0
\$\begingroup\$

I need help solving the following exercise from my book.
enter image description here

I am not very knowledgeable in analyzing biasing circuits but I have understood the following: \$R_1\$ and \$R_2\$ are feedback resistors. This is a voltage amplifier. \$R_L\$ is the load. The output stage of the amplifier is a npn-transistor. the input stage is a differential stage and the peak output voltage is \$v_L = 1.5V\$.

Should I know assume that npn-transistor is a switch and check the two cases? Or how should I proceed. Do I assume anything else about this circuit like that the base currents are 0 and that the base voltages across the differential amplifier stage is zero?

\$\endgroup\$
  • \$\begingroup\$ "The peak output voltage is 1.5V". Use that info, and analyse the circuit for both positive and negative peaks. \$\endgroup\$ – Brian Drummond Aug 3 '17 at 13:11
  • \$\begingroup\$ What is the load current at clipping when PNP is OFF? And the output stage of the amplifier is a PNP-transistor not NPN. \$\endgroup\$ – G36 Aug 3 '17 at 13:17
  • \$\begingroup\$ Also, from which book you have this exercise? \$\endgroup\$ – G36 Aug 3 '17 at 13:26
  • \$\begingroup\$ Unfortunately, I don't know what the load current is there. 1mA = current from feedback resistors + current from load resistor (due to KCL). \$\endgroup\$ – Clone Aug 3 '17 at 14:35
  • \$\begingroup\$ @G36 Book: tinyurl.com/ya3vjpu9 \$\endgroup\$ – Clone Aug 3 '17 at 14:39
1
\$\begingroup\$
  • output stage is the PNP with 1mA sink, not NPN.

  • if Ve =-2.5V and we assume Vbe of 0.6 with IE=0 causing saturation of input left NPN such that Vce =~0

    • then \$V_L* R1/(R1+R2)=Vb\$
    • for \$V_L=+/-1.5V solve for RL
  • if -1.5V and 1mA sink with 11K bias to gnd + Ibe and Iload , is this what is asked? what is Iload max? 1.5V/11k=0.136mA thus -1.5v/0.863mA =1.73K min R load

\$\endgroup\$
  • \$\begingroup\$ Thank you so much for your comment! Is $V_e$ the voltage at the emitter of the pnp transistor? I thought that that should be 2.5V and not -2.5. Why is it negative? \$\endgroup\$ – Clone Aug 3 '17 at 13:39
  • 1
    \$\begingroup\$ it's a bipolar supply +/-2.5V to allow 0V input/output but 1mA Sink is on Vee (-) so I need to correct my error in equation 2 \$\endgroup\$ – Sunnyskyguy EE75 Aug 3 '17 at 13:41
  • 1
    \$\begingroup\$ 2 equal supplies with centre gnd 1 above + and 1 below gnd with - to bottom rail \$\endgroup\$ – Sunnyskyguy EE75 Aug 3 '17 at 13:53
  • 1
    \$\begingroup\$ ground is symbolic only to mean 0V locally and not earth or any other reference. I could be floating as shown but all V ate referenced to this symbol.. My answer is incomplete/incorrect BTW \$\endgroup\$ – Sunnyskyguy EE75 Aug 3 '17 at 14:03
  • 1
    \$\begingroup\$ all gnd symbols are joined , common, "logically" but does Not need to say how. The are usually close but details depend on application. so gnd symbol by definition is 0V \$\endgroup\$ – Sunnyskyguy EE75 Aug 3 '17 at 14:18

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.