transistor output characteristic, question on load resistor

as seen in picture below, what's the purpose of R2, since Ib dictate Ic, why do we still need R2 anyway?

If the print is too small to read,

A better resolution pic can be seen here

or

I1 = constant 10uA, 20uA, 30uA
R1 = 1kΩ
R2 = 100Ω
Q1 = 2N3055
V2 = 0v to 2v step 0.0001v

 

I'm essentially a transistor tyro yet, but...

The Ib current does control Ic, but only during active mode of the transistor. Once Ib pushes the transistor closer to saturation, there is very little resistance in the CE circuit, so we need something to keep the current down. Your simulation is only including Ib's within the active range of the transistor (I assume).

Since the transient in this simulation is V2, going from 0 to 2 volts is forward biasing the CB, which allows the current to flow, hence the Ic growing from 0 to that which the Ib current stipulates the Ic current.

Typically for transistors, you'd be looking at Ib to be transient, and Ic would then vary according to Ib, until the transistor is saturated, then Ic is almost completely dependent on the load resistance.

I'm answering this because I struggled with the idea myself not terribly long ago.

 

Thanks sbixby, your explanation help me out a lot, now I can understand the purpose of the collector resistor!!

So how do I know when a transistor operates in cut-off, saturation or active mode?

Is it controlled by the Ib?

 

Not to differ from WBahn's reply in any way, but from my own perspective of "still looking for clarity", I would answer this as:

The simplest description of cutoff->active->saturation occurs when you can ignore the CE side of the transistor. That is, CE is sufficiently forward biased so that Ic is completely dependent on Ib. This is what we would look at when trying to understand the basic response of a transistor.

In this case:

Ib=0 means Ic=0. (Cutoff)

Ib increasing means that Ic also increases at a ratio of the transistor's Beta. (Active)

At some point, however, increasing Ib no longer has a proportional increase in Ic - that's the "knee" of our Ib/Ie curve - and the transistor goes into saturation mode. This isn't a hard change-over, there's a bit of curvature to the response, so "saturation" isn't an exact Ib value.

Once in saturation, you can continue to increase Ib, but you will no longer see an increase in Ic. (Actually, you will, but that's leakage inherent in semiconductor devices and it's such a tiny increase that it's generally ignored when doing basic analysis and design.)

The final part of this is that every transistor is different in it's response. You need to look at the transistor's data sheet (or lab test it) to have an estimate* of what values of Ib will put the transistor in cutoff, active, and saturation.

(*I say estimate because transistors vary wildly, within and without their own family, so the datasheet is a guideline and designers strive to leave sufficient leeway within that guideline so that replacing a transistor with another one will not significantly affect the operation of the circuit.)

(OK, I'm done pretending I have a clue, for this post. )

 

Hi sbixby, your explanation is actually very easy to understand, thanks

 

As of point 2, I am not quite understand, do you get the saturation voltage from a datasheet? and I think the saturation voltage is different if the Ib is different right?

as of the saturation voltage, do you mean the Vce(sat) here (bc547 datasheet, Page 3)

but I don't seen to understand

 

funny enough, after I read here (another thread in this forum, P2-P3), I understand your explanation now.

even I don't understand 98% of what they are talking about, one of my nerve in my brain must be connected at that point when I read it

now I need to try to understand your last post on how to analyse it Thanks WBahn

 

thanks everyone's input, after you guys' explaination and some simulation on LTSpice, and some breadboarding, I understand BJT transistor a lot better now, appreciated!