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While plotting the output characteristics of a CE configuration, the graph between VCE and IC, it shows that when VCE=0,IC=0 My doubt is whether VCE=0 means we are shorting collector and emitter or is it like from the circuit, VCE=VCC-ICRC . So when VCE=0, it should be like IC=VCC/RC.

NPN transistor, common emitter configuration

output characteristic curve

Source of images: Circuit Globe - Common Emitter Connection (or CE Configuration)

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  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – MiNiMe
    Commented Oct 20, 2023 at 6:15
  • \$\begingroup\$ Vce=0v is unreachable, there is always some drop on transistor CE structure. \$\endgroup\$ Commented Oct 20, 2023 at 7:01
  • \$\begingroup\$ Btw, the output curves does not care about the circuit you connect around. The outer circuit is just the approach how to get transistor to point you want. \$\endgroup\$ Commented Oct 20, 2023 at 7:11
  • \$\begingroup\$ Of course, the condition Vce=0 is "reachable", when for Vcc=0 (and Rc=0). \$\endgroup\$
    – LvW
    Commented Oct 20, 2023 at 7:25
  • \$\begingroup\$ Your circuit is not the way to measure values for the graph. Additionally, that circuit is flawed in that it indicates the output current as a potential difference. \$\endgroup\$
    – Andy aka
    Commented Oct 20, 2023 at 9:23

5 Answers 5

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While plotting the output characteristics of a CE configuration, the graph between VCE and IC, it shows that when VCE=0,IC=0 My doubt is whether VCE=0 means we are shorting collector and emitter or is it like from the circuit, VCE=VCC-ICRC .

I have to say that it was circuit diagrams like the one you first show (not the curves, but the diagram) that drove me crazy trying to understand BJTs many many decades ago. I think I understand BJTs for at least some practical purposes, but your circuit still makes me feel lost when also looking at those curves.

For example, there's \$V_{_\text{BB}}\$ showing there and something called a signal in series with it. The signal will add to \$V_{_\text{BB}}\$ and therefore change the base to emitter voltage. But the curves are for fixed base currents. So how you are you supposed to relate the schematic to the chart? There's a subtle and backing-into-it way, but it sure isn't obvious. I think it is almost designed to confuse, since a student likely expects that the circuit and its signal somehow relates to that chart in a direct (not indirect) way.

(Well, except for the occasional super-genius student who, of course, would not only understand it fully but would then use it to prove as a necessary consequence that the Hubble radius of the universe has to be exactly congruent with the idea that we live inside a black hole.)

So when VCE=0, it should be like IC=VCC/RC.

Let's stay with the circuit you have and not the circuit you don't have. It can be drawn also this way:

schematic

simulate this circuit – Schematic created using CircuitLab

That's the same thing. Just a little less complicated-looking. So let's look at it.

First off, it appears that a voltage is used to drive \$V_{_\text{BE}}\$. Not a current. Yes, the schematic does add a base current there. But it doesn't say that the base current is driving things. And as it turns out, you don't get to specify both the voltage difference and also the current. You can specify one, or the other. But not both at once. So which is it??? What do they intend? And how does this relate to the curves they plotted out?

Let me just tell you something straight out that will take a moment to absorb. But once you absorb it, it will explain a great many different situations. \$V_{_\text{BE}}\$ does, in fact, drive \$I_{_\text{B}}\$. Or, conversely, \$I_{_\text{B}}\$ determines \$V_{_\text{BE}}\$. You can arrange one, or the other, but not both at once. There's even a simple equation that expresses this relationship:

$$I_{_\text{B}}=\frac{I_{_\text{SAT}}}{\beta_{_\text{F}}} \cdot\left[\exp\left(\frac{V_{_\text{BE}}}{\eta\,V_T}\right)-1\right]$$

\$I_{_\text{SAT}}\$ and \$\eta\$ are relatively useful characteristics of some specific BJT. You can work them out if someone hands you a BJT and you perform some tests on it -- particularly when those tests are performed near where you expect to operate the BJT.

\$\beta_{_\text{F}}\$ is only useful when the BJT is in active mode. And in that case, it is another useful characteristic of a specific BJT near some specific operating point. But most circuit designs try to avoid too much dependence upon it, because it varies a lot from one BJT to another.

The above is independent of whether or not the BJT is saturated or in active mode. And it is independent of the Early Effect, as well. (An odd thing to learn about later which increases the collector current slightly with increasing \$V_{_\text{CE}}\$.) Both saturation and also the Early Effect only impact the collector current. Not so much the base current, though.

Note that the equation doesn't relate the base current to the collector current. When in active mode, there is a relatively constant relationship between these two, \$\beta_{_\text{F}}\$. But in saturation you ignore that value because the entire circuit is analyzed differently then.

So what's going on with the charted curves then??

Well, let's modify the circuit and run things in a simulator for a moment.

Here's the simulator schematic:

enter image description here

And here's the plot with the x-axis being \$V_{_\text{CE}}\$:

enter image description here

Note that in cutoff, with \$V_{_\text{CE}}=1\:\text{V}\$, the collector current is close to zero -- only a few tens of picoamps! The curve you have shows far too much collector current for the BJTs found today.

In the above, I've pretty much removed the Early Effect by setting VAF to 1000. So the curves are fairly flat-looking and close to your curves in that sense.

Normally, that's not true. It's more like this:

enter image description here

Notice the tilt?? This is the Early Effect as it impacts the collector current. (But not the base current!)

If you look very closely at these curves, you will see that the saturation point only occurs when \$V_{_\text{CE}}\le 200\:\text{mV}\$, or so. What causes this, in this case, is the voltage drop across the collector resistor.

Normally, in active mode, the collector current times the collector resistor (which is the voltage drop across the collector resistor) subtracts from \$V_{_\text{CC}}\$ in order to create \$V_{_\text{CE}}=V_{_\text{CC}}-I_{_\text{C}}\cdot R_{_\text{C}}\$. As the collector current increases, there comes a point where \$V_{_\text{CE}}=V_{_\text{BE}}\$. This is the point where saturation starts to occur with your schematic. And when \$V_{_\text{CE}}\lt V_{_\text{BE}}\$ then you are just starting into saturation. Full saturation tends to occur when \$V_{_\text{BE}}-V_{_\text{CE}}\ge 500\:\text{mV}\$. This is about when the base-collector junction is enough forward-biased to start to dominate the behavior.

It almost never happens that \$V_{_\text{CE}}=0\:\text{V}\$, though, in a circuit like this. The base current would need to tend towards infinity for that. And in practical circuits, that's not possible. But it can get very close.

That said, if you remove the collector resistor and directly drive \$V_{_\text{CE}}=0\:\text{V}\$ using a voltage source, then you can make it happen. But here the collector current will be less than the base current. But you need to know \$\beta_{_\text{R}}\$, as well, to work out exactly how much less.

All of the above isn't really going to help much, though. I admit it. What you should focus on, instead, is working through a lot of different BJT schematics and seeing how close you can come to working out their DC quiescent points. This is the practical experience you need to start out with. Not all the above stuff. So get out some different circuit examples and practice with them. See how well you can work out their DC quiescent values -- collector current, etc.

Keep in mind that the base-emitter voltage will be about \$60\:\text{mV}\$ higher if the collector current is \$10\times\$ more. That's about how things work out for a small signal BJT operating at room temperature. Figure \$700\:\text{mV}\$ for the base-emitter when the collector current is \$2\:\text{mA}\$. But if it is \$20\:\text{mA}\$ then the base emitter voltage is likely to be \$760\:\text{mV}\$. Like that.

That's about it for now.

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    \$\begingroup\$ I'd like to write nice things about your answer (other than just +1) but it's not tolerated here, and every time I do I get in trouble. The OP's schematic is just awful. I, like you, tried unsuccessfully (as a student and later) to understand what these clumsily drawn schematics were about, but apparently this was not clear to the "authors" (reproducers) themselves. So, with much thought and experimentation over the years, I gradually understood the meaning of it all. But I have moved so far from the usual presentation that I am looked at as an alien when I explain what it is about... \$\endgroup\$ Commented Oct 20, 2023 at 17:09
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It seems that you didn't notice that when Ic=0 and V(CE)=0 then I(b) is also zero which means that V(BE) is not greater than 0.7 volts so there is no I(b). More the value of Ib/VBE more will be the value of Ic in the graph.

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  • \$\begingroup\$ No - Vce=0 and Ic=0 does not automatically mean that also Ib=0. of course, you can bias the BE-junction in this case. \$\endgroup\$
    – LvW
    Commented Oct 20, 2023 at 7:35
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The output characteristics are a set of curves obtained by recording collector current IC for different collector-emitter voltage VCE levels whilst keeping base current IB constant.

  • Start with IB = 0 and VCE = 0. Record the collector current. At every step increase VCE by some amount and record IC again. This goes until VCE is close enough to breakdown voltage, VCEO. This collector current level is called "cut-off current" or simply, leakage, BTW. And cannot be zero, practically.
  • Repeat the same process for several different IB levels.

My doubt is whether VCE=0 means we are shorting collector and emitter or is it like from the circuit, VCE=VCC-ICRC . So when VCE=0, it should be like IC=VCC/RC.

There's no need to use a collector resistor to prepare output characteristics because all you do is to play with forced collector-emitter voltage (short RC in your diagram and play with VCC). You can create your own output characteristics using simulators.

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  • Please note that the output characteristic Ic=f(Vce) applies to the transistor alone (without Rc).
  • And yes - Vce=0 is identical to a short between collector and emitter.
  • However, in this case (Vce=0) the collector current Ic will NOT be zero. In this case, the base-collector junction (in addition to the base-emitter junction) will be forward biased and there will be a current through the collector node (in opposite direction to the "normal" current Ic). The value of this curent depends, of course, on the applied bias voltage (resp. bias current).
  • Therefore, the red curves in your diagram will NOT cross the origin - instead, they will cross the Ic-axis at a small negative value (can be verified in SPICE- smulations).
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Basic idea

2-terminal devices are characterized by two electrical quantities - current can flow through them and voltage can appear across them. The relationship between the two quantities is given graphically by the so-called IV curve.

The BJT output characteristic is not something different; it is simply the IV curve of the output collector-emitter part of the transistor. Similarly, the BJT input characteristic is the IV curve of the input base-emitter junction. So the general idea is to explore a 3-terminal device by splitting it into two simpler 2-terminal "devices".

Experiments

I have long been convinced that the best way to do this is not with sterile textbook clichés but through a series of fun experiments accompanied by simple explanations. I have done it many times with my students in the lab and now I am very pleased to recreate it here through CircuitLab. So let's uncover the "philosophy" of this graphical representation in a natural way by exploring various devices - from the humble resistor to the sophisticated transistor, revealing the relationship between them.

We can do this in two ways - by varying the voltage and measuring the current - I = f(V), or by varying the current and measuring the voltage - V = f(I). Which of the two methods to choose depends on the shape of the device characteristic.

Constant resistor

It is logical to start with the famous Ohm's experiment going back two centuries.

I = f(V): In this case, we drive the resistor R by a variable voltage source V. Note that we only need an ammeter because the voltage source sets the voltage across the resistor, and there is no need of voltmeter.

schematic

simulate this circuit – Schematic created using CircuitLab

At first when the voltage is zero, the current is also zero because the resistor is a passive element (no power source inside it and nowhere to get the current). That is why the resistor IV curve passes through the origin of the coordinate system.

STEP 1.1

V = f(I): With the same success, we can drive the resistor by a variable current source I (however, Ohm did not have a current source and therefore did not conduct this dual experiment). In this case, we only need a voltmeter because the current source sets the current through the resistor, and there is no need of ammeter.

schematic

simulate this circuit

At first when the current is zero, the voltage is also zero for the same reason as above (because the resistor is a passive element, no power source inside it and nowhere to get the voltage). That is why, again the resistor IV curve passes through the origin.

STEP 1.2

Variable resistor

It would be interesting to see how resistance affects the slope of the curve. This observation is very necessary to understand circuits.

schematic

simulate this circuit

As we can see in the graph below, the higher the resistance, the steeper the curve. I have set R as the second parameter with three values - 1, 2 and 3 kΩ, and the three curves are superimposed in the so-called "family of curves". It is closely related to the OP's family of transistor output characteristics.

STEP 2

The resistors above have a constant resistance that does not depend on the voltage across or current through them. The IV curves obtained are linear; they present Ohm's law graphically.

Diode

But there are devices that change their "resistance" at the same time as the voltage or current changes, and this way try to keep them constant. For example, let's see how a diode does this "magic".

I = f(V): In the region of 700 ÷ 800 Ω (see the graph below), it reduces its resistance as the voltage increases, and the current increases sharply. Thus it tries to keep the voltage constant but fails because the voltage is produced by a constant voltage source. The problem is that we have connected two voltage stabilizers in parallel, and each of them is trying to maintain its own voltage resulting in conflict.

schematic

simulate this circuit

It is difficult to get enough points in the vertical part of the curve. So this way of taking the diode IV curve (by voltage) is inconvenient and dangerous for the devices.

STEP 3.1

V = f(I): Then let's get the diode IV curve by current, i.e., to drive the diode by a current source. Now there is no conflict between the two devices.

schematic

simulate this circuit

Their IV curves are almost perpendicular and many points of the curve can be obtained. The curve is mirror rotated 90 degrees because CircuitLab puts the sweep variable on the abscissa.

STEP 3.2

Voltage source

Let's take this opportunity to see the IV curve of a voltage source. While we managed, albeit with difficulty, to test the diode by voltage, here we have no chance because both devices are "ideal" voltage sources. The conflict between them is dramatic, the current is huge, and no intersection point we can get. So the only way to explore the voltage source, is through current.

schematic

simulate this circuit

Here are three curves for 1, 2 and 3 V; they are absolutely horizontal. Note that when the current is zero, the voltage is not zero. Here this is possible since there is a power source.

STEP 4

Transistor

Unlike diodes, transistors increase their "resistance" as the voltage increases, and the current does not change. So their IV curve (output characteristic) must be obtained by voltage. In the schematic below, I have connected a bias current source to the base-emitter junction and, adjusted its current Ib = 6.295 μA so that the same collector current of 1 mA flows as through the resistor above. If we consider the input current source and the transistor as a whole, it is a 2-terminal current-stabilizing element ("diode").

schematic

simulate this circuit

The two IV curves (of Q and Vce) are almost perpendicular and many points of the curve can be obtained. Note that when the voltage is zero, the current is also zero because there is no power source inside the transistor.

STEP 5.1

Ib = param: Then I have set Ib as the second parameter with three values causing Ice = 1, 2 and 3 mA. The three curves are superimposed in a "family of output characteristics" like in the OP's graph.

schematic

simulate this circuit

STEP 5.2

Current source

Above we have obtained the IV curve of a voltage source by a current source. Let's, for completness, get the dual IV curve of a current source by a voltage source.

schematic

simulate this circuit

Here are three curves for 1, 2 and 3 mA; they are absolutely horizontal. When the current is zero, the voltage is not zero since there is a power source.

STEP 6

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