I am working on a simple NPN transistor audio amplifier. I have found several diagrams which show how to build a NPN amplifier. These diagrams differ in that some show the output coming from the collector side of the transistor and some from the emitter. Here is a good diagram showing both:

enter image description here

In this diagram, there are two stages of amplification, each with a single NPN transistor. the second stage NPN is connected to the Emitter of the first, however the final output is connected to the Collector of the second stage's NPN transistor. Does it matter which side the output is acquired from? What are the differences and different applications?

  • 1
    \$\begingroup\$ Only if the polarity or the gain matter. \$\endgroup\$ Sep 12, 2016 at 22:21
  • \$\begingroup\$ I am assuming that if I am driving speakers or headphones, polarity does not matter. How is gain effected? Also, is polarity effected if the signal is biased to 1/2 vcc? \$\endgroup\$
    – Hoytman
    Sep 12, 2016 at 22:39

2 Answers 2


At a high level, the main difference is that output from the collector can give you voltage gain, while output from the emitter gives you current gain, but not voltage gain.

Sometimes it's useful to have the same signal but at a lower impedance. In that case a emitter follower (output from emitter) can be useful. In the end, most "amplifiers" need overall voltage gain, so output from the collector is going to be needed somewhere in there.

Note that in the second stage of your circuit, the collector resistor is 10x the emitter resistor. Since roughly the same current flows thru the collector as emitter, the voltage variations as the collector will be about 10x higher (voltage gain of 10) at the collector than the emitter.

Your circuit actually has even higher gain, due to additional resistor and capacitor in series on the emitter. Since the capacitor blocks DC, only the 1 kΩ resistor matters in setting the DC operating point of the transistor, also called biasing the transistor. The 470 Ω resistor with 10 µF cap in series effectively make the emitter resistor lower at audio frequencies. This makes biasing easier and more predictable, but still gives higher gain at the audio frequencies where we care about gain.

The rolloff frequency of 470 Ω and 10 µF is 34 Hz, so this isn't quite a "HiFi" audio amp. At low frequencies, the gain of the second transistor stage is about 10. Starting at around 30 Hz, the gain goes up so that eventually the emitter resistor looks like (1 kΩ // 470 Ω) = 320 Ω, for a gain of about 30.


Yes, it matters to have either impedance reduction(CC) or voltage gain.(CE)

  • input impedance in the common collector (CC) aka Emitter Follower is defined by \$h_{FE} * (R_e // R_{bias})\$. Then the emitter output impedance is \$\frac{R_{bias}}{h_{FE}} // R_e\$.
  • Then add 2nd stage \$R_{in}\$ to the \$R_e\$.
    • voltage gain is 1.
    • Thus \$R_{in} = 1M//4M7 // (h_{FE}*(100K // (h_{FE_{2}}*(1k//10k))) )\$

If your 1st output impedance is the same as the input impedance of next stage, expect attenuation losses at f.

CE amplifier output impedance is the collector current source (high) in parallel with the \$R_c\$ collector resistor.

Thus your overall gain can be measured in terms here Voltage gain * Current gain or Voltage gain * Impedance gain (in terms of reduction)

AC gain is \$10k/470 \sim 21 \$.

Let \$h_{FE} =200 \$, then Stage 1 \$Z_{out_1}= \frac{830k}{200} \sim 4K\$.

With Stage 2 \$Z_{in}(AC)= 200 *470~ 100k \$ {+10k series } and \$Z_{out_2} = 10k \$

Impedance gain 830k/10k =83
Voltage gain= 21

power gain is the net VI product gain.

  • \$\begingroup\$ I am afraid I do not have enough experence to understand all of your terms and abbreviations. \$\endgroup\$
    – Hoytman
    Sep 12, 2016 at 22:41
  • \$\begingroup\$ Source impedance must be << load impedance, hFe is current gain. This cannot drive a speaker with 10k output impedance. \$\endgroup\$ Sep 12, 2016 at 22:44
  • \$\begingroup\$ impedance gain is a tradeoff in transistor design VS voltage gain using the current gain and possible negative feedback and emitter bypass to assist further. Negative feedback lowers input and output impedance while reducing voltage gain. This is why if often takes at least 6 and perhaps 10 stages to go from 1M input to 100 milliohm output impedance with 60 dB voltage gain to drive 8'Ohms with 7 decades or 140 dB current gain \$\endgroup\$ Sep 12, 2016 at 22:47
  • \$\begingroup\$ The dim lights are starting to come on in my head. \$\endgroup\$
    – Hoytman
    Sep 13, 2016 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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