12v 5v 2a dual power supply

i am having problems producing 2A.

It's important to be sure you're getting 2A from the transformer secondary.
Test with a 6 ohm load (or 8 ohms or 10 ohms or whatever value draws 2A at the voltage produced). Watch to see whether the voltage drops.

Then test for 2A at the output of the diode bridge. Watch to see whether the voltage drops.

Does the 12V regulator get very hot? If so then it may automatically go into shutdown mode. You may need a larger heat sink.
Same with the 5V regulator.

If you want both 12V and 5V supplies to be loaded fully, then the 7812 regulator need to carry 4A (too much even with a heat sink). The transformer must be rated at 12+5=17V *2A =34W; say 40-50W (just to keep some margin).

When the 5V2A output is active, the 5V regulator dissipates (12-5)*2=14W. With an input voltage of 18V at the 12V regulator, the 12V regulator too will dissipate a similar amount. Your transformer must be able to supply all the power...

Automobile lamps (12V; various powers) are handy for these loads.

By the way 7812 can supply max 1.5A (as per the std datasheet)

Hi,
Recently, the 78xx series gives output current of 1A. 7812 cannot output 2A. 7805 cannot output 2A either. Maybe you are going to find a way of safely connecting two of your circuits in parallel.

Your transformer too must be properly rated. Make some rough calculation like this for your transformer:

The ON semiconductor datasheet gives "Output voltage max = 5.25V @ (5.0mA <=Io <=1.0A, Pd <=15W) 7.5Vdc <= Vin <= 20V." And "Pd <= 15W" for 7812.
Using worst case;
7805 Pin = 5.25x1 + 15 = 20.25W;
7812 Po = 7805 Pin + 12×1 = 20.25+12.25 = 32.5W.
7812 Pin = 32.5 + 15 = 47.5W.
This is for one part.
2 x 7812 Pin = 47.5×2 = 95W for both parts.
Add rectifier diodes loss to this 95W. If its a 24V transformer then 95/24 = 3.958 = 4A. The loss may be 1.1V at 4A = 4.4W for one diode and hence 8.8W for two diode. Pin before bridge stage = 8.8 + 95 = 103.8W. Your transformer current rating must be greater than 103.8W/Vsec. If it's 24V, then 97/24= 4.325A. May be 20% margin will do (ie 5.19A). Maybe make use of 24V,6A transformer or 24V,5A transformer.

This method would do for your calculation. Goodluck.

Akanimo said:

...
If it's 24V, then 97/24= 4.325A. May be 20% margin will do (ie 5.19A). Maybe make use of 24V,6A transformer or 24V,5A transformer.
...

Click to expand...

I realized that your simulation shows a secondary voltage of 15V. The calculated power was 103.8W. So for a 15V transformer, the current will be 103.8W/15V = 6.92A =7A or greater.

Remember that this is worst case and your actual required current rating would be less than this.

Hi,

While the voltmeter at the primary side shows "AC Volts" ... the meter at the secondary side is connected to AC, but it's no AC meter.

103.8W/15V = 6.92A =7A or greater.

Click to expand...

I recommend "greater".
The formula calculates average rectified current...
But what's important for transformer rating is the "RMS current".

Don't be surprised when the RMS current is 130% .... 140% of the average current.
Thus I recommend to use a 150VA transformer.

Klaus

Yes Klaus, you're right. I didn't quite observe that.

So for a 15V transformer, the current will be 103.8W/15V = 6.92A =7A or greater.

Click to expand...

Max DC current due to load is 4A (ignore the current via the GND connection and the LED indicator). Because all the components are in series, the max AC current will be 4A RMS (assuming the filter cap is sufficient to give sufficient current at the lowest acceptable voltage for the 12V regulator.

The DC voltage at the filter cap will be 15V*1.4=21V (subtract the diode drops) ~19V; at 4A, the power needed will be 80W.

Max dissipation in the 7805 will be (12-5)V*2A=14W; max dissipation in the 7812 will be (19-12)V*4A=28W (you need to use a pass transistor for both but that will not alter the story). Total dissipation is 42W for an effective output of 34W.

Unless the filter cap is large enough, the regulator will shut down twice every cycle (the capacitor must not go below 14V at any time during the cycle) and the transformer supplies current only during 14V to 21V part of the cycle.

Hi

Because all the components are in series, the max AC current will be 4A RMS

Click to expand...

If the DC current is 4A....
Then the AC transformer secondary RMS current must be higher than 4A...please do a simulation.
Don't mix rectified_average_current (which indeed is 4A) with RMS current.

And for transformer rating you need to use RMS current, because this is responsible for heating in the windings.

Klaus

Even if you are using transformer with high current rating, you won't be able to get the desired output. Both of your IC are not having sufficient current rating.
You can use L78S12CV or similar IC after looking into their datasheets.
Also, when both 12V and 5V supplies are in use ,then your rms current will be more than 4A, so choose your transformer rating carefully for desired output power.