Re: Reducing Source Voltage To Power Supply
- From: John Fields <jfields@xxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 10 Jan 2008 13:26:16 -0600
On Thu, 10 Jan 2008 08:43:50 -0800 (PST), "Dave.H"
<the1930s@xxxxxxxxxxxxxx> wrote:
On Jan 11, 3:37 am, John Fields <jfie...@xxxxxxxxxxxxxxxxxxxxx> wrote:
On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"
<the19...@xxxxxxxxxxxxxx> wrote:
I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?
---
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?
--
JF
The wall wart is rated at 1 amp. Not sure what the current draw is.
---
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
.
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