Dropping 1V from a Regulated 6V Wall Wart

  • Thread starter Watson A.Name - "Watt Sun
  • Start date
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Watson A.Name - "Watt Sun

Guest
I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
On Mon, 29 Dec 2003 09:28:53 -0800, the renowned Watson A.Name - "Watt
Sun, Dark Remover" <alondra101@hotmail.com> wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.
Or just hang a dummy load on the output.. 1K0 or a bit higher (<2K0)
ought to do it.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a5a03ef9d0f170c989ac8@news.dslextreme.com...
I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
Why not a 5volt zener and a resistor?....Just a thought...
 
On Mon, 29 Dec 2003 09:28:53 -0800, Watson A.Name - "Watt Sun, Dark
Remover" <alondra101@hotmail.com> wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.
Try this...

+ From
wall
wart >--+
|
+---+------+
| |
\ |
/ |
20 \ / 2SC2334 or TIP31
ohm | |/ NPN power TO-220
o------| Heatsink optional
| |\
| E\
| |
| |
| |
| |
| |
| o-----> +
| | output
| / to load
| \
| / 2.2K
__|___/ |
/ /\ |
/ \--------o
/ \ |
-------- |
| TL430 /
| \ 2.7K
| /
| |
| |
| |
- >--o----------o------> -

You may need a compensation cap between cathode and control pin.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Mon, 29 Dec 2003 09:28:53 -0800, Watson A.Name - "Watt Sun, Dark
Remover" <alondra101@hotmail.com> wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart,
....


6VDC>------+-----------------+
| |
[R1] |
| |
+-----|+\ C
| | >------B
| +--| / E
| | |
| +----+---------+
| |+ |
[R2] [C1] [RL]
| | |
GND>-------+-------+---------+

R1/R2 = 5

C1 = Whatever you need to keep the thing from oscillating; 100ľF is
probably a good guess.

--
John Fields
 
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
<jfields@austininstruments.com> wrote:


R1/R2 = 5
---
R2/R1 = 5

--
John Fields
 
I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <vft0vvk16bjlvnt019n8mpfeekji1k5ush@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
jfields@austininstruments.com> wrote:


R1/R2 = 5

---
R2/R1 = 5

Not always. (;-)
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
 
On Mon, 29 Dec 2003 19:07:59 +0000, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <vft0vvk16bjlvnt019n8mpfeekji1k5ush@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
jfields@austininstruments.com> wrote:


R1/R2 = 5

---
R2/R1 = 5

Not always. (;-)
---
I'm sure there's a joke in there somewhere... :)

I meant that for 6V in and 5V into Rl, R2 = 5R1.

--
John Fields
 
I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <gjv0vv0rdoh63n3oo331lm5anvddskdgb2@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 19:07:59 +0000, John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <vft0vvk16bjlvnt019n8mpfeekji1k5ush@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
jfields@austininstruments.com> wrote:


R1/R2 = 5

---
R2/R1 = 5

Not always. (;-)

---
I'm sure there's a joke in there somewhere... :)

I meant that for 6V in and 5V into Rl, R2 = 5R1.

OK, let's go on in the same vein. R2 is not always 5R1; it might be 5R6
or even 47K. (;-)
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
 
On Mon, 29 Dec 2003 17:47:06 GMT, this.is.a.mung@example.invalid
said...
"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a5a03ef9d0f170c989ac8@news.dslextreme.com...

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

snip

Why not a 5volt zener and a resistor?....Just a thought...

I'm thinking 4.3V Zener and a series pass transistor.
--
Best Regards,
Mike
 
What about using a 5V low-dropout voltage regulator?

Bob



Watson A.Name - "Watt Sun, Dark Remover" <alondra101@hotmail.com>
wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.
 
On Mon, 29 Dec 2003 20:12:23 GMT, the renowned Active8
<mTHISREMOVEcolasono@earthlink.net,invalid> wrote:

I'm thinking 4.3V Zener and a series pass transistor.
Wrong way.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
On Mon, 29 Dec 2003 20:16:37 GMT, speffSNIP@interlogDOTyou.knowwhat
said...
On Mon, 29 Dec 2003 20:12:23 GMT, the renowned Active8
mTHISREMOVEcolasono@earthlink.net,invalid> wrote:


I'm thinking 4.3V Zener and a series pass transistor.

Wrong way.

Best regards,
Spehro Pefhany

Why?
--
Best Regards,
Mike
 
On Mon, 29 Dec 2003 20:36:46 GMT, mTHISREMOVEcolasono@earthlink.net
said...
On Mon, 29 Dec 2003 20:16:37 GMT, speffSNIP@interlogDOTyou.knowwhat
said...
On Mon, 29 Dec 2003 20:12:23 GMT, the renowned Active8
mTHISREMOVEcolasono@earthlink.net,invalid> wrote:


I'm thinking 4.3V Zener and a series pass transistor.

Wrong way.

Best regards,
Spehro Pefhany

Why?

NEver mind. I tried to cancel this but Gravity thinks I didn't post
this.

Still at TARFU level, I guess.

OK 5.6V zener, solly.
--
Best Regards,
Mike
 
On Mon, 29 Dec 2003 19:54:47 +0000, jmw@jmwa.demon.contraspam.yuk
said...
I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <gjv0vv0rdoh63n3oo331lm5anvddskdgb2@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 19:07:59 +0000, John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <vft0vvk16bjlvnt019n8mpfeekji1k5ush@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
jfields@austininstruments.com> wrote:


R1/R2 = 5

---
R2/R1 = 5

Not always. (;-)

---
I'm sure there's a joke in there somewhere... :)

I meant that for 6V in and 5V into Rl, R2 = 5R1.

OK, let's go on in the same vein. R2 is not always 5R1; it might be 5R6
or even 47K. (;-)

But Mr. Wizard, this *package* says 502 and this one's on
psychedelics.

Why won't this magnet pick up this floppy disk?
--
Best Regards,
Mike
 
On Mon, 29 Dec 2003 19:54:47 +0000, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <gjv0vv0rdoh63n3oo331lm5anvddskdgb2@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 19:07:59 +0000, John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <vft0vvk16bjlvnt019n8mpfeekji1k5ush@4ax.com>) about
'Dropping 1V from a Regulated 6V Wall Wart', on Mon, 29 Dec 2003:
On Mon, 29 Dec 2003 12:18:19 -0600, John Fields
jfields@austininstruments.com> wrote:


R1/R2 = 5

---
R2/R1 = 5

Not always. (;-)

---
I'm sure there's a joke in there somewhere... :)

I meant that for 6V in and 5V into Rl, R2 = 5R1.

OK, let's go on in the same vein. R2 is not always 5R1; it might be 5R6
or even 47K. (;-)
<GROANNNN...>

OK, it arteried: "In order to assure a voltage of 5VDC across Rl the
resistance of R2 should be five times the resistance of R1." ?^)

--
John Fields
 
Watson A.Name - Watt Sun, Dark Remover wrote:
I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+
|
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\
| |
| |
+----+-----+
|
|
+------> +
output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.
You almost had it- put the Vbe multiplier inside the feedback loop and
buffer like so:

Please view in a fixed-width font such as Courier.


6V >---+------+---------+
| | |
| / |
| 22 |
| / |
| \ c
| | |/
| +-------| TIP31
| | |\
| | e----+--> 5V
| | |
=== | +----+
| c | |
| \| / |
| |-----> \ |
| /| / ===
| e \ 1000U
| | | |
| | | |
| | | |
GND>--+------+---------+----+--> GND
 
On Mon, 29 Dec 2003 09:28:53 -0800, Watson A.Name - "Watt Sun, Dark
Remover" <alondra101@hotmail.com> wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

[snip]

See also "4V-Regulator.pdf" on the S.E.D/Schematics Page of my
website, for a similar application.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Mon, 29 Dec 2003 20:46:53 GMT, Active8
<mTHISREMOVEcolasono@earthlink.net,invalid> wrote:


But Mr. Wizard, this *package* says 502 and this one's on
psychedelics.

Why won't this magnet pick up this floppy disk?
---
It will, Timmy, you just have to know how to do it.

For a clue, check out "Magnet lifting floppy" on a.b.s.e.

--
John Fields
 
Fred Bloggs wrote:
Watson A.Name - Watt Sun, Dark Remover wrote:

I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
a 3 amp rectifier, but it varied by more than .2V over a range of
loads. So I tried this: (view with courier font)

+ From
wall
wart >--+ |
+---+------+
| |
| |
400 \ / 2SC2334 or TIP31
ohm / |/ NPN power TO-220
WW \<-----| Heatsink optional
pot / |\
| E\ | |
| |
+----+-----+
| |
+------> + output
to load
- >--------------> -
From wall wart

This has some advantages and disadvantages. It's simple and cheap,
and keeps the output at 5V within a tenth of a volt over a current
range. But it has a minimum current below which it loses regulation
and the output starts to go up to 6V, because the transistor is not
conducting and the current is being supplied thru the ww pot. This
circuit is sometimes used in the bias circuit for the output
transistors in high powered amplifiers. Also Win Hill showed us here
how to use a similar circuit to maintain the voltage steady for a
current regulator circuit used on four AA cell rechargeable batteries.

I'm thinking about putting a 5.1V zener on the output so that if the
voltage climbs above that, it just shunts the excess current. Oh,
yeah, I set the pot to various values to see what the output voltage
was with various loads. The two resistances were 120 ohms for the
upper and 280 ohms for the lower. I suppose the 400 ohms total could
be raised to a higher value, but the transistor needs enough base
current to do its job. There's only 1V available minus the .6V E-B
voltage, so even at 400 ohms, that's not a lot of current.



You almost had it- put the Vbe multiplier inside the feedback loop and
buffer like so:

Please view in a fixed-width font such as Courier.


6V >---+------+---------+
| | |
| / |
| 22 |
| / |
| \ c
| | |/
| +-------| TIP31
| | |\
| | e----+--> 5V
| | |
=== | +----+
| c | |
| \| / |
| |-----> \ |
| /| / ===
| e \ 1000U
| | | |
| | | |
| | | |
GND>--+------+---------+----+--> GND
Then you can work a common 5.1V zener into the equation like so:
Use a 2N3906 for the pnp for Vbe,max=6V rating.

Please view in a fixed-width font such as Courier.





6V >---+--------+----------+-------------+
| | | |
| / / |
| 22 51 |
| / / |
| \ \ c
| | | |/
| +----------|-----------| TIP31
| | | |\
| | | e----+--> 5V
| c | |
| \| | pnp |
| npn |--+-----|-------- c e---+
| /| | | \ / |
| e | | ---- |
=== | | | 180 | |
| | | +----/\/\---+ |
| | | | | |
| | | | | |
| | / _/ / ===
| | 1K /^ 5.1v 1.2K 1000U
| | / - / |
| | \ | \ |
| | | | | |
GND>--+--------+----+-----+-----------+------+--> GND
 

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