# Clipping of a negative feedback amplifier

I need help solving the following exercise from my book.

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?

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

• 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 • 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? – Clone Aug 3 '17 at 13:39