I'm trying to understand an exercise reported in Microelectronic Circuit Design by R. C. Jaeger & T. N. Blalock (4th ed, p.237). $$\beta_F = 50, \beta_R = 1$$. enter image description here

I don't agree with the discussion they make. enter image description here

Figure 5.19 displays the results of simulation of the collector current of the transistor in Fig. 5.18 versus the supply voltage V CC . For V CC > 0, the collector-base junction will be reverse-biased, and the transistor will be in the forward-active region. In this region, the circuit behaves essentially as a 1-mA ideal current source in which the output current is independent of V CC . Note that the circuit actually behaves as a current source for V CC down to approximately −0.5 V. By the definitions in Table 5.2, the transistor enters saturation for V CC < 0, but the transistor does not actually enter heavy saturation until the base-collector junction begins to conduct for V BC ≥ +0.5 V

Why does the fact Vcc > 0 automatically implies reverse bias on collector base junction? What if Ic is sufficiently large to produce a voltage drop (on the resistor) bigger than Vcc? In this case the collector base junction will be forward biased, so that the saturation region starts at Vcc clearly different from 0...

In fact, it is essentially what happens if I try to reproduce in SPICE simulation the circuit.. the dark blue line represents Ic which strangely doesn't agree with Fig 5.19! enter image description here Spice schematic: enter image description here

I'm a beginner in electronics and SPICE, so I'd like to know where I am wrong.

  • 2
    \$\begingroup\$ At least your lower voltage source has different polarity when compared to the textbook image. I cannot see actual numeric values reliably, onlt the plus and minus signs. Check them! \$\endgroup\$
    – user136077
    Dec 9, 2017 at 10:55
  • \$\begingroup\$ @user287001 updated question with better pictures... Could you check them? I think the polarity is correct. \$\endgroup\$ Dec 9, 2017 at 11:20
  • \$\begingroup\$ What models for BJT are you using? If you are not using any model, your simulation does not make sense. \$\endgroup\$
    – sarthak
    Dec 9, 2017 at 11:37
  • \$\begingroup\$ @sarthak I'm using (as the book suggest) the builtin model. However I'm not complaining about small differences.. I don't agree on the assumption, reported in the blockquote "for V CC > 0, the collector-base junction will be reverse-biased" that isn't sufficiently explained... \$\endgroup\$ Dec 9, 2017 at 11:50
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    \$\begingroup\$ @Surferonthefall - You're right about the emitter/collector voltage. My bad. Sorry. (Damn those dyslexic fingers!) \$\endgroup\$ Dec 10, 2017 at 2:55

4 Answers 4


Actually you are right. The text in the book has an error. There is written Vcc instead of Vce. Vcc must be more than 3,8V to make the Vce > 0 and the transistor to be fully active.

I must admit that I also thought wrongly the working of the circuit in my previous answer version. I did a DC sweep simulation. I placed GND to the emitter to get Vce easily without math, Vce= V(NODE1)

enter image description here

enter image description here

The voltage of BAT1 goes from 0 to +9V. Vce does not go negative at all. The proper function of the transistor needs Vce > the saturation limit which is about +200mV in this case.

  • \$\begingroup\$ Maybe you're right.. however if Fig 5.19 is Ic against Vce, the negative current should plateau at a shorter height than positive current.. Am I correct? \$\endgroup\$ Dec 9, 2017 at 13:01
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    \$\begingroup\$ @Surferonthefall I started to doubt my fast calculations. I'll update my answer as soon as possible. \$\endgroup\$
    – user136077
    Dec 9, 2017 at 13:28

Your question: "What if Ic is sufficiently large to produce a voltage drop (on the resistor) bigger than Vcc?"

This is impossible. Using the given DC supply values (+- 9V) a voltage drop larger than Vcc would make the emitter potential positiv. In this case, the B-E pn junction is not positively biased and the transistr would be off. Hence, no current. That means: Your primary assumption (large Ic) is obsolete.

You should (a) know the "secrets of negative feedback (caused by R2) and (b) realize that such a negative DC feedback allows an automatic Ic adjustment because there are two equations (rules) which must befulfilled:

(1) Ohms law for R2 and (2) Shockleys exponenrtial equation describing the Voltage-current relation across the B-E junctin.

Finally, for Vcc>0 one of the preconditions for using the BJT as an amplifier is met (C-B junction reverse-biased). The second condition (B-E junction forward-biased) is also met because the base potential (zero) is always larger than the negative collector potential.

  • \$\begingroup\$ thanks a lot for the answer.. I'm thinking about it. However I clearly see, in the simulation, a work point with Vcc = 3.5V, voltage drop on Rc of 4.3V, both of the junction forward biased, Ic = 1mA.. How does it comply with your answer? \$\endgroup\$ Dec 9, 2017 at 11:56

Assuming Vbe=0.7V, we will have: $$ I_{e}=\dfrac{-0.7V+9V}{8.2k\Omega}=1.012mA $$ Calculating the Ic: $$ I_c=\dfrac{\beta_f}{\beta_f + 1}\cdot I_e=0.992mA $$

$$ V_c=V_{cc}-4.3k\Omega\cdot I_c= Vcc-4.26V $$

Calculating Vcc when Vc becomes -0.5V or Vce=0.2V,we will have: $$ -0.5V=Vcc-4.3k\Omega \cdot 0.992mA \\ Vcc=3.76V $$ The result agrees with the simulation value.


For \$V_{CC} \gt 0\$, the collector-base junction will be reverse-biased, and the transistor will be in the forward-active region.

Let's start here. To understand this point, you need to see a few details.

  1. The base is bolted to ground (\$0\:\textrm{V}\$).
  2. The emitter is pulled down with a resistor towards a voltage that is significantly negative with respect to the base.
  3. There is a (hidden) diode in the BJT diagram that goes from base to collector. That diode's anode is at the base, which is at \$0\:\textrm{V}\$ by definition. The collector is the cathode. To make that diode forward biased, the collector would necessarily have to be negative with respect to the base.

Since the statement is talking only about the case where \$V_{CC} \gt 0\:\textrm{V}\$, this is also by definition more positive than the base. Not more negative.

Think of this as a simpler question:


simulate this circuit – Schematic created using CircuitLab

I've intentionally left the base-emitter diode unconnected here, so that you will ignore it for now. Just look at the remaining circuit. If \$V_{CC} \le 0\:\textrm{V}\$ then the BC diode is forward biased to some degree and a current can flow through \$R_C\$. But if \$V_{CC} \gt 0\:\textrm{V}\$, then the BC diode must be reverse biased. So that part of the statement, the part saying "the collector-base junction will be reverse-biased," would appear to be on its face correct from this highly simplified perspective.

Of course, you now argue that, "Yes, I agree so far. But also, there is a collector current that cannot be explained by that simplistic schematic. And this collector current might appear to be any value at all. So this would mean that your example is useless."

Good point. My counter to that is that the collector current cannot be any current. That current is limited to a maximum determined roughly by \$I_{C_{MAX}}\approx \frac{V_{CC}}{R_C}\$. So let's see why.

We can change the current in the emitter by changing the value of \$R\$. The BJT is in a mode of being an emitter follower and so the emitter current will be determined by:


You'll note here that \$R_C\$ doesn't show up here. So it would appear that the emitter current can be set arbitrarily. "Doesn't this mean that we can make the collector current any value we want, then?" And the answer is "NO, you cannot do that."

"So where does the emitter current go to or come from, then?", you ask.

"The difference appears at the base," I reply.

As you lower the value of \$R\$, and while the BJT is in its active and not saturated mode, the collector current increases just as you'd expect. But at some point determined by your choice of values for \$R_C\$ the voltage drop across \$R_C\$ gets very close to the point of forward-biasing the BC junction. Right at the point where \$V_{BC}=0\:\textrm{V}\$, the BC diode remains reverse-biased, just as the text says it does. But if you now continue to increase the collector current by decreasing \$R\$ still further (or, alternately, try to increase the voltage drop across \$R_C\$ by increasing its value), you will just start the very, very early stages of forward-biasing the BC junction. The moment you do that, is the moment that the BC junction just begins to conduct some forward current from the base.

Think closely about that. As you increase the voltage drop across \$R_C\$ sufficiently to begin to forward-bias the BC junction, the base must now begin to supply an additional current for that newly forward-biased junction. This added base current does not increase the collector current but does become part of the emitter current.

[Think of this additional current as "going from the base into the collector region and then falling into the emitter." This happens inside the BJT, though, and does not appear as a measurable current on the wire that represents the collector pin of the BJT. (It can't.) This happens inside the BJT, so it cannot impact the collector pin itself. All it does is add to the emitter current without changing the collector current (much.)]

Once the BC junction has moved into a forward-biased condition, the BJT is said to be "saturated." At first, this is a very shallow saturation as the BC junction is only very lightly forward-biased and the added current is quite small. So from the outside looking in, the \$\beta\$ still appears to be "pretty good." But as you try to increase the emitter current further (or increase the value of \$R_C\$ to try and increase its voltage drop), this BC junction becomes further forward-biased and this means the current flowing through it dramatically increases.

Remember, every \$60\:\textrm{mV}\$ increase in the forward biasing of a BJT diode will increase the current through it by about a factor of 10. So even small changes in the forward-biased voltage of BC yield huge changes in the base current needed. Very, very quickly the BJT goes into deep saturation and the base starts pulling very large currents.

Eventually, in the extreme, almost all of the emitter current is coming from the base, not the collector; with the collector current limited (capped) by a collector voltage that cannot get much closer to the emitter voltage than about \$V_{CE_{SAT}}\approx 200\:\textrm{mV}\$.

In short, while the BJT is in its active mode (and not into deep saturation), the collector "acts like" a current source/sink. And when the BJT is in deep saturation, the collector acts like a voltage source with a voltage very close to the emitter voltage. In the remaining state, in shallow saturation, it is transitioning somewhere in between these two other states.

For your Spice simulations, just take note of the base currents as well as the other two currents. If you see the BC junction going into a forward-biased situation, then this also means you should start to see an increasing percentage of the emitter current as coming from the base and not the collector. Before that point, you should be a roughly constant relationship between base and collector currents. (Or, put another way, you should see \$\beta=\frac{I_C}{I_B}\$ declining as the BC junction becomes increasingly forward-biased.)


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