1
\$\begingroup\$

I have just begun with transistors and came upon the common base configuration.

There it's given that output current is less than input current which makes sense considering some electrons are lost during the recombination process in the base region but I don't understand this particular line:

The base-emitter junction JE at input side acts as a forward biased diode. So the common base amplifier has a low input impedance (low opposition to incoming current). On the other hand, the collector-base junction JC at output side acts somewhat like a reverse biased diode. So the common base amplifier has high output impedance.

Therefore, the common base amplifier provides a low input impedance and high output impedance. what does this exactly mean it's a npn transistor btw though if base-collector is reverse biased then it sould make sense that no electron flow to collector......

Source

\$\endgroup\$
  • \$\begingroup\$ consider a bipolar operating at 1mA, and Vearly of 20 volts. We know the CommonBase config has 'rin' of 0.026 / Ie = 0.026 / 0.001 = 26 ohms. The Vearly does no provide useful 'rout' for CommonBase, because Vearly is related to 'hre', not 'hrb'. However the \$\endgroup\$ – analogsystemsrf Nov 3 '19 at 21:15
0
\$\begingroup\$

consider a bipolar operating at 1mA, and Vearly of 20 volts. We know the CommonBase config has 'rin' of 0.026 / Ie = 0.026 / 0.001 = 26 ohms. The Vearly does not provide useful 'rout' for CommonBase, because Vearly is related to 'hre', not 'hrb'. However the 'rout' of Common Emitter is Vearly/Ie = 20v/0.001 = 20,000 ohms slop of the I-V output plot for the Common Emitter.

Usually the CommonBase is much flatter, that is, 'rout_common_base' is much higher than 20,000 ohms. This compares to 'rin' of 26 ohms.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ What is rin rout and Vearly \$\endgroup\$ – Aladdin Nov 4 '19 at 9:12
  • \$\begingroup\$ hre a d hrb too. \$\endgroup\$ – Aladdin Nov 4 '19 at 10:33

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.