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From: Invisible
Subject: Idle curiosity
Date: 15 Aug 2008 10:10:28
Message: <48a58e54$1@news.povray.org>
What *is* the difference between a BJT and a FET?

Are they different designs for doing the same thing, or do they actually 
"do" something different?

How do you use them anyway?

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*


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From: Darren New
Subject: Re: Idle curiosity
Date: 15 Aug 2008 12:04:58
Message: <48a5a92a$1@news.povray.org>
Invisible wrote:
> What *is* the difference between a BJT and a FET?

I think I can describe the difference (a little) with an analogy.

Think of a BJT like two pools of water, with a low spot between, hoses 
connected. To make the water flow, you pump enough water into the low 
spot that the water from the hoses can flow across. Of course, this 
washes away the water you pumped in, so you have to pump in more.

A FET is like a balloon wrapped around a hose. To make the water stop 
running, you pump water into the balloon, which squeezes off the hose. 
But since it's in the balloon, it really doesn't leak as such, so the 
efficiency is much better.

Of course, these are *way* simplified descriptions. If you really care 
how the semiconductors work at the quantum level, I can go into that, 
but it didn't sound like the kind of question you were asking.

Look at the wiki pages for each. In the BJT, the current flows from the 
blue, thru the green, and into the other blue. That's the "bipolar 
junctions", the two boundaries.

In the FET, there's the "gate" and the "oxide" that keeps the gate from 
actually conducting the electricity you put on it. But when you charge 
up the gate, the "static electricity" pulls/pushes away the charges that 
actually conduct between the two "N" pads there. So, FETs are used in 
things like wrist watches, which are expected to run for a year on a 
tiny little battery, because they work on voltage rather than current, 
and batteries are better at supplying voltage than current.

> Are they different designs for doing the same thing, or do they actually 
> "do" something different?

The do "the same thing" in different ways. The BJT is less efficient but 
it can handle much larger currents (i.e., the "hose" can be bigger 
because you don't need to wrap a balloon around it).

> How do you use them anyway?

They're basically amplifiers of electricity. If you want to make binary 
circuits, you over-drive them so they're either completely off or 
completely on.

-- 
Darren New / San Diego, CA, USA (PST)


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From: Orchid XP v8
Subject: Re: Idle curiosity
Date: 16 Aug 2008 05:05:13
Message: <48a69849$1@news.povray.org>
Darren New wrote:

>> Are they different designs for doing the same thing, or do they 
>> actually "do" something different?
> 
> The do "the same thing" in different ways.

OK. So you'd wire them both up in essentially the same way?

Is there any major overall difference in operating characteristics? (I 
seem to recall that whichever one you pick there's about 50,000,000 
models to choose from anyway!)

>> How do you use them anyway?
> 
> They're basically amplifiers of electricity. If you want to make binary 
> circuits, you over-drive them so they're either completely off or 
> completely on.

So why is it that (say) a simple NAND gate involves 8 transistors, 3 
diodes and 25 resistors? That doesn't make any sense to me... Logically 
it looks like any possible 2-input gate should require 1 or 2 switches 
and nothing else.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*


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From: Darren New
Subject: Re: Idle curiosity
Date: 16 Aug 2008 14:06:00
Message: <48a71708@news.povray.org>
Orchid XP v8 wrote:
> OK. So you'd wire them both up in essentially the same way?

You're getting a bit past my expertise, but yes, physically there's 
three connections with the same functions on each. Electrically, of 
course, there's different voltages, switching times, etc.  It's like 
asking if you drive a camper essentially the same way as you drive a VW bug.

Note that you also wire up vacuum tube triods in "essentially the same 
way", so I'm not sure how helpful that answer is. ;-)

> Is there any major overall difference in operating characteristics?

Well, yes. I thought I'd been clear on that. They take different 
voltages and handle different currents and so on.

> So why is it that (say) a simple NAND gate involves 8 transistors, 3 
> diodes and 25 resistors? That doesn't make any sense to me... Logically 
> it looks like any possible 2-input gate should require 1 or 2 switches 
> and nothing else.

To make a switch, you have to either switch to + or to -, so that's at 
least two switches right there. I.e., you need two transistors, one to 
amplify the + voltage, and one to amplify the - voltage, using CMOS FET, 
just to make a NOT gate.  Depending how you do it, it may take diodes 
and resistors and such.

-- 
Darren New / San Diego, CA, USA (PST)


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From: scott
Subject: Re: Idle curiosity
Date: 18 Aug 2008 03:57:27
Message: <48a92b67@news.povray.org>
> OK. So you'd wire them both up in essentially the same way?

Essentially, yes, but you need to bear in mind the differences between the 
two when designing your circuit.  The main difference is that a FET is 
controlled by input voltage, and the BJT by input current.  So in most 
designs with BJTs you'll see some resistors dotted around to control the 
current.

> Is there any major overall difference in operating characteristics? (I 
> seem to recall that whichever one you pick there's about 50,000,000 models 
> to choose from anyway!)

Good point about FET is that they have almost infinite input impedance, 
which is often useful for low-power operation or when measuring very low 
powered signals.  Good point about BJT is that the output resistance is 
higher, so good for driving higher powered devices rather than pure digital 
signals.

> So why is it that (say) a simple NAND gate involves 8 transistors, 3 
> diodes and 25 resistors? That doesn't make any sense to me... Logically it 
> looks like any possible 2-input gate should require 1 or 2 switches and 
> nothing else.

A NAND gate needs 4 FETS and no resistors:

http://en.wikipedia.org/wiki/Image:CMOS_NAND.svg

Inside an IC, resistors are HUGE compared to FETs, there is no way you could 
make a modern CPU if you needed to include resistors for each gate.


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From: Invisible
Subject: Re: Idle curiosity
Date: 18 Aug 2008 05:09:59
Message: <48a93c67$1@news.povray.org>
>> OK. So you'd wire them both up in essentially the same way?
> 
> Essentially, yes, but you need to bear in mind the differences between 
> the two when designing your circuit.  The main difference is that a FET 
> is controlled by input voltage, and the BJT by input current.  So in 
> most designs with BJTs you'll see some resistors dotted around to 
> control the current.

See, in theory the way a transistor works is supposed to be simple. 
However, every transistor circuit I've ever looked at seems so damned 
complicated...

> Good point about FET is that they have almost infinite input impedance, 
> which is often useful for low-power operation or when measuring very low 
> powered signals.  Good point about BJT is that the output resistance is 
> higher, so good for driving higher powered devices rather than pure 
> digital signals.

...and this is why I like to stick to logic circuits. You don't have to 
worry about obscure concepts like impedance or capacitance. (I mean, 
unless you're designing "high performance" circuits anyway.)

>> So why is it that (say) a simple NAND gate involves 8 transistors, 3 
>> diodes and 25 resistors? That doesn't make any sense to me... 
>> Logically it looks like any possible 2-input gate should require 1 or 
>> 2 switches and nothing else.
> 
> A NAND gate needs 4 FETS and no resistors:
> 
> http://en.wikipedia.org/wiki/Image:CMOS_NAND.svg
> 
> Inside an IC, resistors are HUGE compared to FETs, there is no way you 
> could make a modern CPU if you needed to include resistors for each gate.

Even so, you can make a NAND gate out of just 2 electromechanical 
relays. Why does it require twice that number when you use transistors? 
(Additionally, I can't make any sense out of that diagram. The wiring 
seems to have no reason or rhyme to it.)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*


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From: scott
Subject: Re: Idle curiosity
Date: 18 Aug 2008 05:38:14
Message: <48a94306$1@news.povray.org>
> See, in theory the way a transistor works is supposed to be simple. 
> However, every transistor circuit I've ever looked at seems so damned 
> complicated...

You mean something like this - a simple amplifier circuit?

http://www.cjseymour.plus.com/layin/BJTAMP.jpg

The reason it looks complicated is because you are using a 
current-amplifying device (the BJT) to amplify voltage.

The two resistors on the left are to bias the input voltage to somewhere in 
the middle of the operating range of the transistor.  Otherwise you cannot 
amplify negative input voltages (quite common in audio signals...).

The capacitor on the left allows the input voltage to be added to the bias 
voltage.

Resistor R3 causes the amount of current flowing out of the emitter to be 
(almost) proportional to the input (base) voltage.  Capacitor C2 helps to 
boost the gain by allowing AC signals to see a lower resistance to ground, 
but still keeping the DC operating point.

Resistor R4 generates the output voltage proportional to the amplified 
current flowing through the collector/emitter.  Capacitor C3 removes any DC 
component of the signal.

Note that the circuit inverts as well as amplifying...

> ...and this is why I like to stick to logic circuits. You don't have to 
> worry about obscure concepts like impedance or capacitance. (I mean, 
> unless you're designing "high performance" circuits anyway.)

Even with logic circuits, you cannot connect a single output to 5000 inputs, 
there is always a limit due to internal resistances.  At some point your 
circuit is going to fail, and for some technologies the number is much lower 
than 5000.

> Even so, you can make a NAND gate out of just 2 electromechanical relays.

How?

> Why does it require twice that number when you use transistors? 
> (Additionally, I can't make any sense out of that diagram. The wiring 
> seems to have no reason or rhyme to it.)

Out is either connected to Vss (low) or Vdd (high) depending on the inputs A 
and B.  If A or B or both are low, then Out gets connected to Vdd because 
one (or both) of the top FETs will start conducting.  Only if both A and B 
are high will there be a link between Out and Vss.


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From: Invisible
Subject: Re: Idle curiosity
Date: 18 Aug 2008 06:34:53
Message: <48a9504d$1@news.povray.org>
>> See, in theory the way a transistor works is supposed to be simple. 
>> However, every transistor circuit I've ever looked at seems so damned 
>> complicated...
> 
> You mean something like this - a simple amplifier circuit?
> 
> http://www.cjseymour.plus.com/layin/BJTAMP.jpg

Yeah, pretty much.

The books say "a transistor is an amplifier" - and then should you a 
*working* amplifier which actually consists of almost a dozen 
components, of which the transistor is only one.

> The reason it looks complicated is because you are using a 
> current-amplifying device (the BJT) to amplify voltage.

Uh... does it matter? I mean, the two are inherantly linked anyway...

>> ...and this is why I like to stick to logic circuits. You don't have 
>> to worry about obscure concepts like impedance or capacitance. (I 
>> mean, unless you're designing "high performance" circuits anyway.)
> 
> Even with logic circuits, you cannot connect a single output to 5000 
> inputs, there is always a limit due to internal resistances.  At some 
> point your circuit is going to fail, and for some technologies the 
> number is much lower than 5000.

Yeah, but I'm never likely to build anything that even *contains* 5,000 
components in total, am I? ;-)

>> Even so, you can make a NAND gate out of just 2 electromechanical relays.
> 
> How?

Let's assume that a positive potential represents logic high, and a zero 
potential represents logic low. Take a pair of relays connected up so 
that by default the circuit is closed, and energising the magnet opens 
the circuit. (This is easy to arrange with 3-pin switch type relays.) 
Connect the relays in parallel between the positive rail and the signal 
output terminal. Connect each input to one side of one magnet coil, and 
connect the other sides to the negative rail. Done.

>> Why does it require twice that number when you use transistors? 
>> (Additionally, I can't make any sense out of that diagram. The wiring 
>> seems to have no reason or rhyme to it.)
> 
> Out is either connected to Vss (low) or Vdd (high) depending on the 
> inputs A and B.  If A or B or both are low, then Out gets connected to 
> Vdd because one (or both) of the top FETs will start conducting.  Only 
> if both A and B are high will there be a link between Out and Vss.

It probably doesn't help that I'm not understanding the symbols in the 
diagram, but OK...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*


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From: scott
Subject: Re: Idle curiosity
Date: 18 Aug 2008 07:06:14
Message: <48a957a6@news.povray.org>
>> The reason it looks complicated is because you are using a 
>> current-amplifying device (the BJT) to amplify voltage.
>
> Uh... does it matter?

Well yeh, that's why the resistors are needed.  Generally you want to 
amplify voltages, not currents.

> Let's assume that a positive potential represents logic high, and a zero 
> potential represents logic low. Take a pair of relays connected up so that 
> by default the circuit is closed, and energising the magnet opens the 
> circuit. (This is easy to arrange with 3-pin switch type relays.) Connect 
> the relays in parallel between the positive rail and the signal output 
> terminal. Connect each input to one side of one magnet coil, and connect 
> the other sides to the negative rail. Done.

I don't understand what is connected to the negative rail?

Don't forget that the output must be connected to either the negative or 
positive rail (and not both!) under all input conditions.  It sounds to me 
like in your circuit, the output will just be floating (ie not connected to 
anything) when you charge both relays.


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From: Invisible
Subject: Re: Idle curiosity
Date: 18 Aug 2008 07:22:47
Message: <48a95b87$1@news.povray.org>
>>> The reason it looks complicated is because you are using a 
>>> current-amplifying device (the BJT) to amplify voltage.
>>
>> Uh... does it matter?
> 
> Well yeh, that's why the resistors are needed.  Generally you want to 
> amplify voltages, not currents.

...because you don't want to actually push too many electrons around and 
warm up the wiring?

>> Let's assume that a positive potential represents logic high, and a 
>> zero potential represents logic low. Take a pair of relays connected 
>> up so that by default the circuit is closed, and energising the magnet 
>> opens the circuit. (This is easy to arrange with 3-pin switch type 
>> relays.) Connect the relays in parallel between the positive rail and 
>> the signal output terminal. Connect each input to one side of one 
>> magnet coil, and connect the other sides to the negative rail. Done.
> 
> I don't understand what is connected to the negative rail?
> 
> Don't forget that the output must be connected to either the negative or 
> positive rail (and not both!) under all input conditions.  It sounds to 
> me like in your circuit, the output will just be floating (ie not 
> connected to anything) when you charge both relays.

Ah, OK. If you want bipolar then yes, you'll need more relays. I'm used 
to playing with TTL circuits where if you connect it to V+, that's 1, 
and if you connect it to thin air, that's 0. (After all, a digital 
circuit is just a circuit that has two possible states. It doesn't 
matter especially what those states are, so long as everybody agrees...)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*


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