Lab 10 RS

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Lab 10 Unregulated power supply


Use a transformer for the experiment.

here is a description of the transformer.

File:TF EIM 241 transformer.pdf

File:IN5230-B-T DataSheet.pdf

Half-Wave Rectifier Circuit

1.)Consider building circuit below.

TF EIM Lab10 HW Rectifier.png

Determine the components needed in order to make the output ripple have a ΔV less than 1 Volt.

The output ripple can be found by ΔV=IΔtC

Taking AC signal from outlet equals to 60 Hz my input pulse width is Δt=160 sec=17 ms and using say C=2.2 uF I need my current to be:

I1 V2.2 uF17 ms0.129 mA

Taking R100 kΩI=12 V100 kΩ0.12 mA

that satisfy the condition above for current so my output ripple becomes less than 1 Volts.



List the components below and show your instructor the output observed on the scope and sketch it below.

I have used the following components:

[math]R = 96.9\ k\Omega[/math]
[math]R_L = 98.7\ k\Omega[/math]
[math]R_{scope} = 1\ M\Omega[/math]
[math]C = 2.2\ uF[/math]
[math]\mbox{Zener}\ \mbox{Diode}\ 4.7\ V[/math]

and the following input parameters:

[math]\Delta t = 17\ ms[/math]
[math]V_{in} = 12\ V[/math]


The current through the circuit can be found as I=VinRtot

where Rtot=R+|RL1jωCRL+1jωC|=R+(RL1+jωCRL)(RL1+jωCRL)=R+(R2L1+(ωCRL)2)

=96.9 kΩ+(98.7 kΩ)21+(2π 60 sec1)2(2.2 uF)2(98.7 kΩ)2=96.9 kΩ+1.2 kΩ=98.1 kΩ.

And the current becomes I=12 V98.1 kΩ=0.122 mA

So my output ripple becomes ΔV=0.122 mA17 ms2.2 uF=0.9V


Tek00047.png

As we can see from the sketch above my output voltage has ripple about ΔV=0.710 V<math>,rippletimeisthesameasinputtime<math>Δt=16.6 ms, and the output DC voltage is about Vout=0.6 V

Full-Wave Rectifier Circuit

TF EIM Lab10 FW Rectifier.png


Determine the components needed in order to make the above circuit's output ripple have a ΔV less than 0.5 Volt.


The output ripple in this case can be found by ΔV=I(Δt/2)C

Because we are now using (Δt/2) instead of (Δt) than in previous case for half-wave rectifier theoretiacally we will be able to make ΔV two times less then in previous case just by using exactly the same ellements and input parameters as before:


List the components below and show your instructor the output observed on the scope and sketch it below.

I have used the following components:

[math]R = 96.9\ k\Omega[/math]
[math]R_L = 98.7\ k\Omega[/math]
[math]R_{scope} = 1\ M\Omega[/math]
[math]C = 2.2\ uF[/math]
[math]2\ \mbox{Zener}\ \mbox{Diode}\ 4.7\ V[/math]

and the following input parameters:

[math]\Delta t = 17\ ms[/math]
[math]V_{in} = 12\ V[/math]


Because now I need to replace Δt by (Δt)/2 my output voltage now becomes two times less:

ΔV=0.9 V2=0.45 V


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