Day 17 1st Order Circuits contc

We started the day with talking about first order op amp circuits. Capacitors and inductors apply three useful properties in  electric circuits. Caps store energy, Caps oppose abrupt changes in voltage, Inductors oppose abrupt changes on current, this is useful for spark and arc suppression and for changing pulsing dc voltage into a smooth dc voltage. Lastly they both are frequency dependent, which in turns make them frequency discrimination for ac applications.

We were asked to show the algebra of an Integrator Op-amp. White board work below.

White Board work of an Integrator Op-amp. use Ir = Ic and Ic = -CdVo/dt. Gain is -1/RC and integration of Vin which is -Vin/Tao.

Next we were asked to another example with values and show the signal going in of the Op-amp.

White Board work: We take an input signal and take the derivative and negative 1.

White board work of the Differentiator. Not the same integrator, since output is -RC(derivative) and not -1/RC(integrate)

White board work: showing the output real. It is the gain(-1) times the derivative of input signal.

Afterwards we move onto the lab of the day: Inverting Differnetiator.
Prelab: Is done on the white board. C1 = .1uF, R = 1M ohm, RC = -1
Vout = -1/RC dVin/dt Vin = Acos(wt) w = 2f,  Vout = 1/(1.04)(w)sin(wt)

White Board work: Showing our circuit with values on the Op amp. Note R of 10M is 9.6M

White board work of voltages in at the different freq. 2khz,1khz and 500 hz.

White board work: The lab asked for 1V offset but we used 5mV,5mV and 2mV instead, don't remember why. The order is 2khz, 1khz, and 500 hz.

We then took the amplitude reading of the values. Channel 2 is Vin, Channel 1 is Vout.

White Board work: 500hz amp in 2.7mV, 2khz amp in 2.68mV, and 1khz amp in 2.69mV. Their respected Amplitude out are: 3.3V,2.56V and 3.48V. This is the tablet for post lab. Percent error is further down. There are difference between expected and measured, the differences believe are because of using the wrong formula and/or not the right input voltage/different RC.

Lastly our comments on the lab, Vout is Pi/2 out of phase of Vin and if frequency goes up Amplitude goes down.

White board work: Our conclusion of the lab, Vout is pi/2 out of phase and frequency goes up Amplitude goes down.

Here are the pictures required of the post lab: Pics of different frequency 2khz,1khz, and 500hz

Picture of the 1khz signal, yellow is output signal, cosine, blue is input signal cosine plus phase shift.

Picture of the data from the 1khz signal: Vout =3.2 V, Vin = 2.68 mV

Picture of 2khz signal, Yellow is Vout, Blue is Vin. Ch1, Ch2.

Picture of the 2khz signal, Yellow is cosine, blue is cosine +phaseshift. Vin, Vout.

Picture of the 2khz data: Vout = 1.6V, Vin= 2.68 mV

Picture of 500hz signal. blue is Vin, Yellow is Vin.

Picture of the 500 hz signal: Yellow is Vout, Blue is Vin.

Picture of Data for 500hz: Vout = 3.3V, Vin = 2.68 mV.

Here is a picture of the circuit used.

Picture of the circuit used: Op amp, .1uF and 10M (9.6 M real) resistor.

For the conclusion of the lab: POST LAB
5*cos(2*pi*f) w1 = 12566,w2 = 6283, w3 = 3141 1/RC = .96 Vin 2.68mV
-1/RC *Vin*w1*sin(w1) = 2.68 V, real is 2.56 %diff = 3.7%
-1/RC*Vin*w2*sin(w2) = 4.7 V, real is 3.48 %dif = 25%
-1/RC*Vin*w3*sin(w3) =4.5 V, real is 3.3 %dif = 26.6%

I do not feel confident is these percent error's,  I tried using different input voltages and the numbers calculated do not add up, I do believe we recorded the right numbers. I think I must be using the wrong formula :(

 We then moved onto switching functions! We started with the delta and what it represents. An area of 1.

White board work of delta, delta = 1!

We then tried an example of using switching formula.

White board work of the switching example. Ic = CdVct/dt. The derivative of u is delta the derivative of rt is u.

 Vc at 5 at0.

More white board work shown.

White board work: Ic, the impulse is .6 at .5.

Next we move onto the step response of RC circuit.White board work below.

White Board work of the step respond. V/R times e of -t/tao times the step respond.

We then tried another example by using what we learned and did on the white board.

White board work: Vs = 5u, Vintial = 0, Tao = 5. Tao is Req times C.

More white board work shown. The final formula needed. Vinf + [Vnot - Vinf]e to the -t/tao.

White board work with the equation in standard form.

White board work with numbers plugged in.

White board work: Cdv/dt - .5(-5/4)(-1/5)e^-t/Tao.

We then move onto the Step Respond of a RL circuit. i = in+if. In = Ae^-t/Tao, Tao =  L/R. We tried an example shown on the white board.

White Board work. We needed to find Inot, then Req for Tao. Then we used node voltage to find Iinf.

We came to the conclusion formula of i(t) = 5/6(1+e^-t)A

In summary we started with op amps. One being the integrator and the other being the Differentor. The integrator integrated Vin, and the differenetor differentiated Vin by the op amp. We then moved onto the lab of the which was the differentor. We found our numbers to not match the expected values with percent different errors up to 26% but we believe that is due to not following the instruction in the lab with the right input voltages are the right resistors. We do think we got resonalbe data. It seemed from the pictured that the Vout was 90degress out of phase from the input which makes sense and the higher the frequency the lower the amplitude was recorded. Lastingly for the day we talked and derived step response logic for both the RC and RL circuit. For RC the initial capacitor voltage is 0, final is Vinf and then find Tao. For RL the initial inductor current is 0, final current if Iinf and then find Tao. last thing we talked about was delay circuits and relays.