power supply question

power supply spike filter 7805

I want to control a single phase induction motor(used in a water pump) with my 89S51 uC. The problem is that as the motor draws a very large current at start it causes a momentary voltage drop on the main supply. As I want to use a power supply made with a transformer,bridge,capacitor and a 7805,connected to the main supply, to give power to the uC I am worried that it might reset the microcontroller when the motor is switched on. Also I am using a 12V relay so I have to use a 7812 as well.

I plan to use a 220V/25V transformer in my supply. Will it work without resetting the uC?
Are there any precautions that I should Observe?

7805 input 30v

for this you can use large filter capacitor in rectifier output.
With 24-25 Volt input , if your circuit draw current more than
about 40-50mA it is better to use switching regulator.
for example MC34063
or LM2575 or LM2576 ..

Regards,
Davood.

power supply 7812 470uf

Hi,

Why you want to use 25VAC to made low voltage supply. In this case you will get voltage over 30VDC, which you want to decrise to 5V => you will get lot of heat using 7805.

Use 12VAC, bridge, capacitor and 7805. To avoid problems with short voltage drops, I always put ceramic condensator of 100nF as near as I can to mcu, or use some tantal condensator of 4.7uF or more very near to power pins of mcu.


Regards,

Mr.Cube

7805 heat at 500ma

Why you want to use 25VAC to made low voltage supply. In this case you will get voltage over 30VDC, which you want to decrise to 5V => you will get lot of heat using 7805.

Use 12VAC, bridge, capacitor and 7805. To avoid problems with short voltage drops, I always put ceramic condensator of 100nF as near as I can to mcu, or use some tantal condensator of 4.7uF or more very near to power pins of mcu.


Regards,

Mr.Cube

Click to expand...

I am using 25V AC because as I mentioned I am using a 7812 as well for the 12V relay and the minimum required input voltage is 14.7V. After the capacitor I will get greater than 30V ! i didnt concider that (thanks for pointing it out) what value of AC voltage do you think I should apply at the bridge in this case?

Also does the heat generated by the 7805 depend on the input voltage of 7805 ??

for this you can use large filter capacitor in rectifier output.
With 24-25 Volt input , if your circuit draw current more than
about 40-50mA it is better to use switching regulator.
for example MC34063
or LM2575 or LM2576 ..

Regards,
Davood.

Click to expand...

The problem is that my circuit is drawing more than 500mA current !!! and I only want to use 3 termal regulators. I am ready to use a large heatsink.

Is there any formula to calculate the size of the heatsink ?

7812 220uf

230/25V is an over kill and you will need a large heatsink on the regulator. Use 230/12V. This will give you around 16V DC (12 x 1.4) which is enough for the 7812 and use 1000uF filter capacitor. Near the VCC pin of the uC use 0.1uF capacitor. The voltage drop across 7812 will be 16-12 = 4. The dissipation will be 4 x 0.5 = 2 watts. You can use a P48 heatsink but mind you, the heatsink will need air circulation. If you have totally enclosed it in a box, then the heat buildup will be more.

Cheers

Ravi

lm2576 vs 7805

what value of AC voltage do you think I should apply at the bridge in this case?

Click to expand...

As I told you, use 12VAC voltage on secundar of transformer.

12VAC*1.41-1.4V=15.5V which are surely enough for making 12V with 7812.

You can make 5V from this 12VDC but be carefull becouse of spikes of voltages when relay is turning off (use lot of capacity).

Also does the heat generated by the 7805 depend on the input voltage of 7805 ??

Click to expand...

Of course it does!!!!

Meaning:

Lets suppose that you have your 30VDC input voltage in your 7805, and that your circuit sink 500mA.

In this case you have (30V-5V)*0.5A => 12.5W which is to much, and you need biiiiig heatsink.



Regards,

Mr.Cube

7812 filter capacitor

As I told you, use 12VAC voltage on secundar of transformer.

12VAC*1.41-1.4V=15.5V which are surely enough for making 12V with 7812.

You can make 5V from this 12VDC but be carefull becouse of spikes of voltages when relay is turning off (use lot of capacity).

Click to expand...

You calculated the output dc voltage from the formula (12*1.41 - 1.4) = 15.5V DC. This is the no load voltage right? Now when 500mA will be drawn from this supply the voltage will decrease?, although it will still remain above the minimum
threshold of 7.3V for the 7805 but it will become very close to the minimum
threshold of 14.6V for the 7812(it might even go below it !!)

and remember that my primary concern was to avoid the effect of voltage drop which occurs due to starting of the motor.That was the reason I was planning to use 220/25V transformer so that during switching, evem if the voltage drops to say 18V or 20V, it remains safely above the threshold limit of both 7805 and 7812 so that they continue to provide constant output voltages uninterrupted.

Also, you suggested 0.1uF capacitors near to the microcontroller to avoid spikes. Will these capacitors bear the voltage drop due to starting of the motor(can starting of the induction motor be concidered as a spike ?,i mean is the time duration short enough to be concidered as a spike ?).

7812 with heat sink

7805 Can't handle 30 Volts or more on its input. Maximun is 25 Volts to my knowledge. Best is to use a 12 Volts transformer and a 220uF C direct after the diode bridge. Then the 7812 for your relay. Use a diode in reverse across your relay coil. Then a diode from the 220uF C to your 7805 input, and place an additional C of 470uF on the input. Use the 5 Volts for the CPU only and use a different 7805 direct from the 220uF C for other 5 Volt devices that don't reset.

Good luck/

The Data Sheet of the 7805 says that it can handle upto 35V of input voltage. I am attaching a table of the data sheet which specify the absolute maximum ratings.