Transistor questions

[apakhira]

I've read many articles on transistors 2 get a solid concept, but the more i read, the more confused i get. So here r a few newbie question which i'd b happy if u'd answer:
1. Talking in terms of conventional current, in a PNP bipolar juntion transistor why doesn't the collector current get out thru the base since the base has lower potential, and instead goes thru the emitter?

2. Why is it said dat the magnitude of current thru the emitter-collector depends on base current and not base voltage (i though and I r interrelated?? OK i know transistors dont obey Ohms law, but its better if u xplained a bit )

3. Do i need a resistor b4 the base if the current is low?

4. In 1 article ( http://www.williamson-labs.com/480_xtor.htm) its said:

Voltage to Current Convertor
First, you must convert the input voltage to a current by using a Voltage to Current Convertor--a resistor.

Current to Voltage Convertor
Next, you convert the output current into a voltage by using a Current to Voltage Convertor in the collector circuit--you guessed it--a resistor.

Wats that supposed 2 mean?

Thanx
Thats all 4 now. I'll add more when they come into my mind.

 

[Papabravo]

1. Using Kirchoff's Current Law (KCL) the sum of currents at a node is zero. If you think of the insides of the transistor as a node the the sum of emitter current, base current, and collector current must be zero. For a PNP tranistor the emitter current has the largest magnitude. The base current and the collector current each have an aritmetic sign which is opposite to the arithmetic sign of the emitter current.

To help understand what is happening with current in a transistor think of a vertical pipe with water flowing into a T-junction(emitter). The diameter of the horizontal section(base) is 1% of the diameter of the input(emitter). A small part of the flow goes out horizontally(base) while the rest continues vertically(collector). This mechanical model does not account for amplification, only how the current is divided up.

2. It's not just said, it's what actually happens. A small change in base current results in large changes in emitter current. This is acomplished by changing the potential barrier seen by the charge carriers in the emitter. When the base current has the transistor in it's cutoff region the potential energy barrier is large and the carriers do not have sufficient energy to climb the barrier and flow down the other side. A small amount of base current will lower the potential barrier so that some of the carriers can get over the potential barrier. A large amount of base current will drive the transistor into saturation where any Tom, Dick, or Harry charge carrier with any energy at all will go sailing over the barrier and down the other side.

3. You need a base resistor whenever the voltage measured between the base and the emitter would be greater than about 0.7V for a silicon transistor or 0.2V for a germanium transistor. The base resistor lets the other side go wherever it wants to go while the voltage called Vbe is at 0.7 Volts or less.

4. It means that if you know either the voltage or the current and the value of the resistor you can compute the other quantity.
Examples of this principle are computing the value of a resistor to set the current of a LED to 10 mA from a 12 VDC source. Another example is the 4-20mA current loop used in process control. A resistor of 250 ohms will convert a 4-20 mA current into a 1VDC to 5VDC voltage.

 

[apakhira]

1. U r explaining Q1 using Kirchhoff's Current Law, coz u "know" that Ie=Ic+Ib, or Ie-Ic-Ib=0. But how can just say that Ic and Ib r -ve? OK, maybe there's some gr8 math deducion nd all, but i wanted 2 know wats happening physically. Based on ur water-pipe analogy, i understand that a small amount does go thru the base. Am i right? Why is it so? Because Vce>Vcb?

2.+3. Fine. i understand. But how do i increase Ieb without increasing Vbc? If i decrease the resistance towards the base 2 increase the current, wont Veb change 2, and get more than 0.7V (I knew current flows between bodies @ diff potentials)?

4. So its just Ohm's law. ok.

 

[Papabravo]

1. In a typical PNP circuit with the transistor in the "ON" state. Let Ve be 12VDC since that is a typical supply voltage. Vb will be at 11.3 VDC or 0.7 volts below Ve and the base emitter junction will be forward biased, and Vbe will be -0.7 volts. Not forwad biased is the same as reverse biased in terms of behavior. A voltage rise has a negative sign.

Vc will be at 11.8 volts for a Vce of -0.2V. This means that Vbc is at -0.5V and thus is not forward biased. Inside the silicon in the absense of potential energy on any of the three leads there are drift currents as the charges move back and forth across the two pn junctions. When voltage is applied the situation changes. Now there are potential barriers to the flow of carriers. In the p-type material the majority carriers are holes. A hole is the absence of an electron in the valence band of the p-type doping material. P-type dopants come from column III of the periodic table. In the n-type material the majority carriers are electrons which come from the valence band of the n-type doping material. N-type dopants come from column V of the periodic table. As the potentials(voltages) on the external nodes change, potential barriers to the flow of carriers(holes and electrons) are raised and lowered. When the potential barriers are raised there is no current flow. When they are lowered current flows more freely, but not as freely as it would in a metal. These are semi-conductors: not insulators and not conductors. So there are actually two currents in a transistor, and they are called the hole current and the electron current and they are in balance. When an electron moves from the valence band of one atom in the silicon crystal lattice it moves into a space that was previously a hole, and it leaves a hole behind. If electrons move to the left, then holes move to the right.

The greek letter beta or the symbol hfe is a number which characterizes how much collector current you get for a given amount of base current. Like all physical properties beta will have a mean and a variance for any given manufactured part.

2. The symbol Ieb has no meaning. Currents only have one subscript, voltages have two. In an amplifier Ib is normally established by a DC biasing circuit. Then an AC voltage is coupled onto the base which increases and decreses Ib. This change in Ib produces corresponding, but much larger changes in Ie and Ic. As long as the transistor remains in its linear region Vbc will increase and decrease according to Ie and Ic. The physics of the pn junction will not allow the magnitude of Vbe to exceed 0.7V If you look at the characteristic curve of a silicon diode you will note that the slope of the I-V curve is nearly vertical at V=0.7V. It means a small change in voltage (millivolts) will result in a large change in current(Amps) unless there is some limiting in the external circuit. Try connecting a really big power diode to a lantern battery and see if you can make the voltage exceed 0.7V

I don't know if all of this is going to be helpful but it's the best I can do in a post. If it works, fine; if not then maybe someone else can take a crack at it.

 

[AllVol]

Good job, Mr. Papabravo

 

[Papabravo]

Thanks, I did put some effort into this one. If anybody is interested I might be persuaded to write a more detailed paper on transistor operations.

 

[Roff]

I only have one comment, and that is the fact that, in a saturated (ON) BJT, the collector-base junction is forward biased. If this were not the case, the collector voltage would be very susceptible to changes in beta (and other temperature-dependent parameters), supply voltage, and load current.
Other than that - nice explanation!

 

[Papabravo]

You're right. When a BJT is in saturation the base collector junction is slightly forward biased, but always less than the threshold of 0.7V. The difference is Vce which can get small but cannot be zero. That was the reason for the statement about slightly forward biased being similar in behavior to reverse biased, in the sense that the slope of the I-V characteristic is reasonably flat as you approach the threshold and changes dramtically once you get there. I apologize for any confusion I may have created.

 

[lord loh.]

1. Talking in terms of conventional current, in a PNP bipolar juntion transistor why doesn't the collector current get out thru the base since the base has lower potential, and instead goes thru the emitter?

The base region is narrow and is the least doped of the three regions. So base does not have much charge carriers. Hence
1. the base current is less.
2. The charge carriers that enter from the Collector does not have any opposite charge carriers to recombine with. Since these have come from another region, they are minority carriers in the base and are swept by the reverse bias field of the base emitter region. Swept from base to emitter.
So a majority of the charge carriers from collector enter the emitter.

3. Do i need a resistor b4 the base if the current is low?

It is good to have one. The base emitter region has a certain voltage rating. If you use a resistor in series with the base, it forms a potential divider. A potential diver of the Base-Emitter diode and your resistor. However in biasing circuits, resistors are not used in series with the base. The chief purpose of a transisitor is to amply signals. So it is biased in the active region with the help of a potential divider network. And the signal is applied to the base. This signal that enters the base in an AC(your signal) riding on a DC(the bias current).

 

[hyedenny]

why doesn't the collector current get out thru the base since the base has lower potential, and instead goes thru the emitter?

In other words, the transistor works (and is manufactured) much the same as a diode - TWO diodes, actually.

As you probably know, an ideal diode allows currrent to flow only in one direction.

 

[apakhira]

So i guess i cud picture the case-emitter region this way:
the base-emitter r like the terminals of a short-circuited cell, but with a little resistor between the two, which keeps the PD across the terminals(Vbe) constant at 0.7V by varying its resistance in response 2 current thru the terminals (which is controlled by the biasing circuit). BUt the resistance, tho affecting PD CAN't affect current thru the terminals (wierd!). Am i right?

Now cud ne1 just tell me y this happens in the be region but doens't in the bc region?

Since these have come from another region, they are minority carriers in the base and are swept by the reverse bias field of the base emitter region. Swept from base to emitter.

"Reverse" w.r.t the base-collector region, right?

In other words, the transistor works (and is manufactured) much the same as a diode - TWO diodes, actually.

As you probably know, an ideal diode allows currrent to flow only in one direction.

Could u really xplain it that way? I mean it xplains why there's very low base-current, but it also means current would not flow even 2 emitter (if u try 2 make it look like 2 real diodes back2back)

 

[lord loh.]

the base-emitter r like the terminals of a short-circuited cell, but with a little resistor between the two, which keeps the PD across the terminals(Vbe) constant at 0.7V by varying its resistance in response 2 current thru the terminals (which is controlled by the biasing circuit). BUt the resistance, tho affecting PD CAN't affect current thru the terminals (wierd!). Am i right?

I am unable to follow what you mean by short sircuited cell... I look at it like a potential divider, where one element(forward biased diode) has a constant voltage drop irrespective of the Vcc or the value of other resistors in the network. This by no means means that you can have Mohms of resistance. You have to bother about the current as well. Over voltage causes breakdown.

Since these have come from another region, they are minority carriers in the base and are swept by the reverse bias field of the base emitter region. Swept from base to emitter.

"Reverse" w.r.t the base-collector region, right?

No. Reverse in it's own absolute sense. Once the electrons are in the base, it has nothing to do with the collector anymore. Since the base emitter region is reverse biased, the charge carriers in this region are swept from base to collector.

Now I would suggest you to play with some simulators... there is one called sWCAD which can be freely opbtained from linear.com

Though some senior members like Nigel Goodwin would frown at the suggestion

 

[apakhira]

I am unable to follow what you mean by short sircuited cell... I look at it like a potential divider, where one element(forward biased diode) has a constant voltage drop irrespective of the Vcc or the value of other resistors in the network. This by no means means that you can have Mohms of resistance. You have to bother about the current as well. Over voltage causes breakdown.

Actually, i was thinking of the forward biased diode as the wire with the small resistor connected 2 terminals of the imaginary cell. The "constant voltage drop irrespective of Vcc" i imagined 2 be due 2 the dynamically varying small resistor, which can change resistance when current increases (4 eg) 2 keep the voltage drop constant, but in the process it doesn't alter the current thru the terminals (coz its a "special" resistor). I guess it makes it more confusing. Ur POV is better . So u look at the base-emitter region as a potential divider consisting of he fw-bias diode and the biasing resistor or resistor network. The PD across this resistor or network is the thing which changes in response 2 change in base current. Now hav i understood?

Since the base emitter region is reverse biased, the charge carriers in this region are swept from base to collector.

I thoght the base emiiter region was fw biased, and the base collector region was reverse-biased? Isn't p connected 2 emitter of a PNP transistor? Pls explain.

 

[lord loh.]

I thoght the base emiiter region was fw biased, and the base collector region was reverse-biased? Isn't p connected 2 emitter of a PNP transistor? Pls explain.

Sorry messed up...

Actually, i was thinking of the forward biased diode as the wire with the small resistor connected 2 terminals of the imaginary cell. The "constant voltage drop irrespective of Vcc" i imagined 2 be due 2 the dynamically varying small resistor, which can change resistance when current increases (4 eg) 2 keep the voltage drop constant, but in the process it doesn't alter the current thru the terminals (coz its a "special" resistor).

Why didn't you think of it as a zener diode instead?

A forward biased diode also lets all current pass irrespective of the Vcc(as long as it is >0.7v) and gives a constant voltage drop of 0.7v (Si).

Everyone requires a different explaination to have the basic transistor jig-saw puzzle to lock into their brains. You have to find your own. A combination of explainations from different people helps to see the same problem from various points of view and understand better. You can do the same by readng the basic transistor theory from several books...

And the forward bias and reverse bias of base collector and emitter regions depend on the mode you want to use it in - Switching / Amplifier.

Are you following the N. N. Bhargava text book?

A few links

http://www.101science.com/transistor.htm
http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/trans2.html#c4 (Good Link)
You might have gone through them ?

 

[apakhira]

i guess i didn't think of the diode explanation u talked about coz i was looking 4 a silicon-less explanation--- I imagined the diode 2 b a sp kind of resistor. But i guess its no use. But y r u suggesting a zener, not a normal 1? 4 the reverse biased section?

Ne way, i have no such thing as a textbook 2 refer 2, coz i'm not doing a course in electronics. In terms of books, i hav my father's old ones, the one i'm using now being "Electronic Engineering" by Charles Alley and Kenneth Atwood. I'm trying 2 combine this with the net info, 2 get a clear picture. And yeah, i had gone thru the 1st and 3rd links u gave. The 1st realy helped. But sometimes, the net info just confuses u more. Take 4 eg:

If (+)voltage is applied to the p-type, to the base wire, while a (-) voltage polarity is applied to the n-type, to the emitter wire, then electrons in the n-type are pushed towards the holes in the p-type. The insulating layer becomes so thin that the clouds of electrons and holes start meeting and combining. A current therefore exists in the base/emitter circuit. But this current is not important to transistor action. What's important to notice is that the *VOLTAGE* across the base/emitter has caused the insulating Depletion Layer to become so thin that the charges can now flow across it.The depletion layer is a voltage-controlled switch which "closes" when the right polarity of voltage is applied.

See? That led 2 1 of the q's i asked b4. But maybe he's just saying, we need a PD across 2 terminals 4 current 2 flow?? What do u say? I'd say, current is more fundamental than voltage. Like, u 2 light a bulb, u need a specified amount of current thru. Put a series resistor in the circuit, and increase the PD 2 compensate 4 the extra voltage drop, so that current remains same, and the bulb'll light up as brightly as b4. The voltage rating in the bulb, is just so that we can put the rated PD across it and not worry about nething else--- otherwise they had 2 specify current requirement and internal resistance. Am i right? (Please, please let me be right this time, i've been wrong every time i wrote that sentence)

btw, wat made u think i was following N.N Bhargava's textbook?

 

[lord loh.]

N. N. Bhargava text book is a basic text book used as one of the first course in electronics in a big part of India...

The "Art of Electronics" book explains transistor with a resistor analogy...

And there is something called "Power".

A bulb requires power... Only current or voltage shall not light it... SO is the case with a loud speaker.

P=IxV

So then comes the problem of impedance matching... Maximum power is delivered when impedances are matched... However the efficiency of a matched impedance only 50%. Half the power is dissipated in the source while half in the load

To draw max current, you should short the power supply. This shall give you a zero volt and hence power 0.

To get max voltage, you have an infinite ohm load... So your current is zero and again power is zero...

If you are not studying electronics as a course subject, are you a hobbyist? What is your age?

 

[apakhira]

Hey, i didn't mean u dont require voltage 2 light a bulb. A PD is always required 2 make a current flow. I said its just a tool 4 making the current flow.
Now Power is the work done in unit time (ie rate of doing work):
P=VI=(W/Q)x(Q/t)=W/t

Like, if u have a turbine in a pipe, and u need a certain pressure of water (which is like electric current, rate of charge flow ) 2 rotate it, if u put a constriction at some point in the pipe b4 the turbine (ie a resistance), u need 2 increase the force of water in the pipe (PD across the load and resistance), so that the turbine experiences same amout of pressure as when there wasn't any constriction. This is what i meant by saying "Put a series resistor in the circuit, and increase the PD 2 compensate 4 the extra voltage drop, so that current remains same, and the bulb'll light up as brightly as b4."

Now, suppose u have a 100W-220V bulb. What i think is that this says, the bulb will deliver 100W power at 220V PD. So I=P/V=100/220A. So at rated Pd, bulb consume 100/220A curr. , which is the rate of charge flow.

I think, the main reason why an electronic gadget works is due 2 electron flow, ie charge flow. More's the charge flown per unit time (ie current), more is the work done per unit time, ie more is the power. Charge flow is due 2 PD, because at higher PD, electron concentration is more, and hence they move 2 lower concentrations 4 reaching equilibrium.
Of course, thats why mathematically, P=VI

Now coming back 2 the bulb, the only thing remaining constant is resistance(assuming temp coefficient of resistance is very low). If at 110V PD, we would get 1/2 the power than b4 (P=110x100/220=50W). But we could keep the power constant if the resistance was decreased 2 (R=V^2/P=110^2/100). So even at reduced PD, power is same,and thats because current remains constant.

How do u like my alternative viewpoint? I rambled a bit, 2 many redundancies i guess. Hope u cud tolerate it.

And yes, i am a hobbyist. I'm in class 11 right now, but please dont disregard me coz "i'm 2 young 4 electronics and stuff" And i will study electronics as a course subject when i graduate from school.

 

[lord loh.]

Yeah... you seem to have understood the current and voltage concept quiet well... I understood it this well in 2nd year B.Sc.

I do not see it as an alternate viewpoint... It is the same as what I was trying to say...but in different words...

And No one shall disregard you here... As long as you are working hard to find your answers, and not solely depending on the forum for spoon feeding, people shall help you....

And which part of India are you from? And are you in CBSE?

11th and 12th is a crutial time... Isn't anyone asking you to concentrate more on your studies?

 

[ljcox]

2. Why is it said dat the magnitude of current thru the emitter-collector depends on base current and not base voltage (i though and I r interrelated?? OK i know transistors dont obey Ohms law, but its better if u xplained a bit )

Certainly Ic = hFE Ib is a useful formula, but if you study how a transistor operates, the collector current is controlled by the base - emitter voltage.

From memory, it is Ic = Io(e^qV/(kT) - 1)

Where Io is a constant, q = the electron charge, V is the base emitter voltage, k is Boltzman's constant and T is the temperature in degrees Kelvin.

 

[apakhira]

@lor.loh:

Yeah... you seem to have understood the current and voltage concept quiet well... I understood it this well in 2nd year B.Sc.

Lucky me! They dont call me the best in Physics in my class 4 nothing (I'm not conceited, it just feels good sometimes)
Yeah, u cud say its not an alternate viewpoint, its just that while writing the other post u thought i believed we dont need both PD and I 4 making a bulb work. I wonder why u ppl use terms like impedance while talking about bulbs and stuff, when there's no reactance involved (i as talking about a bulb in an DC circuit) 2 me ts easier 2 understand with resistance. ( u cud put V=IR anywhere u like if u take resistance )
I'm from Kolkata (WB). And yep, I'm in CBSE (y do u ask?)
Yes again, I'm constantly being told 2 concentrate on studies. Its just that i cant stop "knowing things i want 2". I agree i wud've done better in exams w/o these "distractions". Anyway, i've just completed CBSE class 10 exam, and its my holidays b4 skool starts 4 11. Well, how much did u study during 11-12?

@evry1:

Certainly Ic = hFE Ib is a useful formula, but if you study how a transistor operates, the collector current is controlled by the base - emitter voltage.

From memory, it is Ic = Io(e^qV/(kT) - 1)

Where Io is a constant, q = the electron charge, V is the base emitter voltage, k is Boltzman's constant and T is the temperature in degrees Kelvin.

See? Thats what i was trying 2 speculate somewhere in the thread--u cant hav current w/o PD. But instead of using that super-big formula, can i just do something like this?:
1. Take the value of Ib as required
2. Make a potential divider such that v=Vcc/2 (coz its convenient)
3. Connect the potential divider 2 base with a series resistance Rs such that (Ib*Rs + 0.7)=Vcc/2. Its better if R2 of potential divider be very less, compared 2 Rs, so that potentialdivision does not change much due 2 the load (going 2 base).