Lab 10 RS

Go Back to All Lab Reports

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.

Determine the components needed in order to make the output ripple have a [math]\displaystyle{ \Delta V }[/math] less than 1 Volt.

The output ripple can be found by [math]\displaystyle{ \Delta V=\frac{I\Delta t}{C} }[/math]

Taking AC signal from outlet equals to [math]\displaystyle{ 60\ Hz }[/math] my input pulse width is [math]\displaystyle{ \Delta t = \frac{1}{60\ sec} = 17\ ms }[/math] and using say [math]\displaystyle{ C = 2.2\ uF }[/math] I need my current to be:

[math]\displaystyle{ I \le \frac{1\ V \cdot 2.2\ uF}{17\ ms} \le 0.129\ mA }[/math]

Taking [math]\displaystyle{ R \ge 100\ k\Omega \Rightarrow I = \frac{12\ V}{100\ k\Omega} \le 0.12\ mA }[/math]

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

The current through the circuit can be found as [math]\displaystyle{ I = \frac{V_{in}}{R_{tot}} }[/math]

where [math]\displaystyle{ R_{tot} = R + \left|\frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}} \right| = R + \sqrt{ \left(\frac{R_L}{1 + j\omega CR_L}\right)\left(\frac{R_L}{1 + j\omega CR_L}\right)^* } = R + \sqrt{ \left(\frac{R_L^2}{1 + (\omega CR_L)^2}\right) } }[/math]

[math]\displaystyle{ = 96.9\ k\Omega + \sqrt{\frac{(98.7\ k\Omega)^2 }{1 + (2\pi\ 60\ sec^{-1})^2(2.2\ uF)^2 (98.7\ k\Omega)^2} }= 96.9\ k\Omega + 1.2\ k\Omega =98.1\ k\Omega }[/math].

And the current becomes [math]\displaystyle{ I = \frac{12\ V}{98.1\ k\Omega} = 0.122\ mA }[/math]

So my output ripple becomes [math]\displaystyle{ \Delta V = \frac{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V }[/math]

As we can see from the sketch above my output voltage has ripple about [math]\displaystyle{ \Delta V = 0.710\ V }[/math], ripple time period is the same as input time [math]\displaystyle{ \Delta t = 16.6\ ms }[/math], and the output DC voltage is about [math]\displaystyle{ V_{out} = 0.6\ V }[/math]

Full-Wave Rectifier Circuit

Determine the components needed in order to make the above circuit's output ripple have a [math]\displaystyle{ \Delta V }[/math] less than 0.5 Volt.

The output ripple in this case can be found by [math]\displaystyle{ \Delta V=\frac{I (\Delta t/2)}{C} }[/math]

Because we are now using [math]\displaystyle{ (\Delta t/2) }[/math] instead of [math]\displaystyle{ (\Delta t) }[/math] than in previous case for half-wave rectifier theoretiacally we will be able to make [math]\displaystyle{ \Delta V }[/math] two times less then in previous case just by using exactly the same elements and input parameters as before. But here I realized by experiment that my Zener diode has low power limit and for this kind of circuit I need to use more powerful diode. Instead of 4.7 Zener Diode which has power limit 0.5 W I used rectifier diode 1N4001 which has power limit 50 W.

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]\displaystyle{ R = 96.9\ k\Omega }[/math]
[math]\displaystyle{ R_L = 98.7\ k\Omega }[/math]
[math]\displaystyle{ R_{scope} = 1\ M\Omega }[/math]
[math]\displaystyle{ C = 2.2\ uF }[/math]
[math]\displaystyle{ 2\ \mbox{1N4001}\ \mbox{Diode}\ 50\ W }[/math]

Because now I need to replace [math]\displaystyle{ \Delta t }[/math] by [math]\displaystyle{ (\Delta t)/2 }[/math] my output voltage now becomes two times less:

[math]\displaystyle{ \Delta V = \frac{0.9\ V}{2} = 0.45\ V }[/math]

Now my DC output voltage becomes about [math]\displaystyle{ V = 5.2\ V }[/math] and ripple you can not even see for that scale.

To see ripple I used the AC mode of scope and they are plotted below. As we can see from this sketch my output voltage has ripple about [math]\displaystyle{ \Delta V = 68.4\ mV }[/math], and the ripple time period now is two times less than input time period and is about [math]\displaystyle{ \Delta t = 8.4\ ms }[/math].

Now I have almost perfect output DC current with small ripple.

Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement