# URGENT HELP RQ'D!!!!!!!!!

#### Matt_B

Joined Dec 5, 2008
2
I am currently a 2nd year HNC electronics student and am investigating unregulated non-linear power supplies which convert 230v AC I/P supply into 10v D.C supply by means of a transformer, bridge rectifier, reservior capactior and a low pass filter (smoothing circuit), as per the below diagram, except with a resistor in place of the inductor and a Load Resistor placed on the D.C terminals

Under simulation on a software package, changing the Load resistor in decade steps from 1Ω to 10KΩ gave the following results:

1 Ω................................ 99.38 mV.................. 9.38 mA .............60 mV
10 Ω ..............................920.5 mV ..................92.05 mA ...........440 mV
100 Ω .............................5.472 V ....................54.72 mA........... 460 mV
1000 Ω ..........................11.39 V ......................11.4 mA ............100 mV
10000 Ω......................... 12.57 V ....................1.269 mA ............10.8 mV

Now i understand what these results are telling me although i cannot understand or explain why the ripple (A.C component within the D.C supply) increases than decreases again.

#### bertus

Joined Apr 5, 2008
19,952
Hello,

What value has the resistor, that is in place of the inductor?
This is of influence of the output voltage.

Greetings,
Bertus

#### jpanhalt

Joined Jan 18, 2008
7,699
Perhaps you should consider the ripple as a percentage of the load voltage. Your load current follows the same pattern. Thus, the left side (including the resistor mentioned by Bertus) is current limiting.

John

#### Matt_B

Joined Dec 5, 2008
2
Thank you both for your prompt replies.

The values are as follows:

Reservior capacitor: 100μF

Filter: 100Ω and 100μF

These figures stay constant and the only value that changes is the load resistor.

Thanks again.

#### bertus

Joined Apr 5, 2008
19,952
Hello,

That figures it out.
At 100 Ω load and 100 Ω filter resistor you get about half the input voltage.
When you lower the filter resistor the voltage break will come later.

Greetings,
Bertus