Question

Design 1. Design a non-inverting Schmitt Trigger with switching points of VHL 0V and VH 5V 2. Design an op-amp integrator that uses no more than 1mA of input current and changes its output by 5V in 10usec when driven by the Schmitt Trigger designed in (1) When the integrator is connected to the Schmitt Trigger in the feedback arrangement below it will produce a relaxation oscillator that outputs a 5V triangle wave with a frequency of 50kHz. +9V C1 R1 U1 U2 LF35 R2 out LF35 Ri Rf

Answer 1

For a non-inverting Scmitt trigger V_HL and V_LH formula are as given below:

$V_{LH} = (1+\frac{R_{2}}{R_{1}})V_{-} + \frac{R2}{R1}V_{sat}$ ---------(i)

and

$V_{HL} = (1+\frac{R_{2}}{R_{1}})V_{-} - \frac{R2}{R1}V_{sat}$ ---------(ii)

Where V_sat is the positive rail saturation voltage and V_- is the voltage at inverting pin of the op amp. ( Refer the figure below)

o Vou

Given V_HL = 0 V and V_LH = 5 V. Adding equations (i) and (ii) we get

$V_{HL} + V_{LH}= 2\times (1+\frac{R_{2}}{R_{1}})V_{-}$

substituting V_HL and V_LH we get

$0 + 5= 2\times (1+\frac{R_{2}}{R_{1}})V_{-} = 5$

$(1+\frac{R_{2}}{R_{1}})V_{-} = 2.5$ -------------(iii)

subtracting equation (ii) from (i) we get

$V_{LH} - V_{HL} = 2\times\frac{R2}{R1}V_{sat}$

$5 - 0 = 2\times\frac{R2}{R1}V_{sat} = 5$

$\frac{R2}{R1}V_{sat} = 2.5$

Assuming V_sat = 8 volts ( considering the drop in the op amp) we get

$\frac{R2}{R1} = \frac{2.5}{8} = 0.3125 = \frac{5}{16}$

Thus if R2 = 5 kilo Ohms R1 = 16 kilo Ohms.

Substituting in equation (iii) we get

$1.3125\times V_{-} = 2.5$

or

$V_{-} \approx 1.9 \;V$

We use resistor divider to get the required V_- voltage. If R3 and R4 make up the resistor divider as shown

R3 R4

Then V_- is given by

$V_{-} = 9 \times\frac{R4}{R4+R3} = 1.9$

$\frac{R4}{R4+R3} = \frac{1.9}{9} = 0.211$

$\frac{R4+R3}{R4} = \frac{9}{1.9} = 4.737$

$R4+R3 = 4.737\times R4$

$R3 = 3.737\times R4$

$\frac{R3}{R4} = 3.737$

Thus if R4 = 1 kilo ohms R3 = 3.7 kilo ohms ( approximately ).

With these values the non-inverting schmitt trigger circuit looks as follows

V2 R3 3.7k U1 Vout V3 R2 LM741 1K V1 ŞINE(O 9 100) 5K tran 100m R1 16K

Note that the input source is just a test sine wave of amplitude 9 volts. The output of the circuit is given below:

VIvin) ANA 18ms 24ms 27ms 30ms 33ms 36ms 39ms 42ms 45ms 48ms

Next we design an op- amp integrator.

in out

By the concept of virtual short, current through resistor R is

$I = \frac{V_{in}}{R}$

In this case we want I < 1 mA. V_in comes from the schmitt trigger with amplitude 8 volts thus

$I = \frac{8}{R} \leq 1mA$

$\frac{8}{10^{-3}} \leq R$

So R should be greater than or equal to 8 kilo ohm. Lets take R = 10 kilo Ohms.

We want a change of voltage of 5 volts at the output of integrator when an input of 8 volts drive it in a time of 10 us. The output of integrator is given by

$V_{out} = -\frac{1}{RC}\int V_{in}dt$

We want the output to change by 5 volts in 10 us when input is 8 volts. Using these values

$(0-5) = -\frac{1}{10\times 10^{3}\times C}\int_{0}^{10\times 10^{-6}} 8 \;dt$

$C = \frac{1}{10\times 10^{3}\times 5}\times {10\times 10^{-6}}\times 8$

$C = 1.6 \times 10^{-9} = 1.6 \;nF$

Rigging up the circuit along with the schmitt trigger we get the circuit given below.

Note that in the schematic LM741 is used but it is not recommended for such a high frequency application. For demonstration the frequency is lowered by using 1.6 uF capacitor instead of 1.6 nF. The plot below depicts the output. The purpose of using LM741 was its a common op amp and is available everywhere.

C1 1.6n R3 3.7k V2 R5 10k tran 100m U1 U2 out 9V V3 R4 R2 LM741 LM741 Vin 1K 5K 9V R1 16K

5.4 3.6 585ms 576ms 567ms 558ms 549ms 540ms 531ms 522ms 513ms 504ms 495ms 486ms

I hope it helps. For any doubts or queries that you have, leave a comment below, I'll be happy to help. All the best. :)