Reducing Source Voltage To Power Supply

You can do it with a resistive divider, but what you really want to do is use a voltage regulator.

Radio Shack has strayed far from their roots, but if you're in the US you may want to go to the components section of one and see if they still have the little "Experimenter's Notebooks" -- you want the one that covers linear power supplies. Look in there for directions on making a regulated 6V supply, and while you're in the store pick up the parts you need.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
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Trying to use a resistive voltage divider to drive an active load is not likely to work out well, because then the load itself is part of the voltage divider. When the load increases and decreases, the voltage changes. In order for a resistive voltage divider to supply a somewhat steady voltage while supporting an active (varying) load, the impedance of the divider has to be very low, which means you would have to use chunky, low-value power resistors and waste lots of current. Just a bad way to do it.

I think your best bet is to use a LM7809 voltage regulator after the filter cap.

Trying to use a resistive voltage divider to drive an active load is not likely to work out well, because then the load itself is part of the voltage divider. When the load increases and decreases, the voltage changes. In order for a resistive voltage divider to supply a somewhat steady voltage while supporting an active (varying) load, the impedance of the divider has to be very low, which means you would have to use chunky, low-value power resistors and waste lots of current. Just a bad way to do it.

I think your best bet is to use a LM7809 voltage regulator after the filter cap.

To eliminate the heat problem, you can use a non-polarised capacitor between the transformer secondary and the rectifier bridge. This will drop some voltage similar to a resistor. I have always picked the value by trial and error, someone may speakup with the math required to calculate the proper value. I have found the speaker crossover caps to be useful in this application. WAG, 10uf to 100uf just give you a start. Also note the output voltage of your 9VAC transformer will drop down from your measured 12.5-13 VAC when you load it to its rated current. Mike

I just checked Radio Shack's web site. They don't carry the 7809, but you can use the adjustable LM317T and set the output voltage to 9. Here's the datasheet

Of course, using the 7809 would be a bit simpler but you would have to order it and have it shipped from an online supplier like Digikey.com or a miscellany house like allelectronics.com. To avoid shipping charges on a part that costs about a dollar, it makes sense to stop in at RS for the LM317T. Radio Shack's web site will even tell you whether the Radio Shack nearest you has the part in stock.

I don't live in the US, I live in Australia, and it just so happens that Dick Smith Electronics sells the 7809 regulator (Z6550). I would be interested in using this and need a diagram (simple how to wire it up. The power supply schematic has the transformer connected to the bridge rectifier, connected to the DC output with a cap or two across the output.

An LM317 is an alternative to the LM7809 others have been recommending. It is a variable regulator, in which the voltage is set by a voltage divider on the output. Radio shack does carry these.

BTW, your 9VAC transformer is just right for your device, which is 9VDC. Once you regulate it down, you'll be right on target, and the voltage won't dip (like it would with a 6VAC transformer).

Regards, Bob Monsen

--
You may want to consider something like one of these:
 
http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.


In any case, how much current does the radio draw and what current
is your wall-wart rated for when it\'s outputting 9V?

I'm a bit confused as to what the LM7809 regulators actually are. I've heard they're integrated circuits, but isn't the TO-220 case for power transistors or am I wrong?

The wall wart is rated at 1 amp. Not sure what the current draw is.

Yes, the 78xx family of voltage regulators are integrated circuits - but they only have three external connections (input, output, and ground), so a three terminal TO-220 package is quite suitable. They may also have to dissipate significant power, so they need to be in a heat-sinkable case.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

--- OK, here's what you've got: (View in Courier)

FWB +-----+ | +|----+--[7809]--+--->+9VDC MAINS>----P||S--|~ | +| | | R||E | | [BFC] | [LFC] MAINS>----I||C--|~ | | | | | -|----+-----+----+--->GND +-----+

Since your wall-wart puts out 9VRMS when it's loaded with 1 ampere, that's 12.73V peak. You'll lose about 1.4V of that across the diodes in the bridge, leaving you with about 11.3V into the regulator.

The 7809 has a worst-case dropout of 2.5V, so the bad news is that for a 9V output you'll need 11.5V into it, and you've only got

11.3V. :-(

The good news is that your wall-wart's regulation is poor and with a no-load output voltage of 12.5 it may be able to supply the voltage needed by the 7805. Let's see...

For a 1A load we'll have an output, from the bridge, of:

Vout(1) = (VRMS * sqrt(2)) - 1.4V

= (9V * 1.414) - 1.4V = 11.3 volts

For no load we'll have:

Vout(2) = (12.5V * 1.414) - 1.4V = 16.27 volts

That means that from no load to full load we'll have a voltage change of about 5 volts per ampere of current change.

So what?

Well, since the 7805 needs 11.5 V minimum into its input to provide

9V out, that means that (assuming the transformer's voltage VS current change is linear) the load can never draw more than:

Vout(2) - Vdo 16.27V - 11.5V Il(max) = --------------- = ---------------- = 0.954 ampere. 5 5V

Realistically, though, we haven't even considered ripple and we need to do that in order to choose the value of the BFC.

In order to do that we need to know how much current your radio draws, but you don't know how much that is.

No matter, we can work something out and then plug in your numbers when you find out.

Let's say your radio's running on a watt. Then the current into it will be:

P 1W I = --- = ---- = 0.111A = 110mA E 9V

And the voltage out of the bridge will be:

/ 5V * 0.111A \\ Vout = Vout(2) - | ------------- | \\ 1A / = 16.27V - 0.555V ~ 15.72V

Since the 7809's dropout voltage is 11.5V and we have 15.72V available at the output of the bridge, the difference (4.22V) is the ripple we're allowed.

The filter cap's capacitance, then, would be:

Idt 0.111A * 0.01s C = ----- = ---------------- = 2.63E-4F = 263µF dV 4.22V

Since the bridge will be supplying current to the load as well as to the cap when it's charging, the voltage available from the transformer will drop somewhat during that time so it would be a good idea to increase the value of the capacitor in order to compensate for that.

Without going into it analytically, I'd guess that doubling the cap would do it. Even better, throw 1000µF in there; they're cheap!

Only one thing left to do and that's to check on whether you've got enough headroom to get that 11.5V with low mains, and you now ought to have enough information to be able to do that. :-)

-- JF

ly

er

Okay, a basic flow chart goes like this:

xformer-->bridge rectifier-->filter cap-->regulator-->radio If you have any confusion about how to connect the bridge and the cap, just ask. How big a filter cap you need depends on the radio's current draw. If you have a junk box, just grab a cap with at least 1000 uF, the more the better. Up to a point. You could need as much as 10,000 uF, but it sounds like the radio is pretty small so I really doubt that. Now, on to the regulator. The 7809 regulator has three terminals: in, ground, and out. Ground is common; the rectifier's output, the cap and the radio all share the same ground, or negative. You connect the rectifier's positive output to the regulator's input terminal. Connect the regulator's output terminal to what used to be the radio batteries' positive terminal. Three-terminal regulators are not hard to use, even for a beginner. Good idea to read the datasheet anytime you are using an unfamiliar device:

figure 7, page 22.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

Thanks, but it all sounds so complicated to me, I'm thinking it would be easier to keep it on batteries

--
Oh, well...

Those six D cells give 9V fresh, over 6V until they're dead. And the output impedance is low (similar to a regulated power supply). So, look for a wall wart with regulated DC output in the 6 to 9V range. It'll work.