Question

The following circuit, containing two comparators and one LED as well as 4 resistors, is excited by a triangle input voltage. a) Find the value of the output voltage over the same range of the input. Plot this graph against time right below the input graph with the same scale. Use vertical dashed lines between the input and output graph to show corresponding time points. b) In above graph, indicate where the LED will be ON. Circuit: +5V UIA LM339AN TL 307CK UB LM339AN Input Voltage:

Answer 1

a)

+VY UA VIT24 LM339AN Vout2 LTL307CK VNT2 +5V Vs UIB VIT16N LM339AN 1 Vout1 vs th

From the given circuit VNT2 and VIT1 are constant.

VNT2 =

$5\times \frac{(1+1)}{(1+1+2)}=2.5V$

VIT1 =

$5\times \frac{1}{(1+1+2)}=1.25V$

By the principle of Comparators, when VNT VIT , then Vout = +VS ; when VIT VNT , then Vout = -VS.

For t < 1.25, VIT1 > VNT1  $\because$ VNT1  = Vin.

$\therefore$  Vout1 becomes -Vs ,i.e. Vout1  = 0V.

$\therefore$  Vout  =0V(Ground), irrespective of the voltage Vout2 , for t < 1.25.

For 1.25 < t < 2.5, VNT1 > VIT1 ,

$\therefore$  Vout1 becomes +Vs ,i.e. Vout1  = +5V.

And Vout2  is already +5V, $\because$   VNT2 > VIT2 for 0 < t < 2.5.

$\therefore$ Overall Vout  = +5V for 1.25 < t < 2.5.

For 2.5 < t < 7.5, VIT2 > VNT2 , Vout2 = Vout =0V.

For 7.5 < t < 8.75, Vout1 = Vout2  = Vout = +5V

And for 8.75 < t < 10, Vout1 =0V, $\therefore$ Vout = 0V.

Vin 5v 3.75V 2.5V 1.25V JO 1.25 2.5 3.75 5 6.25 7.5 8.75 10 Vout

b)

Since in the circuit, the cathode of the LED is connected to the output of the Comparator,

$\therefore$  The LED will turn on for Vout = 0V, and turn off for Vout = +5V.

$\therefore$ LED will be ON for 0 < t < 1.25 , 2.5 < t < 7.5, and 8.75 < t < 10.