beta "β", the current gain is also sometimes called "Hfe",

in the example below, lets assume that we have a base current of 1 milliamp, and a collector current of 100 milliamps, the ratio would be 100:1 so the current gain of this transistor would be 100, so in this case β = 100, here's the formula:

β = Ic / Ib

where:
β = current gain
Ic = collector current
Ib = base current

ok now lets apply 2 milliamps base current, with a current gain of 100 that would yield a collector current of 200 milliamps,

so now it can be seen how it is able to amplify a signal, because that a small amount of base current causes a much larger amount of collector current to flow, just a 1 milliamp increase in base current caused a 100 milliamp increase in collector current,



β = Ic/Ib

Vcc = Collector Supply Voltage
Vbb = Base Bias Supply Voltage
Ib = Base Current
Ic = Collector Current
Ie = Emitter Current
Vbe = Base Emitter Voltage, or Voltage measured across the Base and Emitter
Vce = Collector Emitter Voltage, or Voltage measured across the Collector and Emitter
Vcb = Collector Base Voltage, or Voltage measured across the Collector and Base
Rb = Base Bias Resistor, used to limit the base current
RL = Load Resistor, used to limit the collector current
β = Current Gain, or HFE

Offline Lazarus

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Re: Bipolar Junction Transistors "BJT"
« Reply #1 on: January 03, 2015, 12:19:14 AM »
Here is a PNP type, a PNP works exactly the same as an NPN in every respect except all the polarities are reversed, therefore it is an exact polar opposite of an NPN,



by now some of you have probably noticed the arrows showing the direction of current flow and are saying hey wait a minute, what's going on here, well contrary to popular belief, current flows from negative to positive, after all electrical current is the flow of electrons, and electrons are negatively charged, therefore they flow toward a positive charge, so therefore electrical current flows from negative to positive, and it just chaps my hide when i see tutorials plastered all over the internet with illustrations showing it backwards, in fact it really kind of ticks me off,

Offline Lazarus

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Re: Bipolar Junction Transistors "BJT"
« Reply #2 on: January 03, 2015, 05:35:47 AM »
A Practical Application:

to start off with we would need to select a transistor that has a high enough voltage and current rating for the job, so then we would need to know a few details about the load that the transistor would be powering, so lets assume that our load needs 10 volts at 1 amp, we would need to select a transistor with a Vce rating of at least 10v, and a collector current rating of at least 1 amp right? wrong! it is good design practice to pick a transistor rated at least 30% higher than we will ever need, and by doing so the transistor is likely to never fail, so a transistor with a Vce rating of 15v or higher, and a max collector current of 1.5 amps or higher, would do nicely,

ok now lets assume that our transistor has a current gain of 100, so to get started lets start with the collector current, we already know that our load is going to have 10 volts dropped across it, and a current of 1 amp flowing through it, so our collector current is obviously going to be 1 amp,

now lets calculate the load resistance using this formula:

Vcc / Ic = RL

where:
Vcc = collector supply voltage
Ic = collector current
RL = load resistance

so the voltage dropped across the load divided by the current flowing through the load equals the load resistance, so:

10v Vcc / 1A Ic = 10Ω RL

so RL resistance is 10 ohms,

now lets calculate the value for or base bias resistor, since our transistor has a current gain of 100 then we already know that our base current should be 10 milliamps, because 1 amp or, 1000 milliamps collector current divided by the current gain of 100 = 10, here is the formula:

Ic / β = Ib

where:
Ic = collector current
β = current gain
Ib = base current

now lets assume that our base bias supply is also 10 volts, and remember that it takes approx 0.6 volts on the base to turn the transistor on, so that 0.6 volts will be subtracted from the base bias supply voltage, now we can find the value needed for our base bias resistor by using the formula:

(Vbb - Vbe) / Ib = Rb

where:
Vbb = base bias supply voltage
Vbe = base emitter voltage
Ib = base current
Rb = base bias resistor value

so:

(10v Vbb - 0.6v Vbe) / .01A Ib = 940Ω Rb

so our base bias resistor Rb comes out at 940 ohms,

ok i think that just about does it for this little project, you could add a switch in series with the base bias supply to turn it on and off if you want, and it doesn't have to be a big switch either, in fact it would only need to be able to conduct 10 milliamps of current, the transistor is taking care of the heavy 1 amp of current that the load is drawing, which was really the purpose of using the transistor in the first place, so that a very small amount of current flowing through the base can be used to control a much larger amount of current flowing through the collector,

:)