using a 7809 voltage regulator

[catfarm]

I have a KA7809 voltage regulator. I have a couple of buffering caps from
the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout.
if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the Vout.

Ive tried 2 different makes of regulator with the same results. What would
account for this strange drop in Vout?

FWIW I put my meter on a 9V battery and got... indeed... 9V.

 

[CFoley1064]

Subject: using a 7809 voltage regulator

From: "catfarm" [email protected]
Date: 10/23/2004 12:57 PM Central Daylight Time
Message-id: <9Ywed.303627$D%.121731@attbi_s51>

I have a KA7809 voltage regulator. I have a couple of buffering caps from
the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout.
if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the Vout.

Ive tried 2 different makes of regulator with the same results. What would
account for this strange drop in Vout?

FWIW I put my meter on a 9V battery and got... indeed... 9V.

First, check your pinout. That seems to be the most common problem. Assuming
you've got a TO-220 package (metal back, hole in copper for heat sink screw)
this should be the view from the front, plastic side facing you (view in fixed
font or M$ Notepad):

.---------.
| |
| o |
| |
| |
'---------'
| |
| |
| |
| |
| |
'-o--o--o-'
| | |
| | |
| | |
| | |
I G O
N N U
D T

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Once you've checked that, you want to put your meter on AC 20V range and read
the input for AC. If you've got a high AC component with no load on the 7809,
you've probably got a messed up filter cap in the wall wart. You didn't
mention the filter cap with the transformer, so I'm assuming you don't have
one. You need a good sized filter cap before the regulator to keep the minimum
input voltage to the 7809 below 12.5V, otherwise it won't work right. The wall
wart (it IS DC, right?) is supposed to have a big cap built in to keep the
ripple on the DC output low.

Try checking the wall wart and replacing it if there's a lot of AC on the
output.

If you're using the transformer with rectifier, put a 1000 uF 35V cap from the
input terminal of the 7809 to GND.

See if these work first. They're the most common problems.

Good luck
Chris

 

[John Fields]

I have a KA7809 voltage regulator. I have a couple of buffering caps from
the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout.
if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the Vout.

Ive tried 2 different makes of regulator with the same results. What would
account for this strange drop in Vout?

1. Inadequate smoothing (too much ripple) on the wall wart output
for the load, causing the 7809 to go out of regulation when it
runs out of headroom.

Fix: if your load current is less than 400mA, place an electrolytic
cap large enough to keep the wall-wart's output ripple valleys to

11V from the input terminal of the regulator to ground. If the

load current is higher than 400mA, get a new wall wart.

2. Too large a load for the wall wart to support, causing it to fall
below 11V, the 7809's typical dropout voltage.

Fix: Get a new wall wart.

3. If not filtered, (or not filtered enough) dropout when the voltage
out of the XFMR/bridge combo falls below 11V.

Fix: Figure out how large a filter cap you need from:

IdT
C = -----
dV

Where C = the required capacitance in farads
I = the average output current in amperes
dt= the period of the ripple frequency in seconds
dV = the permissible ripple voltage in volts


For a full wave rectifier, dT will be 1/120 = 0.0083s and dV will be
the output voltage from the bridge minus 11 volts. The output voltage
from the bridge will probably be something like the 23 volt RMS out of
the transfromer multiplied by 1.414 to get the peak voltage, then from
that, 1.4 volts subtracted for the diode drops in the bridge. That
comes to


Vout = (VRMS * sqrt2) - 2Vf = (23 * 1.414) - 1.4 ~ 31V

Subtracting 11V from that to get the permissible ripple voltage gets
us 20V for dV so, with the exception of the load current, here ya go:



IdT I * 0.0083s
C = ----- = ------------- = ???
dV 20V

Once you get the capacitance, don't forget to take into consideration
the tolerance of the capacitor, its voltage rating, and its allowable
ripple current rating at the ambient temperature in which it'll be
running.

Just to be on the safe side of everything, increase the capacitance
you get by about 50%, get a cap rated at 50VDC, and make sure the
ripple current rating is about twice your load current rating.

BTW, it wouldn't hurt to change the output cap to 0.1µF, and connect
it and the input caps as close to the regulator package as you can.

 

[John Fields]

First, check your pinout. That seems to be the most common problem. Assuming
you've got a TO-220 package (metal back, hole in copper for heat sink screw)
this should be the view from the front, plastic side facing you (view in fixed
font or M$ Notepad):

.---------.
| |
| o |
| |
| |
'---------'
| |
| |
| |
| |
| |
'-o--o--o-'
| | |
| | |
| | |
| | |
I G O
N N U
D T

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Once you've checked that, you want to put your meter on AC 20V range and read
the input for AC. If you've got a high AC component with no load on the 7809,
you've probably got a messed up filter cap in the wall wart. You didn't
mention the filter cap with the transformer, so I'm assuming you don't have
one. You need a good sized filter cap before the regulator to keep the minimum
input voltage to the 7809 below 12.5V, otherwise it won't work right. The wall

^^^^^ ^^^^^
above 11V typ.

 

[petrus bitbyter]

I have a KA7809 voltage regulator. I have a couple of buffering caps from
the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout.
if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the
Vout.

Ive tried 2 different makes of regulator with the same results. What would
account for this strange drop in Vout?

FWIW I put my meter on a 9V battery and got... indeed... 9V.

First get the datasheet of the 7809. That things sometimes contain usefull
information .

As for the wallwart: A DC wallwart uses to have a smoothing capacitor
inside. But you may have a less common type. A 1uF smoothing capacitor is
much to small for a normal load. 500uF - 1000uF is a common value. Another
thing is the load. Some regulators sometimes need some load for good
regulation. Look what happens if you connect a resistor of let's say 1k to
the output.

As for the transformer/bridge combination: You sure need to add a smoothing
capacitor.

petrus bitbyter.

 

[CFoley1064]

Subject: Re: using a 7809 voltage regulator

From: John Fields [email protected]
Date: 10/23/2004 3:10 PM Central Daylight Time
Message-id: <[email protected]>
wall
^^^^^ ^^^^^
above 11V typ.

How did I write that??? Jeez -- I must have gremlins in the keyboard.

Let's see -- I added 9 + 2.5 and got 12.5, and then used below instead of
above?? Senior moment. You're right, of course.

Thanks.
Chris

 

[Bill Bowden]

I have a KA7809 voltage regulator. I have a couple of buffering caps from
the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout.
if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the Vout.

Ive tried 2 different makes of regulator with the same results. What would
account for this strange drop in Vout?

FWIW I put my meter on a 9V battery and got... indeed... 9V.

Well, the first thing you can do is hook the 7809 input to
a 12 battery and see if it works. If it does, you have filtering
problem with the wall wart. Try a larger capacitor across the
input. The size of the capacitor depends on the load current.
It's about 8000uF per amp, per volt. So, if the wall wart
produces 18 volts rms, the peak will be 1.414 times more, or 25
volts. The 7809 needs a minimum of about 12 volts to deliver 9,
so the wall wart voltage can fall from 25 to 12 (13 volts total).
if the load current is 500mA, the capacitor will be
(8000/13)/2 = 307uF minimum. You can use a larger value for
more current.

-Bill

 

[catfarm]

Thank you to everyone who responded. I put a 1000uF el cap in the line and
everything straightened itself out.

 

[Stan Blazejewski]

First get the datasheet of the 7809. That things sometimes contain usefull
information .

As for the wallwart: A DC wallwart uses to have a smoothing capacitor
inside. But you may have a less common type. A 1uF smoothing capacitor is
much to small for a normal load. 500uF - 1000uF is a common value. Another
thing is the load. Some regulators sometimes need some load for good
regulation. Look what happens if you connect a resistor of let's say 1k to
the output.

As for the transformer/bridge combination: You sure need to add a smoothing
capacitor.

petrus bitbyter.

Just a thought, are the usual small capacitors fitted either side of
the regulator? If not it 'could' be oscillating, particularly with
the long leads from the wall wart.


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