120V AC to 1.5V DC

[Peter Bennett]

On Tue, 20 Jan 2004 16:10:20 GMT, "Jonathan Mohn"
<j.c.mohn@worldnet.att.net> wrote:

At the risk of wearing out my welcome, I have one more simple question that
I hope y'all might answer.

My main question is on the pin-out of the pot. This is one of those blue,
rectangular pots with a screw on one side for moving the wiper. At first I
connected pin 1 to ground, and the middle pin to the adj pin of the LM317
and to the 240 ohm resistor, as shown in the depiction. But I'm having
problems with pin 3. At first, I hooked it up to the Vout side of the
LM317. It seemed to work just fine, but I grew worried about doing it that
way. So then, I hooked up pin 3 to the positive side of my 9V power source,
and, again, it worked fine. Finally, just out of curiousity, I disconnected
pin 3 altoghether, leaving pin 1 hooked up to ground, and the middle pin
hooked up as described above. To my amazement, everything still worked
fine! What gives?

I finally settled on hooking pin 3 up to the positive side of my 9V power
source, but can anyone explain what the appropriate pin-out is and why?

You should either connect pins 2 and 3 together, or leave pin 3
unconnected.

In this application, you are using the pot as if it was a single
resistor (and, it could be replaced by a single resistor, if you knew
the correct value), so you should treat the pot as a two terminal
device. If you connect pins 2 and 3 together, you provide some
protection against high voltage if the wiper happens to go open
circuit as you adjust the pot - however, your 10K pot is much too high
a value to really help that way.

You may have some difficulty adjusting the pot for such a low voltage,
as you will be right at one end - you probably need less than 100 ohms
to get 1.5 volts, and that is less than 1% of the pot's total
resistance. Better practice would be to use a 100 or 250 ohm pot
instead of the 10K.


--
Peter Bennett VE7CEI
email: peterbb (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[John Popelish]

Jonathan Mohn wrote:

I did solve for the required value of R2. I wasn't certain what Iadj should
be in the formula, so I first assumed that it is 0, and I found R2 should be
48 ohms to achieve an output voltage of 1.5V. The way the formula is
written, any value of Iadj that is greater than 0 will result in an R2 of
less than 48 ohms. So 48 ohms seems to be the maximum required (again, for
1.5V).

Iadj is the Adjustment Pin Current specified on page 4. It adds a
slight error to the assumption that only the current that passes
through R1, forced by the 1.25 volts between the output pin and the
adjustment pin has to exit through R2. In addition to this current,
there is a small current that comes out of the adjustment pin. This
current is pretty small, and almost constant (see the next spec down,
Adjustment Pin Current Change)

So the total output voltage is a sum of the drop caused by the 1.25
output across R1 and R1's current plus Iadj dropping voltage across
R2.

I used a 10K ohm pot because it was all that they had available at the Radio
Shack in my neighborhood. Eventually I'll update the circuit with a 5K.
Will I improve on its performance by going with a 5K pot?

It will improve the adjustability by using the whole adjustment range
to swing through the output voltage range. You will reach full output
voltage with less than 1/2 of the adjustment range of the 10k pot.

I'll stick with 5K or up because I like this regulator so much that I want
to be able to use it for other apps, so will want the full range of
adjustment (1.5V - 25V).

Thanks for the answer on hooking it up. That seems to make more sense. The
way I had it hooked up, I think that I was uselessly consuming about 0.9
amps through the pot.

You were contaminating the adjustment pin voltage with ripple current

from the unregulated supply.

"John Popelish" <jpopelish@rica.net> wrote in message
news:400D5ABF.ECC354EF@rica.net...
Jonathan Mohn wrote:

At the risk of wearing out my welcome, I have one more simple question
that
I hope y'all might answer.

I've assembled the 1.25V - 25V adjustable regulator depicted on the
first
page of the LM317 spec sheet --
http://cache.national.com/ds/LM/LM117.pdf --
only I used two 1 uF tantalum capacitors, one for the input side and one
for
the output side, and a 10K cermet potentiometer. If I understand how
the
capacitors are being used in this circuit, it should not matter that I
used
1 uF instead of the recommended 0.1 uF on the input side, no?

My main question is on the pin-out of the pot. This is one of those
blue,
rectangular pots with a screw on one side for moving the wiper. At
first I
connected pin 1 to ground, and the middle pin to the adj pin of the
LM317
and to the 240 ohm resistor, as shown in the depiction. But I'm having
problems with pin 3. At first, I hooked it up to the Vout side of the
LM317. It seemed to work just fine, but I grew worried about doing it
that
way. So then, I hooked up pin 3 to the positive side of my 9V power
source,
and, again, it worked fine. Finally, just out of curiousity, I
disconnected
pin 3 altoghether, leaving pin 1 hooked up to ground, and the middle pin
hooked up as described above. To my amazement, everything still worked
fine! What gives?

I finally settled on hooking pin 3 up to the positive side of my 9V
power
source, but can anyone explain what the appropriate pin-out is and why?

Thanks a bunch.

The three pins on the pot are the two ends of a resistive element, and
one sliding contact. Since this schematic calls for a two terminal
variable resistor, you should use one end terminal and the sliding
contact, only. I question the resistance or the pot. I think a 5000
ohm pot would be better (you will not use half of the 10k pot's
rotation to reach full output), and if your input voltage is less than
27 volts, a 2000 ohm pot might be even better (for a 1.2 to ~12 volt
adjustment). Did you solve the formula for the required value of R2?

The capacitors should be fine.

--
John Popelish

--
John Popelish

 

[Jonathan Mohn]

Thanks for explaining the Iadj, and that portion of the circuit. I had
scanned page 4 of the spec sheet, but I somehow missed that completely. It
still seems like Iadj varies with R2, though. As I increase the resistance
through R2, doesn't Iadj decrease?

Thanks for the info re ripple current. That is yet something else I need to
study up on. I don't anything about it, yet it seems to pop up in circuits
and must be accounted for.

-Jonathan


"John Popelish" <jpopelish@rica.net> wrote in message
news:400DC47C.C922524B@rica.net...

Jonathan Mohn wrote:

I did solve for the required value of R2. I wasn't certain what Iadj
should
be in the formula, so I first assumed that it is 0, and I found R2
should be
48 ohms to achieve an output voltage of 1.5V. The way the formula is
written, any value of Iadj that is greater than 0 will result in an R2
of
less than 48 ohms. So 48 ohms seems to be the maximum required (again,
for
1.5V).

Iadj is the Adjustment Pin Current specified on page 4. It adds a
slight error to the assumption that only the current that passes
through R1, forced by the 1.25 volts between the output pin and the
adjustment pin has to exit through R2. In addition to this current,
there is a small current that comes out of the adjustment pin. This
current is pretty small, and almost constant (see the next spec down,
Adjustment Pin Current Change)

So the total output voltage is a sum of the drop caused by the 1.25
output across R1 and R1's current plus Iadj dropping voltage across
R2.

I used a 10K ohm pot because it was all that they had available at the
Radio
Shack in my neighborhood. Eventually I'll update the circuit with a 5K.
Will I improve on its performance by going with a 5K pot?

It will improve the adjustability by using the whole adjustment range
to swing through the output voltage range. You will reach full output
voltage with less than 1/2 of the adjustment range of the 10k pot.

I'll stick with 5K or up because I like this regulator so much that I
want
to be able to use it for other apps, so will want the full range of
adjustment (1.5V - 25V).

Thanks for the answer on hooking it up. That seems to make more sense.
The
way I had it hooked up, I think that I was uselessly consuming about 0.9
amps through the pot.

You were contaminating the adjustment pin voltage with ripple current
from the unregulated supply.

"John Popelish" <jpopelish@rica.net> wrote in message
news:400D5ABF.ECC354EF@rica.net...
Jonathan Mohn wrote:

At the risk of wearing out my welcome, I have one more simple
question
that
I hope y'all might answer.

I've assembled the 1.25V - 25V adjustable regulator depicted on the
first
page of the LM317 spec sheet --
http://cache.national.com/ds/LM/LM117.pdf --
only I used two 1 uF tantalum capacitors, one for the input side and
one
for
the output side, and a 10K cermet potentiometer. If I understand
how
the
capacitors are being used in this circuit, it should not matter that
I
used
1 uF instead of the recommended 0.1 uF on the input side, no?

My main question is on the pin-out of the pot. This is one of those
blue,
rectangular pots with a screw on one side for moving the wiper. At
first I
connected pin 1 to ground, and the middle pin to the adj pin of the
LM317
and to the 240 ohm resistor, as shown in the depiction. But I'm
having
problems with pin 3. At first, I hooked it up to the Vout side of
the
LM317. It seemed to work just fine, but I grew worried about doing
it
that
way. So then, I hooked up pin 3 to the positive side of my 9V power
source,
and, again, it worked fine. Finally, just out of curiousity, I
disconnected
pin 3 altoghether, leaving pin 1 hooked up to ground, and the middle
pin
hooked up as described above. To my amazement, everything still
worked
fine! What gives?

I finally settled on hooking pin 3 up to the positive side of my 9V
power
source, but can anyone explain what the appropriate pin-out is and
why?

Thanks a bunch.

The three pins on the pot are the two ends of a resistive element, and
one sliding contact. Since this schematic calls for a two terminal
variable resistor, you should use one end terminal and the sliding
contact, only. I question the resistance or the pot. I think a 5000
ohm pot would be better (you will not use half of the 10k pot's
rotation to reach full output), and if your input voltage is less than
27 volts, a 2000 ohm pot might be even better (for a 1.2 to ~12 volt
adjustment). Did you solve the formula for the required value of R2?

The capacitors should be fine.

--
John Popelish


--
John Popelish

 

[John Popelish]

Jonathan Mohn wrote:

Thanks for explaining the Iadj, and that portion of the circuit. I had
scanned page 4 of the spec sheet, but I somehow missed that completely. It
still seems like Iadj varies with R2, though. As I increase the resistance
through R2, doesn't Iadj decrease?

Iadj is essentially independent of the internal workings of the chip.
The voltage drop of R2 is affected by Iadj. but Iadj is not affected
by R2.

Thanks for the info re ripple current. That is yet something else I need to
study up on. I don't anything about it, yet it seems to pop up in circuits
and must be accounted for.

The unregulated supply is the result of rectifying an AC waveform into
pulsing DC, and smoothing that somewhat with a storage capacitor. But
if you draw current from the capacitor, its voltage will decay and be
subsequently pumped back up by the rectified AC. This variation in
voltage at twice the line frequency is the source of ripple riding on
top of the average DC out or the unregulated supply.

http://www.electronics-tutorials.com/basics/power-supply.htm
--
John Popelish

 

[Jonathan Mohn]

How does the voltage regulator eliminate this problem? I figured that the
capacitor might do the trick, but it sounds like the capacitor discharge is
not long enough in duration to fill the gap between DC pulses.

-J

"John Popelish" <jpopelish@rica.net> wrote in message
news:400DDE19.F5C49847@rica.net...

Jonathan Mohn wrote:

Thanks for explaining the Iadj, and that portion of the circuit. I had
scanned page 4 of the spec sheet, but I somehow missed that completely.
It
still seems like Iadj varies with R2, though. As I increase the
resistance
through R2, doesn't Iadj decrease?

Iadj is essentially independent of the internal workings of the chip.
The voltage drop of R2 is affected by Iadj. but Iadj is not affected
by R2.

Thanks for the info re ripple current. That is yet something else I
need to
study up on. I don't anything about it, yet it seems to pop up in
circuits
and must be accounted for.

The unregulated supply is the result of rectifying an AC waveform into
pulsing DC, and smoothing that somewhat with a storage capacitor. But
if you draw current from the capacitor, its voltage will decay and be
subsequently pumped back up by the rectified AC. This variation in
voltage at twice the line frequency is the source of ripple riding on
top of the average DC out or the unregulated supply.

http://www.electronics-tutorials.com/basics/power-supply.htm
--
John Popelish

 

[Jonathan Mohn]

Also, is there a way to detect the DC pulses and measure its frequency with
my multimeter?


"John Popelish" <jpopelish@rica.net> wrote in message
news:400DDE19.F5C49847@rica.net...

Jonathan Mohn wrote:

Thanks for explaining the Iadj, and that portion of the circuit. I had
scanned page 4 of the spec sheet, but I somehow missed that completely.
It
still seems like Iadj varies with R2, though. As I increase the
resistance
through R2, doesn't Iadj decrease?

Iadj is essentially independent of the internal workings of the chip.
The voltage drop of R2 is affected by Iadj. but Iadj is not affected
by R2.

Thanks for the info re ripple current. That is yet something else I
need to
study up on. I don't anything about it, yet it seems to pop up in
circuits
and must be accounted for.

The unregulated supply is the result of rectifying an AC waveform into
pulsing DC, and smoothing that somewhat with a storage capacitor. But
if you draw current from the capacitor, its voltage will decay and be
subsequently pumped back up by the rectified AC. This variation in
voltage at twice the line frequency is the source of ripple riding on
top of the average DC out or the unregulated supply.

http://www.electronics-tutorials.com/basics/power-supply.htm
--
John Popelish

 

[John Popelish]

Jonathan Mohn wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:400DDE19.F5C49847@rica.net...

The unregulated supply is the result of rectifying an AC waveform into
pulsing DC, and smoothing that somewhat with a storage capacitor. But
if you draw current from the capacitor, its voltage will decay and be
subsequently pumped back up by the rectified AC. This variation in
voltage at twice the line frequency is the source of ripple riding on
top of the average DC out or the unregulated supply.

http://www.electronics-tutorials.com/basics/power-supply.htm

How does the voltage regulator eliminate this problem? I figured that the
capacitor might do the trick, but it sounds like the capacitor discharge is
not long enough in duration to fill the gap between DC pulses.

The regulator discards excess voltage leaving a quite fixed remainder

at its output. It needs to have at least 1.5 volts to discard in
order to function properly. As long as the storage capacitor ripple
voltage does not allow the voltage to dip below 1.5 volts more than
the desired output, the regulator will give a quite smooth output.
Once the storage capacitor voltage dips below 1.5 volts more than the
intended output voltage, the output voltage dips also, maintaining
about a constant 1.5 volt difference between regulator input and
output.

The small capacitor shown on the regulator data sheet is not the
charge storage capacitor (which will a much larger value) but serves
an additional function. Its job is to reduce the effect of the wiring
inductance and the storage capacitor's internal inductance so that the
regulator process is stable, and does not turn into a power
oscillator. If the storage capacitor is a low inductance type and is
close to the regulator, neither of these sources of series inductance
exists, so the small input cap can be eliminated, safely. But it is
cheap insurance.
--
John Popelish

 

[John Popelish]

Jonathan Mohn wrote:

Also, is there a way to detect the DC pulses and measure its frequency with
my multimeter?

If you connect your meter set to an AC voltage scale, and put a

capacitor in series with it, you get an approximate measure of the AC
ripple voltage component of the total DC waveform. Some digital
meters don't need the external capacitor. If you get the same reading
with an without the cap, then you can eliminate it. This is an RMS
reading, which if the ripple were a pure sine wave would be .707 of
the peak positive and negative excursions of the wave. If the
waveform is not a pure sine wave, some other factor applies (hence the
'approximate measure').

A much better way to evaluate the ripple voltage situation is to look
at the raw DC voltage with an oscilloscope and see how low the dips
go, in comparison to the regulated output voltage, to verify that
there is always at least 1.5 volts of headroom for the regulator to
work with. If the dips go too low, you will also be able to see the
output of the regulator dropping out of regulation and following the
bottoms of the dips down. There is just nothing else as good as being
able to see voltage versus time for this sort of thing.

--
John Popelish

 

[Jonathan Mohn]

Thanks for the tips, and for the discussion on voltage regulators (prev
e-mails). Very interesting stuff.

Regards,

-Jonathan




"John Popelish" <jpopelish@rica.net> wrote in message
news:400EAB56.CB364BB4@rica.net...

Jonathan Mohn wrote:

Also, is there a way to detect the DC pulses and measure its frequency
with
my multimeter?

If you connect your meter set to an AC voltage scale, and put a
capacitor in series with it, you get an approximate measure of the AC
ripple voltage component of the total DC waveform. Some digital
meters don't need the external capacitor. If you get the same reading
with an without the cap, then you can eliminate it. This is an RMS
reading, which if the ripple were a pure sine wave would be .707 of
the peak positive and negative excursions of the wave. If the
waveform is not a pure sine wave, some other factor applies (hence the
'approximate measure').

A much better way to evaluate the ripple voltage situation is to look
at the raw DC voltage with an oscilloscope and see how low the dips
go, in comparison to the regulated output voltage, to verify that
there is always at least 1.5 volts of headroom for the regulator to
work with. If the dips go too low, you will also be able to see the
output of the regulator dropping out of regulation and following the
bottoms of the dips down. There is just nothing else as good as being
able to see voltage versus time for this sort of thing.

--
John Popelish