I learn some knowledge of common emitter from this website: https://www.electronics-tutorials.ws/amplifier/amp_2.html

But i am confused about this formula : voltage gain \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=-\frac{R_L}{R_E}\$,here is the schematic of common emitter below

enter image description here

First,if we know the output and the input wave,we can easily understand why is \$\frac{V_{out}}{V_{in}}=-\frac{R_L}{R_E}\$,not \$\frac{V_{out}}{V_{in}}=+\frac{R_L}{R_E}\$,but if we don't know common emitter acts like an inverter amplifier,how can we know \$\frac{V_{out}}{V_{in}}=-\frac{R_L}{R_E} ?\$,because the opposite voltage means the opposite current direction,but i can't understand why \$I_C\$ and \$I_E\$ are opposite,is it because the \$I_C\$ flows "into" bjt and \$I_E\$ flows "out of" bjt,so they are opposite?

Next is why is \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{RE}}\$,not \$\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{R2}}\$,i think the reason of it is because \$I_B\$ is a input current of bjt,and \$V_{RE}=I_E\times R_E=(I_B+I_C)\times R_E\$ ,and \$V_{R2}\$ has no relation with \$ I_B\$ ,so we can't say \$\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{R2}}\$,however,i don't know the reason i think is correct or wrong,if wrong,can anyone tell me the reason of it??

  • \$\begingroup\$ Ic and Ie are not opposite, the increase in the base voltage will increase collector current, as Ic increases, the voltage drop on the RL will increase, thus the voltage at collector will decrease. So, increasing input will decrese output voltage. \$\endgroup\$ – muyustan Jun 10 '20 at 13:20
  • 1
    \$\begingroup\$ The given result is a (rather rough) approximation/simplification. The correct formula is GAIN=-RL/(1/g+RE) with transconductance g=Ic/Vt. However, this applies for very low frequencies only (when RE is not bypassed). When the capacitive resistance 1/wC can be assumed to be zero, the gain is (-g*RL). \$\endgroup\$ – LvW Jun 10 '20 at 13:30
  • 1
    \$\begingroup\$ @Andyaka ok i edit the figure now \$\endgroup\$ – tester_ga Jun 10 '20 at 14:11
  • 2
    \$\begingroup\$ With bypassed emitter resistor, gain is actually gain=-RL/re where re is internal dynamic emitter resistance given by re=thermal voltage/collector current. Only when bypass capacitor is removed does the gain become: gain=-RL/(re+RE). \$\endgroup\$ – Leoman12 Jun 10 '20 at 14:11
  • 1
    \$\begingroup\$ @tester_ga, C2 is bypassing RE. At appropriate frequencies, this C2 essentially shorts out RE. \$\endgroup\$ – Leoman12 Jun 10 '20 at 14:16

The minus sign in equation Vout/Vin = - R_load/R_emitter may appear counterintuitive. To understand its origin, remember that the network analysis here is performed for a linear circuit whereas the transistor circuitry is inherently non-linear. In order to be able to use Ohm's and Kirchhoff's laws, the circuit designer makes a number of idealizations, starting with an assumption that the voltages/currents in the circuit can be written as sums of DC voltages/currents plus (small additions of) AC voltages/currents. Or, alternatively, s/he examines small variations of voltages/currents near their "quiescent" values.

So, to fully understand the network analysis of transistor circuits, I would recommend you to learn from textbooks that use transistor models with current controlled current sources (CCCS)/voltage controlled current sources (VCCS) etc., the said models used to calculate biasing, gain, other parameters of small signal transistor amplifiers.

As for the common collector amplifier of the electronics-tutorials.ws tutorial, notice that for all practical purposes we can assume that C1 capacitance is sufficiently high, so that the AC component of V_base voltage is equal to Vin (Vin is an AC voltage source, I assume). In the Vin source's frequency region in which the C2 capacitance is sufficiently small to neglect current flowing thru a capacitor branch, an AC component of the current I_emitter is practically equal to an AC component of the V_emitter voltage divided by the R_emitter resistance. In the small signal approximation, V_base - V_emitter is constant (usually about 0.7V), therefore ΔV_base = ΔV_emitter, and ΔI_emitter = ΔV_base / R_emitter. Within a 1 percent error (beta = 100) I_collector = I_emitter; the AC components of emitter and collector currents are also equal (ΔI_collector = ΔI_emitter).

Vout is a voltage at the collector output of the transistor, as well as its AC component ΔV_load. Ohm's law gives us a voltage across the load resistor R_load measured relative to a reference point which is +Vcc pin of power supply (this means current flows out of R_load resistor); hence the minus sign in the equation

ΔV_load = - ΔI_collector * R_load.

We arrive at equation ΔV_load/ΔV_base = - R_load/R_emitter.

  • \$\begingroup\$ A bit wordy, but correct. In common emitter circuits -Rl/Re dominates Vo no matter the state of Q1, which is assumed to be not fully On or fully OFF, as this introduces severe distortion. \$\endgroup\$ – user105652 Jun 12 '20 at 1:40

Edit: In my analysis I assumed IC≈IE and ic≈ie since base current is usually quite small. Also note: my analysis is for ac voltage vout. From your circuit, Vout actually contains both an ac and DC component. DC component is dc voltage at collector terminal.

Before I explain an answer, I need to mention that for the circuit you've shown which is a common emitter amplifier with bypassed emitter resistor RE , the gain is not vout/vin = -RL/RE . That equation is actually an approximation to a common emitter amplifier with no bypassed emitter resistor. I'll explain that also. I'll also mention that collector current IC and emitter current IE are flowing in the same direction, ie: they are in phase. They are also in phase with the base current IB. What causes the negative sign is the output voltage being out of phase with collector current by 180 degrees.

Now onto the answer:

For a common emitter amplifier with bypassed emitter resistor (bypassed with C2):

For the amplifier to work, the NPN transistor must be biased in active region via the network R1, R2, RE, and RL. C1 is coupling capacitor which at mid frequencies can be approximated as a short. C2 is a bypass capacitor which at mid frequencies can also be approximated as a short.To get an understanding of the gain formula we need to look at the ac analysis of this circuit and use an appropriate ac small signal model of the transistor. This is shown below, using the fact that C1 is short, C2 is short, and VCC is ac equivalent to ground. Also using the t model of BJT.

Now, ΔVB and ΔVL are just small variations of VB and VL. That being said, they are essentially ac signals. We can represent ΔVB as vbe and ΔVL as vout or vRL. Now looking at the ac circuit below, it can be seen that small signal input vin equals vbe which is also equal to vre. (Note lower case letters indicate small signal quantities). Here re is the small signal dynamic emitter resistance of NPN. It represents the small signal modeling of the base to emitter diode inside the NPN. This is given by re=25mV/IC.

Now looking at the output: By ohms law and following the passive sign convention we know that vRL=vout= -icRL. Note that current flows out of positive terminal of RL. But since ic= vbe/re=vin/re, then vout= -(vin/re)RL. And gain is then vout/vin=-RL/re.


simulate this circuit – Schematic created using CircuitLab

Now I'll explain when you have unbypassed emitter resistance:

Without the bypass capacitor C2 across RE, we must incorporate RE into the ac small signal circuit analysis. The new circuit is shown below. In this case vin equals vbe + vRE which is just equal to vre+RE (voltage over both resistors).

Now looking at the output: By ohms law and following the passive sign convention we know that vRL=vout= -icRL. Note that current flows out of positive terminal of RL. But now this time we have ic= vin/(re + RE), then vout= -((vin/(re + RE))RL. And gain is then vout/vin=-RL/(re + RE).

For when RE >> re then you may approximate the gain formula as vout/vin=-RL/RE.


simulate this circuit


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.