Transistors

Hi,

I am new to electronics and working with transistors for the first time. I chose 2N4401transistor from Fairchild to begin with. You can find the data sheet at

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I have following questions

  1. The base current Ib controls the collector current Ic. But I am unable to find the base current from the data sheet.

  1. I am also unable to find the maximum and minimum values of Ib too.

  2. Is it ok to determine Ic and then calculate Ib using Ic = beta x Ic.

  1. What should I consider to do regarding DC biasing of the transistor before applying the AC signal to the transistor?

Please advice! John

Reply to
john
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This should probably be sent to sci.electronics.basics. Follow ups to my response are directed there.

The base current will depend on what you are trying to do, the topology used, the BJT of course, and upon the Ic you expect to name a few things.

... why do you want them?

I assume you mean Ic = beta * Ib.

Yes, in some arrangements. If you are using the BJT as a switch, then probably 'no.' It all depends.

But it looks, below, as though you are thinking about an AC signal input and a voltage amplifier stage. Not sure, though. But in such cases, other considerations usually dominate your thoughts, and not so much the beta of the transistor or the exact base current which varies from one to another even within a part number, let alone across BJT families.

Ah. So this is for simple voltage amplification of an AC signal? What are your design constraints (signal, input source, output, etc) and chosen topology, so far?

Jon

Reply to
Jonathan Kirwan

  1. The base current is often determined by a bias circuit that consists of a voltage divider plus some feedback.
  2. Ib MAX is usually not a factor (of course, you can blow the base junction).
  3. BIb = Ic
  4. The center of the load line is the starting point (that gives the best output swing before clipping).

I'll suggest a basic textbook on linear electronics. Malvino is readable (Electronic Principles).

Reply to
Charles

The whole point of transistors is that Ic = Beta*Ib and that you can "program" the base current.

That is, you have control of what to put in for Ib and hence you get a much larger Ic out. Ie = Ib + Ic and Ic = Beta*Ib.

Usually this is done by creating a current source with a voltage and resistor.

V ----R----Trans Base

Then you know Ib = (V - (VE + 0.6))/R

(VB is a diode drop above VE)

If VE is grounded then Ib = V/R and you can easily calculate your base current.

If Ib is to large then Ic != Beta*Ib.

In generally you do not design circuits that depend specifically on Beta...

You basic amplifier has a formula like Beta/(Beta + 1) and for Beta = 50 its

98% and Beta = 200 its 99.5%. So any drastic change in Beta doesn't really effect the result(unless Beta drops to low).

Beta depends on a lot of factors but mainly temperature. This is why its bad to use Beta as a parameter. In transistor switches you just have to saturate the base and you can use minimum value of beta to do that. In the case of switches Ic != Beta*Ib... or at least beta is not constant.

Generally you have to abide by the total power dissipation. Since Ie = Ib + Ic and Ib < Ic with Ib's max rating being comparible to Ic you usually don't have to worry about it... as long as your not running the transistor near is maximum ratings in the first place.

No, as I said before, Beta is not a stable parameter. You can use it but in fact you probably don't know what it is(Look at h_FE in the datasheet and see its range... for the 3904 it ranges from 30 to 300).

Again, circuits are usually designed to use feedback to that Beta is not important... this is a good thing.

Look up load lines. You bias the transistor for maximum AC Swing(well, close to maximum) in a class A amplifier and maximum DC stability. That is, you find the Q point and it is the midpoint on the load line.

Doing a quick search gives,

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which probably can explain thigns better than I can.

Reply to
Jon Slaughter

Approximately zero, up to 200mA (or whatever the maximum is rated for). The practical minimum is in the nanoampere range.

Base current is given on the Ice vs. Vce, Ibe graph and similar plots. Also hFE vs. Ic.

Tim

-- Deep Fryer: A very philosophical monk. Website @

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Reply to
Tim Williams

On a sunny day (Mon, 12 Nov 2007 13:18:52 -0800) it happened john wrote in :

Well, some can be found from the graph on page 3, 'typical pulsed current gain versus collector current.' For example at Ic 100mA at -40C the beta is about 110, so the Ib was .9 mA.

Yes, but beta depends strongly on Ic and temp, as that graph shows.

The idea is to apply an AC CURRENT within the operating DC bias current.

The operating point needs to be stabilised too, there are 2 main ways for common emittor: ----------------------- + | | R1 R3 | |-------------- out | c |--- b - C1-| e | |--- R5 --- | | | R2 R4 C2 | | |

------------------------------ 0V

In this case the divider R2 / R1 + R2 sets the base voltage, and the DC emiter current is about Ur2 - 0.7 / R4 The collector current is about the same.

For AC you can see that a 1V AC signal does not only see R4, abut als the R5 C2 parallel to R4, so for AC there is a bigger emitter current change (for the same emittor voltage change), resulting in a higher gain. So R5 C2 sets the AC gain basically, together with R3. The voltage on R3 is Ic x R3. The input impedance is R1 parallel with R2 parallel with the network (R4 parallel with R5 + C2) x beta. So what is in the emittor appears beta x higher in the base.

A simpler DC biasing system could look like this: +Ub | R2 |------ out --R1-| | | | c

-- C1-----b e |

-------------------

In this case the ACinput impedance in higly non-linear due to the be junction, but if driven from a high impedance source that non-linear effect gets much smaller.

For max outpu swing Vce should be about 1/2 Ub, so Ur1 is 1/2 Ub - Vbe or about 1/2Ub - 0.7. You know IR2(=Ic) for 1.2 Ub drop, you know beta, you then know Ib. and can calculate R1. You could improve linearity by adding an emittor resistor.

Reply to
Jan Panteltje

On a sunny day (Mon, 12 Nov 2007 13:18:52 -0800) it happened john wrote in :

The idea is to apply an AC CURRENT within the operating DC bias current.

The operating point needs to be stabilised too, there are 2 main ways for common emittor: ----------------------- Ub | | R1 R3 | |-------------- out | c |--- b - C1-| e | |--- R5 --- | | | R2 R4 C2 | | |

------------------------------ 0V

PS: In this circuit you can say that approx, for frequencies where C1 and C2 are low impedance: The voltage gain is R3 / R5. The output impedance is R3. The input impedance is beta * R5 in parallel with R1 and R2. The output is 180 degrees out of phase with the input.

Reply to
Jan Panteltje

The base-emitter junction behaves much like a diode. It conducts very little current below the turn-on voltage, then conducts rapidly above that point.

The maximum base current isn't normally a factor. In a typical circuit, the base current will never get anywhere near the rated maximum value.

It depends. The main problem is that you often only have a vague idea of what beta is likely to be, as it varies both between individual transistors of the same type and with temperature.

For a simple class A amplifier, it's often preferable to use a resistor between the emitter and ground to fix the gain via feedback. When the transistor is conducting, Vbe is roughly constant, so the emitter voltage will follow the base voltage. The emitter current (and hence the collector current) will be the base voltage (minus the constant turn-on voltage) divided by the value of the emitter resistor, and independent of beta.

E.g. if you have 1K between the emitter and ground and 10K between the collector and the positive rail, you get a 10x voltage gain. A 100mV change in the base voltage causes a 100mV change in the emitter voltage, a

100uA change in emitter current, a 100uA change in collector current and a 1V change in the collector voltage.

For a class A amplifier, you normally want the output to lie in the middle of the linear region in the absence of a signal.

Reply to
Nobody

That's because setting the base current is your job (or your circuit)

Maximum Ib is way over maximum Ic / HFE. So you wouldn't find it in a data sheet for an amplifier transistor.

Yes and no. Yes, because it's the Ic you're aiming for. No, because you can never trust beta -- expect it to vary over a factor of two range over temperature, another factor of two range with collector current, and yet another factor of two range over the span of all available parts*.

Part of transistor design is making circuits that work well in spite of variations of Hfe.

As recommended by others, I think you need to find a good book on basic transistor circuit design. Answering all the questions you need to ask will take _way_ more than a page.

  • That's too many factors of two, but you get the idea...
--
Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Or maybe less- perhaps minus a bit if you don't mind some beta degradation. ;-)

Note that that when Vce gets relatively low (say less than a few hundred millivolts) Ib doesn't have as great an effect on Ic.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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Reply to
Spehro Pefhany

You supply it. A transistor is a device for letting one current control another.

Hmmm, they should specify it. Minimum is obviously 0. In practice you're not going to let it go above about 2 mA.

It would be in theory. In practice, beta varies tremendously from unit to unit.

Hmmm... Before doing anything else, find an introductory electronics textbook and read about transistor circuits. The data sheet is NOT enough to tell you how to use them.

Reply to
mc

I would recommend backing up just a wee bit further... say to loop and nodal analysis of passive circuits.

Once you can manage that use of Algebra you can easily add active components.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

It's usually better thought of as using a voltage (Vbe) to control a current, with some current required to set a given Vbe.

That's why beta isn't a good parameter--for a given circuit, you need it to have some minimum value, but building a circuit that depends on a specific value of beta is A Bad Idea. Dividing Ic by Ib to get beta doesn't mean that beta is constant--it really depends on both Ic and Vce. Some transistors have beta constant to 10% over 3 decades of Ic, but some have beta varying 3:1 over a single decade. It really really isn't a constant.

Transconductance is a much better parameter--there's some variation between devices and with temperature, but basically, every bipolar transistor follows the Ebers-Moll relation very closely over a huge range of collector current. Two completely different transistors, even made of different semiconductors, will have the same curve of Ic vs Vbe, apart from a single multiplicative constant. You can't say that about Ic vs Ib--not even close.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Only physicists think that way ;-)

[snip]

Learn to bias with miniscule sensitivity to beta.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Your first three words, I'm tempted to agree with ;)

Sure. In the application circuits world we typically do that with emitter degeneration or N*Vbe biasing. Emitter degeneration is simple and robust, but there are an awful lot of people who try putting a 1M pot from the base to the positive supply, and adjusting Ib to set the bias point. That would be very reasonable with a truly current controlled device, such as a current mirror, but not in with a voltage-controlled device like a BJT.

I still want a T shirt that says "Ebers and Moll are my homeboys".

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Hey, it works in Spice!

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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Reply to
Spehro Pefhany

Makes a great temperature sensor, too.

Always fun to set up a differential amplifier on the breadboard, open loop, gain circa 80dB, and touch one of the differential transistors. Weee, the output's saturating again...

Tim

-- Deep Fryer: A very philosophical monk. Website @

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Reply to
Tim Williams

Here's something simple you can do to get a feel for the situation. Assuming you have a 9v or 12V DC supply, and your transistor is NPN small case:

Emitter to V- through a 1K resistor. Collector to V+ through a 1K resistor. Base to V- through a 3K3 resistor. Base to V+ through a 10K resistor.

Attach the battery and measure some voltages.

The key voltage is between the base and emitter. Base will be roughly 0.6V positive above the emitter.

*Rule 1: Whenever the transistor is 'working' (let's not worry about exactly what this means) this voltage, Vbe, will be around 0.6 volts.

*Rule 2. Whenever the transistor is 'working' collector current Ic = base current * 'beta', where beta will be somewhere between 50 and a few hundred, and will vary a lot from device to device.

Now measuring voltages from V-.

Emitter will be at about +2V. Collector will be at about 2 volts less than the supply (+7V?).

How should you view this?

You've fixed the base voltage at about 2.5V with the 10K/3K3 divider (1/4 supply).

So by Rule 1, the emitter voltage Ve will be about 0.6 volts less, or about

2 volts. Or IOW, the voltage across the emitter resistor is 2 volts.

So by Ohms law, the current through the emitter resistor is V/R = 2volts /

1K = 2mA.

Now the collector current Ic = emitter current - base current. BUT by rule

2, since beta is 'big', we can say that the collector current roughly equals the emitter current within a percent or two.

So the voltage across the collector resistor, by Ohms law, will be Ic * 1K =

2 volts.

Note that nowhere here did you consider the base current at all. Rather you set up a system where the transistor took the base current that it needs to satisfy the above rules itself. A great deal of bipolar circuit analysis uses these principles. It's worth taking some time to get the concepts clear.

Good luck

Reply to
Bruce Varley

Hey! This is SED--no fair being helpful!

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

This is a nice little test to give to applicants for tech or engineering positions:

+10V | | | c +5V--------b e | | | 1K | | gnd

It's an average NPN transistor.

What's the emitter voltage? Current?

What's the base voltage?

What's the base current?

What's the collector current?

Will anything get hot?

Any other comments?

Some hot-shot resumes have been deflated by this The base voltage is 0.6

The transistor is saturated, so collector voltage is zero.

The transistor is saturated, so emitter voltage is +10.

I haven't done this in a while, so I don't remember the formulas.

And other weird stuff.

Pitiful.

John

Reply to
John Larkin

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