Continue to Site

# Power Regulation question

Status
Not open for further replies.

#### psecody

##### Member
Hello I was looking for a way to connect power from say 9v battery or 4 AA batteries to a project I'm designing. Since the servo and the microcontroller both use 5V I was going to put an 7805 between the batteries and the circuits so do this and I saw **broken link removed** website and was wondering something about it. He puts the caps in there to smooth out any fluctuation right? Now the real question, how would you decide on those values? What formulas do you use to figure this? Thanks I've just been wondering on things like that with capacitors how you figure out the values needed.

For power supply caps, one general rule is the bigger the better. Of course there is a trade-off at some point, but it works in general. It doesn't mean a super cap is better than an aluminum.

The calculations are on the board here in many threads. Basically, to find the minimum cap size you need, you need to know what your input ripple voltage and output current draw.

The input capacitor has to fill in any voltage ripple from the input (from rectified AC mains usually) and the output filter is to buffer any current draw surges from the device you are powering.

The website you found is wrong.
A little 9V battery quickly drops below 8V which is the minimum input voltage of a 7805 regulator. 6V from four brand new AA cells that slowly drop to 4V will not work.
You need a low-dropout regulator that works perfectly until its input drops below 5.5V or a little less.

The datasheet of a voltage regulator shows recommended or mandatory capacitors.

ok say you were taking from 120ac step it down to say 12v then go through a full wave rectifier and you get a wave

so the peak would be 12v?

well I don't guess the caps before the regulator would be necessary if using batteries right since they're DC there wouldn't be anything to need filtering. Right?

Sorry for the elementary questions, capacitors have always kind of been a hazy area for me. I'm starting to think I jumped in too far in electronics without a strong foundation in the basics so I'm trying to go back and fill in the stuff I have trouble with or aren't as sure about. I'm also trying to figure out more of how to design my own projects from scratch, by knowing what formulas and ways to calculate things instead of just trial and error. Thanks again for the help and I'm sorry again for the elementary-ness of my questions.

Last edited:
If the transformer is 24V center-tapped then the average voltage is about 12V. The peak voltage is 17V.

ok say you were taking from 120ac step it down to say 12v then go through a full wave rectifier and you get a wave

well I don't guess the caps before the regulator would be necessary if using batteries right since they're DC there wouldn't be anything to need filtering. Right?

Wrong! The output from the rectifier is DC, but it's not a constant level. It's essentially a DC pulse and you want a steady DC levl. You can only get that from smoothing with a big parallel connected capacitor.

EDIT: You put up a graphic showing an AC transformer and recitifier, and then asked about a battery. It be clear, your diagram requires a capacitor to convert the pulsing DC to a smooth DC. A battery does not requre anything in the picture nor a smoothing cap.

Last edited:
yeah sorry about that I was using the picture more for the waveform on the right to kind of ask about the voltage ripple. I was thinking that the ripple would be 12v since that would be the max of the wave form. audioguru where did you get 17V from? Could you elaborate on that some more?

The part where I asked about batteries was more of a statement. I was just processing the original circuit and was reading smanches description. The caps before the regulator are for smoothing ripples and the ones after are for current draw surges. It was kind of my way of saying "oh so really IF I was just using batteries as my main source the caps before the regulator aren't necessary." I was just stating it in my message just so if for some odd reason I was wrong then someone would catch the error and point it out to me.

12VAC is 34V peak-to-peak.

The rectifier cuts off half so the result is 17V peak. A filter capacitor charges to about 16V.
The peak voltage of a sine-wave is 1.414 times (the root of 2) the RMS voltage.

At 60 Hz, the general "rule of thumb is" 1000uF per Amp of load as a filter.

For DC, of course, filtering isn't necessary. In both cases, small value caps are often used to catch spikes. Usually, less than 1uF. As far down as .0047uF, depending on the application.

As noted above, 78xx regulators drop 3 volts or so. The higher the supply voltage, the more current through the ground pin(78xx center pin) and higher the dissipated heat.

12 VAC is the RMS value, once rectified, it will be much closer to 17 volts beause you are filtering to the peak voltage.

Back in the day, when tube type transformers were still around, the 6.3 volt filiment leads would provide 8 volts(m/l) once rectified and filtered. Also available was(is?) the same rating for use on "S-100" bus micros. Linear power supplies use lots of copper.... Either was imminently suitable for a +5 volt supply. 7805's up to an amp, LM309's up to 5 amps with heat sinks.

Now, batteries.... A 9 volt MN1604 (transistor radio) battery doesn't have the ampacity to hold its' nominal voltage very long. An option for extended use is a lantern battery, either 6 volt or 12 volt. Find 'em at the BigBlueBox in the camping section. For 12 volts, you would need the regulator. For a 6 volt, 1 (or 2) 1N4006 diode in series will drop the voltage for all but the most sensitive use. If you are using standard TTL, you can operate between 4.2 and 5.1 volts, period! CMOS and LS-TTL are more forgiving. Consumer electronics.... you're on your own there, I would have to see the circuit.

As noted above, 78xx regulators drop 3 volts or so. The higher the supply voltage, the more current through the ground pin(78xx center pin) and higher the dissipated heat.
The current in the ground pin doesn't change much, and does not account for the increased dissipation. Typically 95% or more of the heat is (Vin-Vout)*I, where I is the current flowing through the path from input to output.

Last edited:
Status
Not open for further replies.

Replies
10
Views
619
Replies
2
Views
827
Replies
4
Views
1K
Replies
6
Views
1K
Replies
4
Views
1K