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Anonymous asked in Science & MathematicsEngineering · 10 years ago

I'm having a problem understanding transistors...?

Hello, I'm pretty new to electronics and I have a problem understand how a bipolar transistor (PNP or NPN) works. Every transistor has three legs that connect to the different layers of the transistor. We are supposed to connect the two outer layers of the transistor to the circuit. If electricity passes through the middler layer (the base) then the circuit will work, right?

Now, I have a few questions:

1) How do we pass electricity to the middle layer (the base) if it only has one wire that connects to it? You can't create a closed circuit like that. Please explain.

2) I also read that you can use a transistor as an amplifier, how does that work?

Thanks ahead.

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  • 10 years ago
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    Hey there!

    As far as your first question (in the paragraph), the answer is YES.

    Before I answer some of your questions I would just like to make sure you know some of the terms.

    Collector: The pin that current flows into.

    Emitter: The pin that current flows out of.

    Base: The pin that controls the flow from Collector to Emitter.

    I'll explain a couple of things about the different transistors. NPN transistors are normally CLOSED. That is, if nothing is flowing into the BASE, then the transistor is not allowing any current to flow from the COLLECTOR to the EMITTER. But when current is allowed to flow into the base on an NPN, then current will be allowed to flow from COLLECTOR to EMITTER.. PNP transistors are the opposite (normally OPEN). They will allow current to pass from the COLLECTOR to the EMITTER, until you apply current to the BASE (then it turns off).

    1. The way you are able to pass current through the BASE is because the current actually goes into the BASE then out the EMITTER. The EMITTER is the ground for both the BASE and the COLLECTOR.

    2. Yes, they can be used as an amplifier. The way you do this is to take the small signal you would get from, say a CD player, and apply the signal to the BASE. The BASE would then open and close the 'gate'* from the COLLECTOR and the EMITTER at the same frequency the CD player was putting out. I hope that makes sense.

    There's also something referred to as "Gain." This is how much the transistor will amplify the current at the base. Now it won't amplify it from the base, it will just allow that much to flow from the COLLECTOR to the EMITTER.

    For Example: If a transistor has a gain of 50, then applying 2mA at the BASE will allow 100mA to flow from the COLLECTOR to the EMITTER. The term "gain" does not have a unit, this is because it is a ratio.

    This is probably just 1% of what you could learn about transistors. There is so much to learn, even for me.

    If you can get Billrussell42 to answer your question, you'll know a lot more than I can tell you. He's a freaking genius.

    Sorry if capitalizing the pins was annoying, I just wanted to try and make it a little easier.

    BTW: There are a couple of common cases you will find transistor in.

    TO-220: Black rectangle with 3 pins coming out of the bottom (back plate is usually connected to the BASE pin)

    TO-92: Black half-cylinder with 3 (much thinner) pins coming out of the bottom.

    TO-3: Short, Silver cylinder with 2 (stiff) pins coming out of the bottom. The two pins are the BASE and the EMITTER. While the entire case acts as the COLLECTOR. This one is a favorite for particularly high current projects (>5A or so).

    * = I don't want to confuse you by saying "Gate," this is a term used in MOSFET'S (but it plays the same role as the BASE does in a Bipolar Junction Transistor.

    Source(s): Electronics Engineering student
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  • 10 years ago

    1) We pass current (not electricity) through the wire in the base by having a higher voltage that we have on the emitter (on a NPN transistor). This currents gets out through the emitter as well. So the current coming out of the emitter equals the current coming in through the collector plus the current from the base (which is tiny compared to the collector current). Ec = Cc + Bc

    Technically the Base-Emitter junction works as a diode, so if you have a higher voltage (around 0.7V higher) on the base that on the emitter this "diode" turns on and the transistor starts working. In this case the transistor is working as a switch (which I think is the way you want to use it), if you put voltage on the base whatever is connected to the other side of the circuit will turn on, if you take that voltage away (or lower it past the emitter voltage) the everything turns off on the other side.

    Everything I just said so far is for a NPN transistor, for PNP everything is reversed.

    2) The transistor has a parameter that is called 'beta', which is how much the current on the base gets amplified on the other side (Collector-emitter). Normally beta = 100, which means that if you inject 1milli Amp into the base, you'll have 100milli Amps going through the other side. So you can see how by manipulating the current on the base, you can amplify what's going on the circuit that you have hooked up through collector-emitter.

    Hope this helps.

    Source(s): Electrical Engineer
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  • Julie
    Lv 4
    4 years ago

    For the best answers, search on this site https://shorturl.im/awGHF

    IMHO with a bit of thought the transistor option can be fine for switching low voltages, if relay is on same board as , even with heavy currents *if* the track routing has been done carefully. but understand damage might happen to uC if the diode fails- with cheep uC that's not as big as incentive as it used to be. Notice that with the transistor circuit the microprocessor shares it's ground with the load. This is perfectly OK when instead of driving a relay the transistor circuit is used to interface different logic families. But especially when switching heavy currents it's best to keep signal and power lines as isolated as possible. Also if you're switching mains, consider what would happen if that ground connection was ever lost. What you thought was a safe voltage with respect to ground, suddenly becomes mains voltage with respect to earth. That's potentially dangerous for anyone who is sent to figure out why the system has stopped working, as well as requiring the blown uC be replaced. You could add protection, but... photo isolator then wins out on simplicity/parts count A real potential advantage of the photo isolator is there's no reason why the LED need be in same physical package as the photodetector. Electrically it's the same schematic even if they're on separate boards with a length of fibre optic to direct the light between them. That helps organisation of boards into "low voltage" and "high voltage/power". Sending the signal as light reduces the opportunities for electrical interference to distort the control signal. If the LED uses visible light that can provides a useful confirmation that a activate signal is actually being sent- you just unplug the optical circuit and have a look to see if there's light . In addition it also means that there is no wire that could conduct (potentially high) voltages back to the microprocessor. In the field where you might get an idiot,or someone a little careless/distracted when possible it's usually best to reduce the opportunity for dangerous mistakes to happen

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  • 10 years ago

    1) There must be at least two wires connected to an transistor for it to do any effect.

    2) The more power being sent into a field effect transistor the more power it wil send through, usually more than being sent in. Logic transistor on the other hand theoretically only have two states - let trough and dont let trough.

    Source(s): Experience :D
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  • Anonymous
    4 years ago

    The same question pops up again

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