What is the collector-emitter resistance of NPN Transistor?

the question may look ridiculous since I'm not sure if the collector-emitter resistance exists or not. Here is a simple commom emitter circuit

As I learn that when the Vb increase that will make Ib increase so Ic must increase too. When Ic increase as there is Load Resistor but Vcc is constant and Ic = (Vcc-Vc)/RL (Load Resistor) then Vc must decrease and vice versa. That how common emitter work

Now, what I concern is the Voltage Drop between Vcc and Ground is constant as well as Load Resistor value. Suppose that there is not thing between Emitter and Ground that make Ve = 0 and Vb = 0.6-0.7 while Vc is much larger (that depend on load resistor). So, there must be something that wastes the energy to make Ve=0 that cause voltage drop between collector and emitter. Is there something act like varying resistor between collector and emitter to make that.

In other words, to make voltage drop between collector and emitter there must be something act like resistor between them, right? If no, what make the difference in voltage?

In other configuration does collector-emitter have resistance as well?

• Ideally the collector is connected only to a current source, thus the collector-emitter resistance is infinite. The output voltage is set by the collector resistance. Check here. Usually $h_{re}=0\frac{V}{V}$ and $h_{oe}=0\Omega^{-1}$ – Vladimir Cravero Aug 30 '14 at 16:52

The BJT collector current equation is

$$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CB}}{V_A}\right)$$

where $V_A$ is the Early voltage. But, this formula is often written as

$$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CE}}{V_A}\right)$$

Thus

$$\frac{\partial i_C}{\partial v_{CE}} = \frac{I_S\cdot e^{\frac{V_{BE}}{V_T}}}{V_A} = \frac{i_C}{V_A + v_{CE}}$$

This is clearly a non-linear function of the collector-emitter voltage and collector current so this cannot be interpreted as a conductance.

However, for small changes around some fixed value of collector current $I_C$ and collector-emitter voltage $V_{CE}$, we can write

$$I_C + i_c \approx I_C\left(1 + \frac{v_{ce}}{V_A + V_{CE}} \right) = I_C + \frac{v_{ce}}{r_o}$$

where

$$r_o = \frac{V_A + V_{CE}}{I_C}$$

We call $r_o$ the collector-emitter dynamic, or differential or small-signal resistance.

It is not a true resistance since it is not constant but, instead, varies with the operating point of the transistor as can be seen by the formula.

• I like to add that the transistor is a strongly NON-LINEAR element. Hence - as for each non-linear part - you must discriminate between the static resistance (Rce=VCE/IC) and the differential (dynamic) resistance (rce=ro=d(VCE)/d(IC). Don`t get confused, it is correct, that in the above answer the expression for ro contains only DC values. This is the result of differenciating an exponential function. Note that the static reistance Rce plays no major role in circuit design. – LvW Jul 25 '16 at 7:29

You've got a couple of good answers. I'll try to add some intuitive insight.

When the transistor is biased such that that it is not saturated, it behaves like a current sink (recall that a perfect current sink has infinite impedance), so the collector-load junction looks like a voltage source with a Thevenin equivalent source impedance equal to the load resistor. The voltage is dependent on the base current and beta. This is equivalent to what Alfred wrote, but with an infinite Early voltage. The collector impedance due to the Early voltage is in parallel with the load resistor, so to get an answer that is realistic without the load resistor you must include it, as Alfred did.

When the transistor is saturated, it behaves more like a voltage source of << 1 volt with a fairly low small-signal source resistance.

To answer in simple terms: the collector behaves like a current sink, and the collector voltage settles to whatever value lets that amount of current flow (though it can't go any lower than approximately Ve+0.2V).

In your example circuit, the collector-emitter junction can be thought of as a variable resistance whose value depends on the electronical situation present at the amplifier's output. It also heats up like a resistor: Ic * Vc = the amount of heat generated in Watts, heating up the transistor.

If the supply voltage and the load resistance remain constant, then as the base current varies, the collector voltage and current will vary.

Such being the case, then for any collector current there must be a resistance between the collector and emitter such that:

EDIT:

$$R2 = \frac{E2R1}{E1 - E2}$$

Where R2 is the transistor's collector-to-emitter resistance, E1 is the supply voltage, E2 is the collector-to-emitter voltage, and R1 is the load resistance.

• That answer is a bit dimensionally challenged. The reciprocal of that, n'est ce pas? – Spehro Pefhany Aug 30 '14 at 17:54
• Spehro: The conductance of the channel? – EM Fields Aug 30 '14 at 18:08
• If it's R2 = then the units should be ohms. They're 1/$\Omega$ (volts cancels out, leaving 1/$\Omega$). $\frac{1}{R2} =$ will work too. – Spehro Pefhany Aug 30 '14 at 18:21
• Spehro: Excellent catch! I got the numerator and the denominator bassackwards, aargh... Thanks for the reality check. – EM Fields Aug 30 '14 at 18:40
• Already a simple example reveals the problems associated with the given formula - because it considers dc voltages only. Set E2=E1/2 and we have R1=R2. A result that does not help at all. The collector-emitter path is strongly non-linear and we always must discriminate between static and dynamic (differential) resistances. More than that, the formal definition of a static resistance for the BJT is completey useless. My recommendation to aukxn: Rely on A. Centauri´s answer only. – LvW Aug 31 '14 at 13:25

It is not really the proper question to ask. While a semiconductor does have resistance to the flow of current, so does a capacitor. The way to start is to ask, what is the voltage drop across the transistor. This is a value that is typically published for each component. This way, when you know the particular operating conditions, you can easily calculate the voltage and appropriate resistances to place in the other parts of the circuit.