5
\$\begingroup\$

enter image description here

I am trying to learn some transistor amplification by simulating. However there is something I don't understand why.

enter image description here

Why all waveforms are somehow distorted or bad-looking sinusoidal-ish waves instead of nice sine waves? Only the Vbase waveform looks decent.

\$\endgroup\$
  • 1
    \$\begingroup\$ Read this electronics.stackexchange.com/questions/374083/… \$\endgroup\$ – G36 Dec 2 at 18:57
  • \$\begingroup\$ While the answers pointing to the inherent nonlinearity of the device are correct (irregardless of how one might consider the BJT to be voltage or current driven), the reason you are getting that massive distortion is that you are biasing the transistor in a less than optimal way. Try to put the quiescent point in the middle of the output characteristics and things will go better. For example, if you change Rb=160k and Rc= 360 ohm, you will see much less distortion. \$\endgroup\$ – Sredni Vashtar Dec 3 at 7:04
6
\$\begingroup\$

A very short answer: The input signal at the base (Vin=VBE) is transferred into an output current Ic acording to the transfer characteristic (control function):

Ic=Io[exp(VBE/Vt)-1]

This is not a linear function and, therefore, the output current does not swing in exact proportion to the input voltage (and the same applies to the output voltage at the collector which is caused by the output current swing). And this nonlinearity causes distortions.

Only for very small input voltages we could consider the relation as "quasi-linear".

\$\endgroup\$
  • \$\begingroup\$ Is this also the concept about small signal condition, because when you mentioned non-linearity I just remembered the phrase "small signal range for linear amplification" ? \$\endgroup\$ – muyustan Dec 2 at 18:39
  • 2
    \$\begingroup\$ Yes - that is true! In reality, exact linear amplification is not possible. However, we can do a lot in order to improve the situation - and the keyword is "negative feedback". One method is to introduce a "feedback resistor RE" between the emitter node and ground. So we can reduce non-linear effects down to a level where it can be acceptable in many cases. \$\endgroup\$ – LvW Dec 2 at 18:48
  • \$\begingroup\$ what happens when you use a bypass capacitor across RE ? Does it still provide a reduction in non-linear effects. Because as I know, if not bypassed, RE lowers the voltage gain of a CE amplifier. \$\endgroup\$ – muyustan Dec 2 at 19:16
  • 1
    \$\begingroup\$ Yes - that is the price we have to pay for improving the signal quality (less distortions): Lower gain. A capacitor in parallel to RE increases the gain again - however, also the distortions (but DC feedback ist still present and stabilizes the DC bias point.) Therefore, often a trade-off is envisaged: Only a part of the emitter resistor is shunted with a capacitor. So we have good DC feedback and a certain amount of signal feedback. \$\endgroup\$ – LvW Dec 2 at 21:08
3
\$\begingroup\$

The base emitter region behaves like a forward biased diode (high in non linearities) hence, as you change the base to emitter AC voltage sinusoidally, the AC collector current that results is non linear. This means that the output voltage is also non linear.

It’s a poor amplifier in the real world.

\$\endgroup\$
  • \$\begingroup\$ Hmm, is that non-linearity you mentioned related to his ? zen22142.zen.co.uk/Design/graph/inputchar.gif \$\endgroup\$ – muyustan Dec 2 at 18:17
  • 1
    \$\begingroup\$ Yes it is, certainly. If you apply much smaller AC signal levels it improves. \$\endgroup\$ – Andy aka Dec 2 at 18:18
  • \$\begingroup\$ Is this also the concept about small signal condition, because when you mentioned non-linearity I just remembered that phrase "small signal range for linear amplification" \$\endgroup\$ – muyustan Dec 2 at 18:21
  • 1
    \$\begingroup\$ Yes, small signal analysis is underpinned by making signals so small that non linearities can be largely ignored. \$\endgroup\$ – Andy aka Dec 2 at 20:07
1
\$\begingroup\$

The other answers have glossed over something fundamental that is of vital consideration for BJT Transistors.

BJT transistors are current driven devices. Ultimately you should expect the wave of the transistor's collector current to be most similar to its base current, not it's base voltage.

\$\endgroup\$
  • \$\begingroup\$ While researching on the internet, most times I saw the input of a bjt is referred as "Vbe" and output as "Vce". So they are kind of wrong? \$\endgroup\$ – muyustan Dec 2 at 18:52
  • 1
    \$\begingroup\$ No, they're not wrong, but they're abstracting over the fundamental physical mechanics of BJT transistors. It can be very useful, and easier, to analyze circuits through the lens of voltage. Properties like Vbe are independently important to design, but the gain of the transistor is based on current. en.wikipedia.org/wiki/… \$\endgroup\$ – Shadetheartist Dec 3 at 1:12
  • 1
    \$\begingroup\$ Note that beta isn't constant w.r.t collector current, so the relationship \beta = I_C/I_B isn't actually linear either. See figure 15 in onsemi.com/pub/Collateral/2N3903-D.PDF. \$\endgroup\$ – andars Dec 3 at 4:17
  • \$\begingroup\$ @Shadetheartist...Sorry to say but you are completely wrong! The BJT is a voltage controlled device. Do you really think that two additional charged carriers in the base can release 500 additional carriers arriving at he collector (assuming a beta of 250). Never heard about the transconductance gm=d(Ic)/d(VBE) derived from the well-known Shockley equation? Do you need further explanations or proofs? \$\endgroup\$ – LvW Dec 3 at 12:19
  • 1
    \$\begingroup\$ @LvW I'm not going to pretend to be an expert on semiconductor physics but from what i've read the physics actually do play out such that BJT's are current driven. See the discussion here: physicsforums.com/threads/… Regardless i think analyzing the graphs OP presented in regard to current is more useful. \$\endgroup\$ – Shadetheartist Dec 4 at 1:55

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.