• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

12V PSU Schematic

Status
Not open for further replies.
Hi I am new to schematics and power supplies and I am in the process of designing and simulating a PSU based on the following specifications:

230 V smoothed unregulated power supply
 Output voltage 12 V
 Output current 2 A
 ripple factor < 0.024

I have used Proteus 8 as my simulation tool and I am struggling with getting the correct components for my simulation.

I have attached a schematic that I have been trying to put together, however I am not getting the required values.

Would anybody be able to have a look and point me towards what I have done wrong?

Thanks
 

Attachments

atferrari

Well-Known Member
Please post a graphic file or a .pdf.
 
How would I account for this value as I am struggling with the transformer settings. I am given a coupling value, which I have set at 0.12. The way I have been taught is to look at the winding ratios. Thanks
 

MaxHeadRoom78

Well-Known Member
The resultant DC is 1.4 x the AC supply when using a smoothing circuit, or conversely AC = .707 x DC value.
Max.
 

spec

Well-Known Member
Most Helpful Member
Aside from the values is there anything wrong with the physical design of the psu?
No, it is fine.:)

(1) Off-load the output voltage from a bridge rectifier as you have drawn is.

(1.1) Vout DC= V secondary RMS * 1.414 (root 2) - 1.2V (two off-load diode drops)

(2) The peak to peak ripple voltage across a reservoir capacitor = (I * T)/C

Where
I = current drain in Amps
T = period of mains voltage/2 in seconds
C = the reservoir capacitor value in Farads

The big unknown, in any power supply like this, is the characteristics of the transformer which controls what the output voltage will be under load.

As a very rough rule of thumb, with 50Hz/60H mains transformers, is that the peak voltage of the secondary under the load current specified for the transformer will be equal to the specified off load RMS Voltage of the secondary.

For example, if you had a transformer specified with a secondary of 12V RMS at a current of 1 Amp (12Watt transformer) and you used it in the circuit you posted with a current drain of 1A, the out put voltage would be 12V- 2V (two diode drop) = 10V. The reason why the diodes now drop 2V instead of 1.2V as when off-load is because the diodes are now conducting a heavy current. In fact the diodes charge up the reservoir capacitor with huge bursts of current at the negative and positive peaks of the secondary voltage sine wave.

If the reservoir capacitor was 5,000uF and the mains frequency was 50Hz, then the peak to peak ripple Voltage across the reservoir capacitor would be:

(1A * 10 mili seconds)/ 5,000 micro Farads = 2V peak to peak.

spec

PS: the above is very simplified but it will get you by for the vast majority of power supply designs.
 
Last edited:

spec

Well-Known Member
Most Helpful Member
Only as long as you need the DC common to be a earth ground potential as shown.
Max.
That is true, but it is not an uncommon arrangement. Of course, you can move the earth connection to the positive output of the reservoir capacitor and have a negative supply rail.:)

spec
 

alec_t

Well-Known Member
Most Helpful Member
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top