http://imageshack.us/photo/my-images/827/1110011245.jpg/ A) Show that thresold v

Question

http://imageshack.us/photo/my-images/827/1110011245.jpg/

A) Show that thresold value of Vs in the single ended comparator circuit shown in figure

Vs=(-R1/R2)xVref

B)Assume Vref=-10V; R1=10 ohm and R2=20 ohm sketch Vzero and Vs for -10=<Vs=<10 V


C)REpeat b if Vref = 10 V

Explanation / Answer

a. using superposition theorem, for only contribution of Vs the voltage at the negative input node is=(R2/(R1+R2))Vs for only contribution of Vref the voltage at the negative input node is=(R1/(R1+R2))Vref so total voltage at this node V- =(R2/(R1+R2))Vs+(R1/(R1+R2))Vref the threshold voltage will be the one when the negative end input and positive end input will be same. so, V-=0 and it gives Vs=(-R1/R2)xVref b. for Vref=-10V contribution to the negative node,=Vref*R1/(R1+R2) =-3.333V as the Vs changes from -10V->10V the contribution at the node changes -10*20/(20+10)V->10*20/(10+20)V =-6.667V->3.333V so in the range of the Vs -6.667->3.333 the output will be=-Vcc so in the range of the Vs 3.333->6.667 the output will be=Vcc in terms of Vs it will be=-10V->5V OUTPUT -Vcc 5V->10V OUTPUT +Vcc c. for Vref 10V the contribution will be=3.333V so the range of Vs will be -10V->-5V OUTPUT=-Vcc -5V->10V OUTPUT=+Vcc

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