I need help solving the following exercise from my book.
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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?

  • \$\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
  • 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

  • \$\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
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    \$\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\$ – Tony Stewart Sunnyskyguy EE75 Aug 3 '17 at 13:41
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    \$\begingroup\$ 2 equal supplies with centre gnd 1 above + and 1 below gnd with - to bottom rail \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 3 '17 at 13:53
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    \$\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\$ – Tony Stewart Sunnyskyguy EE75 Aug 3 '17 at 14:03
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    \$\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\$ – Tony Stewart Sunnyskyguy EE75 Aug 3 '17 at 14:18

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