Regulating Voltage with LM78XX

[Don]

I have a project that requires 12V. I have some LM7815's and looked at
the Fairchild pdf sheet and I need some help understanding the formula
to specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo? You can see I am no good in math. I was a
programmer and If I know the formula and meanings of I? I could write a
program and let the computer do it. Most programmers are dumber than Dog
S!@# in math and anything not(simple minded) me, anyway. KISS is my motto.

While I'm here, this circuit is supposed to switch a 12V relay after
being on for 2 hrs, for 120V on/off and I was wondering how to use the
120V instead of buying a transformer like 120/18V or whatever? to drive
the relay.
Any Help is appreciated;
Thanks, Don Bauer

 

[Michael A. Terrell]

I have a project that requires 12V. I have some LM7815's and looked at
the Fairchild pdf sheet and I need some help understanding the formula
to specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo? You can see I am no good in math. I was a
programmer and If I know the formula and meanings of I? I could write a
program and let the computer do it. Most programmers are dumber than Dog
S!@# in math and anything not(simple minded) me, anyway. KISS is my motto.

While I'm here, this circuit is supposed to switch a 12V relay after
being on for 2 hrs, for 120V on/off and I was wondering how to use the
120V instead of buying a transformer like 120/18V or whatever? to drive
the relay.
Any Help is appreciated;
Thanks, Don Bauer

You can go up, but not down in voltage with that regulator.

--
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

 

[Chris]

I have a project that requires 12V. I have some LM7815's and looked at
the Fairchild pdf sheet and I need some help understanding the formula
to specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo? You can see I am no good in math. I was a
programmer and If I know the formula and meanings of I? I could write a
program and let the computer do it. Most programmers are dumber than Dog
S!@# in math and anything not(simple minded) me, anyway. KISS is my motto.

While I'm here, this circuit is supposed to switch a 12V relay after
being on for 2 hrs, for 120V on/off and I was wondering how to use the
120V instead of buying a transformer like 120/18V or whatever? to drive
the relay.
Any Help is appreciated;
Thanks, Don Bauer

Hi, Don. As Mr. Terrell says, you can only increase the output of a
78XX by using the voltage divider trick. You'll have to just use a
7812 and be done with it.

| _____
| + | | +
| o---o---| 7812|---o-----o
| | |_____| +|
| Vin --- | --- +12V Out
| 18VDC --- | ---10uF
| |.33uF | |
| o---o------o------o-----o
| - -
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

As you know from programming, elegant and simple usually isn't easy.
;-) For someone without electrical or electronics experience, it's not
safe and doesn't make sense to try doing this without transformer
isolation.

Good luck
Chris

 

[Don]

Chris; Thanks for the reply
I'll get a transformer and a 7812
Thanks again; Don

 

[James Thompson]

I have a project that requires 12V. I have some LM7815's and looked at the
Fairchild pdf sheet and I need some help understanding the formula to
specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo? You can see I am no good in math. I was a
programmer and If I know the formula and meanings of I? I could write a
program and let the computer do it. Most programmers are dumber than Dog
S!@# in math and anything not(simple minded) me, anyway. KISS is my motto.

While I'm here, this circuit is supposed to switch a 12V relay after being
on for 2 hrs, for 120V on/off and I was wondering how to use the 120V
instead of buying a transformer like 120/18V or whatever? to drive the
relay.
Any Help is appreciated;
Thanks, Don Bauer

If you must use that regulator, the only way to get 12 volt from the 15 volt
regulator is by further dropping the output with some series diodes.
So 15 - 12 leaves 3 volts, and each diode drops .6 to .7 volt so four series
diodes will drop 2.8 volts leaving you an output of 12.2 volt.
If your circuit has a stable current draw you can simply insert a resistor
in place of the diodes to drop the extra 3 volt. So say if the circuit
you're feeding draws .5 amp to drop 3 volts you will divide 3 by the .5 to
get you a 1.5 ohm resistor which consumes 3 * .5 = 1.5 watt so you would use
at least a 2 watt resistor there.

 

[ehsjr]

Chris; Thanks for the reply
I'll get a transformer and a 7812
Thanks again; Don

Since you have to get a transformer, just get a power supply.
A transformer alone won't do it - you need a transformer,
rectifier bridge & filter cap, plus an enclosure, and your
7812 circuit (2 caps and the 7812).

As an example, MPJA catalog #15548 at $6.95 is a 12 volt
regulated supply capable of 2 amps. You need the line cord,
too, catalog #15447 at $1.25. http://www.mpja.com/

Ed

 

[Stanislaw Flatto]

If you must use that regulator, the only way to get 12 volt from the 15 volt
regulator is by further dropping the output with some series diodes.
So 15 - 12 leaves 3 volts, and each diode drops .6 to .7 volt so four series
diodes will drop 2.8 volts leaving you an output of 12.2 volt.
If your circuit has a stable current draw you can simply insert a resistor
in place of the diodes to drop the extra 3 volt. So say if the circuit
you're feeding draws .5 amp to drop 3 volts you will divide 3 by the .5 to
get you a 1.5 ohm resistor which consumes 3 * .5 = 1.5 watt so you would use
at least a 2 watt resistor there.

He said that he IS a programmer, so at least calculate for him Ohm's Law.
No need for computer, just pull out your slide rule.
Dropping 3V at 1/2A gives (on mine 30cm Darmstadt) 6 Ohms, the power
numbers are correct.

(Note: Since when relays started beying ssssoooo sensitive to supply?)

Have fun

Stanislaw

 

[redbelly]

I have a project that requires 12V. I have some LM7815's and looked at
the Fairchild pdf sheet and I need some help understanding the formula
to specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo? You can see I am no good in math. I was a
programmer and If I know the formula and meanings of I? I could write a
program and let the computer do it. Most programmers are dumber than Dog
S!@# in math and anything not(simple minded) me, anyway. KISS is my motto.

While I'm here, this circuit is supposed to switch a 12V relay after
being on for 2 hrs, for 120V on/off and I was wondering how to use the
120V instead of buying a transformer like 120/18V or whatever? to drive
the relay.
Any Help is appreciated;
Thanks, Don Bauer

As you've now learned, if you want 12V then get a 7812. However, it
puzzles me what led you toward the circuit you originally posted. It
looks like a constant-CURRENT drive for some load represented by RL.

To answer your questions: Io is the current delivered by this current
source. You can think of Iq as a "leakage" current from pin 2 of the
regulator. Iq, along with the current through R1, contributes to Io.
The spec sheet probably has a typical and maximum values for Iq (I
think it's around 5 mA for the 7805 , for example), and you might want
to measure it if you wanted some degree of precision in your current
.... IF you were wanting to build a current source, that is, which
you're not.

Regards,

Mark

 

[James Thompson]

He said that he IS a programmer, so at least calculate for him Ohm's Law.
No need for computer, just pull out your slide rule.
Dropping 3V at 1/2A gives (on mine 30cm Darmstadt) 6 Ohms, the power
numbers are correct.

(Note: Since when relays started beying ssssoooo sensitive to supply?)

Have fun

Stanislaw

Yep, i mixed up the watts. hehe.
Thanks for pointing that out. So easy to make dum mistakes and its good to
have many eyes looking it over.

 

[Michael Black]

(Note: Since when relays started beying ssssoooo sensitive to supply?)

I missed that part.

But the regulator may be there more for isolation (though it may not
be the right way to isolate a relay).

Remember, in the old days it was rare to see regulators. They'd be
in test equipment, and lab type power supplies, but rare in consumer
electroncis. At the very most, you'd regulate the plate of the oscillator
tube in that receiver. Most of the time, any regulation was taken care
of by a VR tube, no pass element needed.

When solid state came along, suddenly regulators came into their own.
The early magazines would show diodes feeding massive capacitors, because
they were seeking that really low output impedance. Regulators took away
some of that burden. Then when ttl came along, you'd want to regulate
not because they needed precise voltage (unlike those receiver oscillators
that would shift frequency when the plate voltage shifted), so regulators
became mandatory. At that point, it became easy since the introduction of
three terminal regulators coincided with that ttl boom.

But those three terminal regulators were originally called "card regulators",
the intent being that you'd have a big transformer, diodes and capacitors in
one place, and feed that higher voltage to all the boards. And each board
would have a card regulator, now that they were small and cheap. The
regulation was not in one central point, but distributed over the equipment.
COmpared to previous regulator ICs, like the 723 needing all those external
components and the pass transistor if you needed more current, the 3 termainl
regulators needed nothing more bypass capacitors.

THose 3 terminal regulators got even cheaper. There was little point in not
using them, they likely did provide some isolation between stages if you
put in more than one. They beat the previous cheap arrangement, zener
diodes that you'd have to calculate the dropping resistor, and recalculate
that resistor if the current demand changed.

No, there's little need to regulator voltage to a relay. But it may isolate
the spike when it clunks in from the rest of the circuit. Though again, there
may be better means of providing that isolation.

I was given a used slide rule around the time I got my first three terminal
regulator, which was around the time they first were introduced. Around
the same time, I saw a pocket calculator for the first time, an HP-35 that
a friend had paid big bucks for (though less than others since he was
involved in a group buy at work). Wasn't long before that slide rule
became obsolete, indeed before I had reason to get good at using it.

Michael

 

[John Popelish]

I have a project that requires 12V. I have some LM7815's and looked at
the Fairchild pdf sheet and I need some help understanding the formula
to specify 12 Volt out. (maybe that can't be done)?
------------------
in 18V------o-- 1 -| |- 3 ---o--------o- output
| | LM78XX | | > ^ ^
| ------------------ | < | Vxx
--- | 2 --- R1 > | |
c1 --- 0.33uF | | Co --- 0.1uF | | |
| | | | | | |
| Iq | | | | | v --- Io
|------------v--0-----------------o-----------|-----| |
| v
Vxx > /
Io = ----- +Iq RL </
R1 >
/<
/ |
GND

The only thing I know is R1 = 220 (I have these).
What is Iq/lq and Io/lo?

The method shown can raise the output voltage above the 15
volt normal output of a 7815, but it cannot lower it.

The concept is based on the property of the regulator that
the output voltage is regulated to be 15 volts more positive
than the voltage on pin 2 (the reference pin). You can use
current from the output voltage to produce a drop in
resistor R1 and that drop is applied to the reference pin.
The regulator keeps raising its output voltage (raising the
voltage applied to its own reference pin, till the voltage
of the output is 15 volts more positive than that on the
reference pin, and there it stops.

The formula tells you that the voltage drop across R1 is the
sum of the current that passes through R1 (15 volts/R1, for
the 7815) plus the small and fairly constant bias current
that passes out of the reference pin, Iq. Normally you make
R1 low enough so that its current is many times Iq, so that
small variations in Iq do not much change the output voltage.

You could easily use this method to jack the output of a
7805, 7806, 7808, 7810 or LM317 (a 1.2 volt regulator
designed specifically for this method, with an especially
low and stable Iq), but not to lower the output voltage of
any of them.

(snip)

While I'm here, this circuit is supposed to switch a 12V relay after
being on for 2 hrs, for 120V on/off and I was wondering how to use the
120V instead of buying a transformer like 120/18V or whatever? to drive
the relay.

You will have to be more pedantic in describing the
available signals and what you want to do with them, for me
to help with this.

 

[Don]

puzzles me what led you toward the circuit you originally posted. It
looks like a constant-CURRENT drive for some load represented by RL.

To answer your questions: Io is the current delivered by this current
source. You can think of Iq as a "leakage" current from pin 2 of the
regulator. Iq, along with the current through R1, contributes to Io.
The spec sheet probably has a typical and maximum values for Iq (I
think it's around 5 mA for the 7805 , for example), and you might want
to measure it if you wanted some degree of precision in your current
... IF you were wanting to build a current source, that is, which
you're not.

Regards,

Mark
Mark, I appreciate the explanation of the Io/Iq although I'd have no

idea how to determine them. The circuit just said 12V for the relay and
I have read the other posts and it seems as if relays are not all that
sensitive, however it does have an IC4010 if I remember correctly and I
did not know if running that at 15V would be good or bad. Thanks for all
the posts, I have certainly learned a LITTLE, and that's what counts. I
have worked with several dozen EE's, (I wrote the programs to bill TVA's
power to utilities). They always tried to explain the power factors,
transformer losses, heat loss/gain with the HVAC engineers, but for my
job all I needed was the formula to make everyone happy and come up with
the correct answer. I really liked the HVAC guys, they always included a
C sub d parameter that was always last in the equation. They would see a
bunch of results and say change c sub d to .92874 and run again. Took me
awhile to figure out that it was what you multiplied the formula answer
by in order to come up to the value that THEY wanted. I'm continuing the
learning anyway,
Thanks; Don

 

[Don]

The method shown can raise the output voltage above the 15 volt normal
output of a 7815, but it cannot lower it.

The concept is based on the property of the regulator that the output
voltage is regulated to be 15 volts more positive than the voltage on
pin 2 (the reference pin). You can use current from the output voltage
to produce a drop in resistor R1 and that drop is applied to the
reference pin. The regulator keeps raising its output voltage (raising
the voltage applied to its own reference pin, till the voltage of the
output is 15 volts more positive than that on the reference pin, and
there it stops.

The formula tells you that the voltage drop across R1 is the sum of the
current that passes through R1 (15 volts/R1, for the 7815) plus the
small and fairly constant bias current that passes out of the reference
pin, Iq. Normally you make R1 low enough so that its current is many
times Iq, so that small variations in Iq do not much change the output
voltage.

You could easily use this method to jack the output of a 7805, 7806,
7808, 7810 or LM317 (a 1.2 volt regulator designed specifically for this
method, with an especially low and stable Iq), but not to lower the
output voltage of any of them.

(snip)


You will have to be more pedantic in describing the available signals
and what you want to do with them, for me to help with this.

John; Thanks for the information on increasing voltage with the LM78XX,
I'll file it away for reference. RE:, "pedantic", I had to look that one
up, but I got the drift from the sentence context.
Thanks; Don

 

[John Popelish]

John; Thanks for the information on increasing voltage with the LM78XX,
I'll file it away for reference. RE:, "pedantic", I had to look that one
up, but I got the drift from the sentence context.

I meant the "emphasizing minutia" aspect of "pedantic".

So, tell us, in detail, what you are trying to accomplish
and what you have to work with. There may be a simple
solution that is completely different than what you are
picturing.

 

[Fred Abse]

He said that he IS a programmer, so at least calculate for him Ohm's Law.
No need for computer, just pull out your slide rule.
Dropping 3V at 1/2A gives (on mine 30cm Darmstadt) 6 Ohms, the power
numbers are correct.

Oddly enough, *my* "30cm" Darmstadt, that I've had for nearly 50 years
(Castell 1/54) has a 0-10 *inch* scale on one edge, but no metric at all.

I can confirm your figures

 

[ehsjr]

He said that he IS a programmer, so at least calculate for him Ohm's Law.
No need for computer, just pull out your slide rule.
Dropping 3V at 1/2A gives (on mine 30cm Darmstadt) 6 Ohms, the power
numbers are correct.

(Note: Since when relays started beying ssssoooo sensitive to supply?)

Your point about relays is well made. From time to time
you see people unaccountably straining to provide precise
voltage for them. Still, he may have a circuit that *is*
sensitive to the supply that drives the relay. He mentioned
something about the circuit switching the relay after 2 hours.

Ed

 

[Chris]

I meant the "emphasizing minutia" aspect of "pedantic".

<snip>

Isn't it "minutiae", Mr. Popelish? I believe you implied the plural.
;-)

Cheers
Chris

 

[ehsjr]

As you've now learned, if you want 12V then get a 7812. However, it
puzzles me what led you toward the circuit you originally posted. It
looks like a constant-CURRENT drive for some load represented by RL.

Yes, his diagram is confusing and confused. What he is calling RL
isn't RL. I think the confusion comes from all his labeling and
adding of lines to indicate current. The key is that the variable
R that he mislabelled as RL goes to ground. I think what he has is
this:
------
18 in ---+---|LM78XX|---+------+---> +Vout
| ------ | |
[.33uF] | [R1] [.1uF]
| | | |
| +------+ |
| | |
| P |
| O<---+ |
| T | |
| +----+ |
| | |
Gnd ----+-------+-------------+---> Gnd

The pot and R1 form a voltage divider that raises
the gnd pin on the 78XX to some specific voltage
above ground, and the 78XX keeps its Vout pin at
XX volts above its ground pin. RL would be the
load that gets connected from Vout to gnd, and is
not shown on his diagram - he has mis-labeled the
adjustment pot as RL.

Ed

 

[John Popelish]

<snip>

Isn't it "minutiae", Mr. Popelish? I believe you implied the plural.
;-)

Could be.
My spell checker was happy, so I was happy. I'm just glad
it wasn't happy with the entirely wrong word.

 

[Chris]

Could be.
My spell checker was happy, so I was happy. I'm just glad
it wasn't happy with the entirely wrong word.

Hi, Mr. Popelish. Your spell checker was happy because "minutia" is
singular, and "minutiae" is plural. Talk about minutiae! ;-)

Getting the relevant information from the OP seems to be like pulling
teeth. I'll bet that when he's done, his best solution would be a
simple Time Delay Relay (TDR). Power it up with DC or AC, and you get
a two hour timed relay contact closure. The OP might want to check out
SSAC for some good inexpensive ones that should do the job:

http://www.ssac.com/

But the education is in the journey, not just the destination. You
seem to enjoy this stuff more than just about anyone in the newsgroups.
I'll bet you would have made a heck of a science teacher in another
life, Mr. Popelish.

Cheers
Chris