Discussion in 'General Electronics Chat' started by eminthepooh, Aug 26, 2012.

1. ### eminthepooh Thread Starter New Member

Aug 26, 2012
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I'm designing a tiny controller box which needs DC power for a micro. I want to bring in a single AC power cord in and thru transformer/rectifier/capacitor make it into DC, finally there will be a regulator to give me 5VDC.

I had a transformer which take 120VAC down to 16VAC, and my big Cap connected the output of the recifier is a 10KuF 25V cap. When I plugged in this part of the circuit, the output of the DC charged to 25V and upon unplugging, it stayed at 25V. Luckily, I hadn't connected the 5VDC regulator and logic parts of the controller yet.

My questions are:
1) Why is DC output at 25V and not 16V.
2) What is a safe way to have the big cap drain?
3) Is the Cap too big? what should I use instead?

Thank you,

2. ### cork_ie Member

Oct 8, 2011
348
58
There are 2 reasons why you are getting 25V

1) 16V AC is RMS voltage not peak, when you rectify your AC supply through a bridge rectifier you will always get a higher DC Voltage.

Vpeak=√2 VRMS

Peak voltage, in your case, would be (1.414*16) = 22.63 V
The smoothing capacitor will level out the remaining ripple after the rectifier and you will approach peak voltage , less the forward voltage of the rectifier diodes, normally around 0.7V +&- for silicon diodes.

2) Transformer output voltage generally varies with load , this is called regulation. In general no load voltage will be several Volts higher than when the load is applied to the transformer. Obviously transformer power rating and several other factors such as core design will affect the amount of regulation.

Last edited: Aug 26, 2012
3. ### #12 Expert

Nov 30, 2010
16,337
6,820
When you rectify AC into a capacitor the capacitor charges to the peak voltage of the AC wave which is 1.414 times the RMS value. 16 x 1.414 = 22.6 volts and apparently your transformer was not loaded down with its rated load, so it was supplying the highest voltage it could with no load.

If you want 5 volts, you should start with a much smaller voltage, like 8 volts AC. Use 4 diodes to get twice the frequency of the input and lose about 1.5 volts off the peak value. Then a filter capacitor and a 7805 regulator chip to get to 5 volts DC.

To know the capacitor size, I would have to know how much current you will be using and what country you are in (the frequency of the power line).

edit: haha. Simultaneous posts

4. ### cork_ie Member

Oct 8, 2011
348
58
"Great minds think alike & fools seldom differ" it's old but how true

5. ### eminthepooh Thread Starter New Member

Aug 26, 2012
27
0
Thank you guys for the replies. How do you calculate the capacitor value based on the current and hertz #12?
I'm assuming about 200mA at 60Hz (US) formy system.

Also, how should the capacitor bleed out that charge after the device has been unplugged? It seems dangerous to let that sit charged
Thank you,

6. ### #12 Expert

Nov 30, 2010
16,337
6,820
1.414 C Eripple Frequency = current in amps.

For a 4 diode bridge, the frequency is 120
For a single diode, the frequency is 60

Whatever excess voltage you have after figuring the 5 volts of output and the voltage used up by the regulator, the ripple voltage must be less than that.

and leaving 5 or 10 volts in a capacitor is not all that dangerous. Actually, it's healthier for the capacitor to leave a little bit of charge in it. Keeps it polarized.
You can put a resistor across the filter capacitor to let the leftover voltage run to ground after the power is turned off.

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7. ### eminthepooh Thread Starter New Member

Aug 26, 2012
27
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Thank you #12, I appreciate the explanation.