0
\$\begingroup\$

I'm confused about a couple of things concerning the transistor.. is Vcc the signal to be amplified or Vin? And does the transistor actually amplify the voltage fed to it or does it just regulate how much of that voltage is led out to a say a device that is part of the voltage divider circuit that contains the transistor?

Also why do we use a transistor in the first place? Wouldn't a variable resistor in the divider circuit do the same job?

\$\endgroup\$
  • \$\begingroup\$ There are several different types of transistor. A BJT works in a different way to a FET. The main reason for using transistors is that they are electronically controlled. You couldn't build, for instance, an audio amplifier using just variable resistors. \$\endgroup\$ – Simon B May 18 '16 at 16:05
  • \$\begingroup\$ $V_{cc}$ is the power source that you feed in to a transistor amplifier circuit in order to amplify the weak ac input signal $v_{in}$. \$\endgroup\$ – rsadhvika May 18 '16 at 16:09
  • \$\begingroup\$ Yeah sorry I didn't specify, my question is about bjt transistors \$\endgroup\$ – snix May 18 '16 at 16:10
  • 1
    \$\begingroup\$ I think these fluid analogies are great for getting a physical feel of how a transistor works for beginners reuk.co.uk/OtherImages/transistor-model.gif images.slideplayer.com/18/6192351/slides/slide_12.jpg \$\endgroup\$ – rsadhvika May 18 '16 at 16:16
  • \$\begingroup\$ In particular notice how a small change in base-emitter flow creates a large change in collector-emitter flow \$\endgroup\$ – AgentS May 18 '16 at 16:30
1
\$\begingroup\$

A short answer: The bipolar transistor (BJT) works as a voltage-to-current converter for (small) input voltage variations around a suitable DC bias point. In any case (for all three basic configurations), this input voltage is the voltage (change) between the base node and the emitter node.

The corresponding voltage-to-current relationship can be verified by the exponential transfer characteristic Ic=f(Vbe), which first was described by W. Shockley. The parameter which describes this relationship is the transconductance gm=d(Ic)/d(Vbe).

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

The output of a common emitter amplifier is a voltage signal that is proportional to the input voltage. All of the power for the output signal comes from the power supply VCC. The input signal Vin is just a control signal that determines how power from VCC comes out.

To be clear, a BJT transistor is a current controlled current source. The current through the collector is proportional to the current through the base. A transistor in a common emitter amplifier only outputs a voltage because of the attached resistors.

The base current is related to the base-emitter voltage by...

IB = Is * (e^(Vbe/n/Vt)-1) / (Beta + 1)
IE = (Beta + 1) * IB
IC = Beta * IB

IE is the emitter current
IC is the collector current
IB is the base current
Beta is the transistor current gain, typically on the order of 100
Is is the base emitter diode saturation current
Vbe is the base emitter voltage
n is an ideality factor, usually between 1 and 2
Vt is the thermal voltage, 26mV at 25C
e is 2.71828...

When applying small AC voltage signals to the base, around some DC bias point, we can form a linear approximation treating the base-emitter junction as a small resistor having resistance...

re = (dIC/dVBE) / ICQ = Vt / ICQ
IB = Vin / Beta / Re
IC = Beta * IB = Vin / re

re is the base-emitter resistance used for the linear approximation at IC=ICQ
ICQ is the collector current at the bias point.

The bias point is determined by the resistors attached to the base, and possibly to the emitter.

By adding a resistor RC to the collector, the voltage across RC becomes...

Vout = RC * IC
Vout = RC / re * Vin

The gain of the amplifier is RC/re. That's how the basic common emitter amplifier works.

There are alternate technologies to transistors. One of the most widely used ones in recent history was vacuum tubes.

One could even use a rotary potentiometer controlled by an electric motor to make an amplifier, but the practical frequency ranges would be pretty small. A typical transistor can pass signals at millions of cycles per second. Good luck designing a motor or potentiometer that can change directions that fast.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Quote 1: "To be clear, a BJT transistor is a current controlled current source". Quote 2: "IB = Is * (e^(Vbe/n/Vt)-1) / (Beta + 1) and IC = Beta * IB". Two different (conflicting) statements. Current or voltage controlled? \$\endgroup\$ – LvW May 18 '16 at 20:17
  • \$\begingroup\$ You can look at it either way. Both equations are true. The linear equation is of course much easier to use with linear circuit analysis than the exponential one. \$\endgroup\$ – user4574 May 18 '16 at 21:55
  • 1
    \$\begingroup\$ Yes - the equations are valid. However, don`t you realize that the contence of the first cited sentence does not support the formulas? It is a contradiction! More than that, your gain formula Rc/Re is (a) a rough approximation only and (b) does not apply in case of a capacitor across Re. \$\endgroup\$ – LvW May 19 '16 at 7:06
  • \$\begingroup\$ Of course it was an approximation, but a good one used in many electronics textbooks. The Re is was referring to was not a real resistor, it was a mathematical construct, the slope of exponential VBE vs IE curve. The fact that you suggest adding a capacitor across it implies that you were unaware of that. I know that it is customary to add two external resistors to the emitter, with one bypassed, to stabilize the gain and DC bias point. I was not attempting to go into every detail of a real common emitter design, I was only attempting to show the bare minimum of how it works. \$\endgroup\$ – user4574 May 19 '16 at 20:22
  • 1
    \$\begingroup\$ Nobody could know that the symbol Re you have used was a dynamic resistance (slope of the iE=f(VBE) curve) because in this case, you should always use a symbol like re (not Re). For non-linear parts it is of great importance to discriminate between static and dynamic resistances. \$\endgroup\$ – LvW May 19 '16 at 20:59
-4
\$\begingroup\$

Transistor is an amplifying device.It amplifies the input signal i.e. Vin.Now the voltage divider network you are talking about is for temperature stabilization. As temperature increases the collector current increases which affects the operational point of the transistor.A variable resistor will not amplify the input signal hence it can't be used.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ It is the task of the voltage divider to bias the base node with a DC voltage (as stiff as possible, under the limitataions imposed by DC power consumption and input resistance). Stabilization against temperature changes and other uncertainties (tolerances!) is performed by negative voltage feedback only - in most cases we are using an emitter resistance Re for this purpose. \$\endgroup\$ – LvW May 19 '16 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy