Answers to Analogue Electronics

Ideal Operational Amplifier 

  1. 	E		(1993 PI  Q19)

Inverting Amplifier	
	
  2.a. 	   Yes
    b.i.   -2.5V
      ii.  -10V
      iii. -2V
      iv.  -13V to -15V (amplifier saturated)
	
  3.a. 	Rf/R1 = 14.4/3.6 = 4	
    b. 	Swap the position of the feedback resistor with the 
	input resistor.
	Rf/R1 = 3.6/14.4 = 0.25

  4. 	B, C & D 	
  5. 	B		(1999 PI  Q20)
  6.	C		(1995 PI  Q20)

  7.i.			(1994 PII  Q7ai)
   ii.	Vin = 0.219V	(1994 PII  Q7aii)

  8.i.	Gain = 27			(1997 PII  Q6ai)
   ii.	Vrms = 190.9mV			(1997 PII  Q6aii)
  iii.	Choose a ratio ratio Rf/R1 = 27
					(1997 PII  Q6aiii)
		
  9.i.	Vpk = 0.075V	(1992 PII  Q8ai)
   ii.	Rf = 120kW	(1992 PII  Q8aii)

 10.i.	Voutput = 3.64V	(1999 PII  Q8ai)
   ii.			(1999 PII  Q8aii)

 11.i.	Rfeedback = 1.6kW	(1996 PII  Q8bi)
   ii.			(1996 PII  Q8bii)

 12.i.	Vout = -10V
   ii.	Vout = -13V to -15V (amplifier saturated)

 13.i.	Vout(peak) = 2.55V
   ii.  

 14.	 A		(1994 PI  Q20)

 15.i.	Inverting mode	(1993 PII  Q7ai)
   ii.	Gain = 500	(1993 PII  Q7aii)
  iii.			(1993 PII  Q7aiii)
			
   iv.			(1993 PII  Q7aiv)


Differential Amplifier 

  16.	Rfeedback = 100kW
	V1 = 0.2V
 
  17.i.	The difference between V2 & V1 is calculated.
	Subtraction is the operation.
    ii.	Vout(peak) = 4.1V

  18.i.	Differential mode	(1996 PII  Q8ai)
    ii.	Vout = 0.5V		(1996 PII  Q8ai)

  19.	Vout = 0.2V		(1992 PII  Q8bi)

  20.a.	Vout = +1.5V 	Transistor conducts and bulb is on.
     b.	Vout = -1.5V 	Transistor does not conduct and bulb is off.
     c.	Vout = -4.5V 	Transistor does not conduct and bulb is off.
     d.	Vout = +4.5V 	Transistor conducts and bulb is on.

  21.a	Vout = +4.0V
     b.	Vout = -1.0V
     c.	Vout = +5.0V
     d.	Vout = -5.0V
     e.	Vout = -13V to -15V (amplifier saturated)

  22.i.	0V			(1997 PII  Q6bi)
    ii.	(VY-VX) = 0.018V		(1997 PII  Q6bii)
   iii.				(1997 PII  Q6bii)

  23.i.	Vout = (Rf/R1)(V2-V1)	(1993 PII  Q7bi)
    ii.				(1993 PII  Q7bii)