1
\$\begingroup\$

enter image description here

I have a textbook that details the designing process of a common emitter BJT amplifier. When designing the biasing point, the text says to set the bias collector current so that half of the supply voltage is dropped across the resistors and half is dropped across the transistor. For example, if we had a supply voltage of 10V, we would assume that at the middle point of the load line there would be 5V across the collector-emitter junction and the other 5V would be distributed across the resistors. This supposedly gives the highest room for the signal to "swing" when input is applied.

But how does this make sense, since usually the output is taken to be the voltage between to collector and ground? So wouldn't we want to bias so that the voltage across the transistor AND the emitter resistor is equal to half of the supply? What am I missing here?

\$\endgroup\$
  • \$\begingroup\$ You are right. But the book is talking about the CE amplifier without the RE resistor. But with the RE resistor, we usually set the Vc voltage at (Vcc - Ve)/2. \$\endgroup\$ – G36 Jan 7 at 17:27
  • \$\begingroup\$ But in the book, the picture has the RE resistor. I also found a Youtube video which has it also, but uses a half of the supply across the transistor: youtube.com/watch?v=1IYOGhS2OZc \$\endgroup\$ – S. Rotos Jan 7 at 17:32
  • \$\begingroup\$ Yes - it is correct. The optimum bias point is app. half of the supplx voltage - even for relatively large RE values. \$\endgroup\$ – LvW Jan 7 at 17:44
  • \$\begingroup\$ @LvW But the way the biasing is made here, the voltage at the collector is not half of the supply. \$\endgroup\$ – S. Rotos Jan 7 at 17:47
  • \$\begingroup\$ S. Rotos - see my detailed answer for derivation of the optimum bias point. When you say "voltage at the collector" you mean "referenced to ground". But in most books and other contributions the bias point is described as the pair "Ic and Vce". This is because we can detrmine and verify the bias point in the Ic=f(Vce) set of curves. Remember: You spoke about the voltage "across the transistor" - and that is Vce. \$\endgroup\$ – LvW Jan 7 at 17:49
0
\$\begingroup\$

When we apply Ohms law to the series combination of RC and Re we have:

Ic=(Vcc-Vce)/(RC+RE)

This gives the so-called "gain line" (with negative slope) in the classical output set of curves Ic=f(Vce) for IB=const or Vbe=const.

Here we can see that we have Ic=0 for Vcc=Vce. The maximum of Ic (Icmax) is assumed to be for Vce~0 (neglecting the saturation voltage Vce,sat of app. 0.5volts).

The optimum bias point,therefore, is app. at Vce=Vcc/2 (middle of the gain line) - independent on the RE value.

\$\endgroup\$
  • \$\begingroup\$ I disagree, the optimal Vce will change if you add CE capacitor into the circuit. And it will be equal to \$V_{CEopt} = V_{CC}\frac{R_C+R_E}{2R_C +R_E}+ V_{CEsat}\frac{R_C +R_E}{2R_C+R_E}\approx V_{CC}\frac{R_C+R_E}{2R_C +R_E}\$ and the optimal collector current is \$I_{Copt}=\frac{V_{CC}-V_{CEsat}}{2R_C +R_E}\$ But without a capacitor CE capacitor the Vce optimal is indeed equal to to \$V_{CEopt}\approx V_{CC}*\frac{R_C+R_E }{2(R_C+R_E)}\approx 0.5V_{CC} \$ and \$I_{Copt}=\frac{V_{CC}-V_{CEsat}}{2(R_C+R_E)}\$ \$\endgroup\$ – G36 Jan 7 at 17:55
  • \$\begingroup\$ As far as I can see, the circuit under question did not contain any capacitor. \$\endgroup\$ – LvW Jan 7 at 17:58
  • \$\begingroup\$ Fantastic MathJAX though. \$\endgroup\$ – Transistor Jan 7 at 18:45
  • \$\begingroup\$ @G36....I disagree with your result if CE is added. Not considering the saturation voltage Vcesat your result is Vce,opt=Vcc(Rc+Re)/(2Rc+Re). That means: For increasing Re values the optimum voltage Vce,opt will approach the supply Vcc. Obviously, this cannot be correct. Here is my result: Vce,opt=Vcc(Rc)/(2Rc+Re). For rising Re values, the voltage Vce,opt becomes smaller and smaller (which makes sense to me). \$\endgroup\$ – LvW Jan 9 at 20:37
  • \$\begingroup\$ ....and the same considerations apply when Rc becomes smaller and smaller ..... \$\endgroup\$ – LvW Jan 9 at 20:46
1
\$\begingroup\$

You are correct. The emitter resistor, RE, is usually much smaller than the collector resistor so the voltage across RE will not change very much as the input signal changes. RE provides some negative feedback to help stabilize the amplifier. The output signal is really the voltage that appears across RC, and that is the resistor voltage that will change in proportion to changes in the input voltage. In general, you want to bias the amplifier so that the voltage across RC is about half of the supply voltage.

\$\endgroup\$

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.