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

Thread Starter

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:

Load Resistor Value __Load Voltage (VL) __Load Current__ Ripple (Pk/Pk)
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. :confused:

PLEASE HELP!!!!!
 

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
 

Thread Starter

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
 
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