Summer 2016 – Q.3
3. (a) Draw a neat diagram of 3-input inverting summing amplifier and obtain expression for its sip voltage. [6M]
In this simple summing amplifier circuit, the output voltage, ( Vout ) now becomes proportional to the sum of the input voltages, V1, V2, V3, etc.
$latex \begin{array}{l}I_F=I_1+I_2+I_3=-\left[\frac{V_1}{R_{in}}+\frac{V_2}{R_{in}}+\frac{V_3}{R_{in}}\right]\\\\Inverting\;Equation:\;V_{out}=-\frac{R_f}{R_{in}}\times V_{in}\\\\\therefore-V_{out}=\left[\frac{R_F}{R_{in}}V_1+\frac{R_F}{R_{in}}V_2+\frac{R_F}{R_{in}}V_3\right]\end{array}$
However, if all the input impedances, ( Rin ) are equal in value, we can simplify the above equation to give an output voltage of: Summing Amplifier Equation
$latex V_{out}=-\frac{R_F}{R_{IN}}\left(V_1+V_2+V_3……etc\right)$
| Sr. No. | Parameter | Asynchronous counter | synchronous counter |
| 1. | Circuit complexity | Logic circuit simple. | With increase in number of states, thee logic circuit becomes complicated. |
| 2. | Connection pattern | Output of thee preceding FF, is connected to clock of the next FF. | There is no connection between output of preceding FF and CLK of next one. |
| 3. | Clock input | All the FFs are not clocked simultaneously. | All FFs receive clock signal simultaneously. |
| 4. | Propagation delay | P.D.=n×(td) where n is number of FF and td is p.d. per FF. | P.D.=n×(td)FF+(td)gate. It is much shorter than that of asynchronous counter. |
| 5. | Maximum frequency of operation | Low because of the long propagation delay. | High due to shorter propagation delay. |
De-Morgan’s First Theorem
According to De-Morgan’s first theorem, a NOR gate is equivalent to a bubbled AND gate.
$latex \begin{array}{l}Z=\overline{A+B}\\\end{array}$
For the bubbled AND gate the equation is
$latex \begin{array}{l}Z=\overline{A\cdot B}\\\end{array}$
De-Morgan’s Second Theorem
De-Morgan’s Second Theorem states that the NAND gate is equivalent to a bubbled OR gate.
The Boolean expression for the NAND gate is given by the equation shown below.
$latex \begin{array}{l}Z=\overline{A\cdot B}\\\end{array}$
The Boolean expression for the bubbled OR gate is given by the equation shown below.
$latex \begin{array}{l}Z=\overline{A+B}\\\end{array}$
