Have 5v regulator need 9 volts

[amdx]

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

 

[John Larkin]

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

You can stack regulators: hang the ref pin of one 5V reg on the output of
another.

See U9:U10 here:

https://dl.dropbox.com/u/53724080/Circuits/ESM/ESM_power.pdf

Dang, I could have done that with U11 and U12, too.





--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Phil Allison]

"amdx"

Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my divider
to bring it up to 4 volts.

** All you need is a 3.9V zener in series with the ground lead.



.... Phil

 

[David Eather]

pOn Thu, 24 Jan 2013 12:27:50 +1000, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

Stacking regulators in your case may not work. For the top regulator to
work its quiescent current must go through the load on the 5v rail. If
there is no load on the 5v output then it won't regulate. You can fix this
situation by putting a 470 ohm resistor permanently across the lower
regulator (this provides a continuous load of 11ma which is greater than
the top regulators 8ma maximum quiescent current so it will regulate
properly - see data sheet link below). This may be the best solution if
the wall wart is up to it.

But the wall wart could be a problem since it is unregulated and is rated
at 9v at some specified current which you haven't provided. All regulators
need some voltage 'overhead' to work. From the LM7805 datasheet

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

The TYPICAL drop out voltage is 2 volts so if you want 9 volts out of the
regulator you need 11 volts going into it TYPICALLY - some 7805's may need
more. When you start putting a load on that wall wart the 14 volts output
is going to fall and at 350ma it may be too low. My *guess* is that unless
the wall wart is rated at more than 9v @1Amp you have no chance of making
it work as you have suggested.

There are a few easy solutions to this: get a 9v regulated wall wart or
get a 12v unregulated wall wart but I suspect as much as possible you want
to roll-your-own. This regulator might solve your problem and still use
your wall wart:

http://www.ebay.com/itm/Boost-Buck-Voltage-Module-3-35V-to-2-2-30V-Step-Up-Down-Converter-Regulator-/290833291375?pt=LH_DefaultDomain_0&hash=item43b703d86f

I haven't dealt with this supplier or used this circuit but it seems
reasonable and is from a top rated supplier on ebay so chances are good,
and if you are a hobbyist getting into electronics you will want to get
used to using ebay.

The voltage divider question:

The 78xx series is not a great candidate for this type of modification
because it's quiescent current is so high. It's not impossible but the
LM317 is much better for this (search http://www.fairchildsemi.com for the
data sheet).

The simplest way to jack up the voltage output is to put a zener diode in
series with the gnd pin - for your example a 3.9v diode will do the job
nicely. The quiescent current is typically about 5ma and may be as high as
8ma which is fine for the zener diode. Since the quiescent current only
changes by a maximum of .5ma this will have almost no effect on the zener
voltage and hence the output voltage

A brute force approach for the 7805 with voltage divider would go like
this:

We want to be able to plug any old 7805 regulator into the circuit and be
within 10% of the 9v value so we set the current through the voltage
divider to be much greater than the maximum quiescent current of the
regulator (8ma). To achieve this We make the design decision that the
divider current will be at least 80ma. Therefor the total resistance of
the divider must be no more than 112 ohms. The regulator gnd pin must be
raised by 4 volts so the bottom resistor (Rb) of the divider must be 4v/9v
* 112 and the top resistor (Rt) equals the remainder. Rb = 49.8 ohm Rt=
62.2. The closest E24 values are 51 and 62 ohms (47 and 56 for E12
values). Because of the amount of current involved you should check the
power dissipation in these resistors which is I*I*R which comes to
approximately 0.4 of a watt. In this example you need to use at least 1/2
watt resistors for Rt and Rb (if the circuit is to sit around powered up
for years and be reliable than 1 watt resistors would be a better choice).

This design continually wastes 0.8 watt of power but has the advantage
that you could use it on a production line and get the right result
without adjustment but it is wasteful and inefficient.

An approach more suitable to a designer who will adjust his work is like
this:

The maximum change of the quiescent current in the regulator is 0.5ma
(DqI)so wee want to make that small (less than 10%) relative to the
current in the voltage divider. The top resistor (Rt) will always have 5
volts across it and for 5ma (DqI * 10) has a value of 1k ohm. The value
for Rb has to be selected by testing as quiescent current of the regulator
can vary from about 4ma to 8ma and that plus the current from Rt must
cause a 4v drop in Rb. So the value for Rb will be between about 360 to
440 ohms. In this case the power dissipated in Rt is just 25mW and Rb is
under 50mW. Common 1/4 watt resistors will be perfect and the wasted
power is under 0.1 watt. Just grabbing a 390 ohm resistor would put the
circuit within about a maximum of 11% of the desired output but most
people would trim that to be closer to the 9v.

The reason for not using Rb alone is because of how much the quiescent
current can change between devices (4 to 8ma) and how much it can change
in one device (0.5ma) depending on how it is used.

The zener diode works well with the 7805 but the resistor divider not so
well. The resistor divider works well with the LM317 because of its low
current through the adj pin but also for that reason the simple zener
approach does not work so well with the 317

HTH

 

[Michael A. Terrell]

David Eather wrote:

pOn Thu, 24 Jan 2013 12:27:50 +1000, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

Stacking regulators in your case may not work. For the top regulator to
work its quiescent current must go through the load on the 5v rail. If
there is no load on the 5v output then it won't regulate. You can fix this
situation by putting a 470 ohm resistor permanently across the lower
regulator (this provides a continuous load of 11ma which is greater than
the top regulators 8ma maximum quiescent current so it will regulate
properly - see data sheet link below). This may be the best solution if
the wall wart is up to it.

But the wall wart could be a problem since it is unregulated and is rated
at 9v at some specified current which you haven't provided. All regulators
need some voltage 'overhead' to work. From the LM7805 datasheet

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

The TYPICAL drop out voltage is 2 volts so if you want 9 volts out of the
regulator you need 11 volts going into it TYPICALLY - some 7805's may need
more. When you start putting a load on that wall wart the 14 volts output
is going to fall and at 350ma it may be too low. My *guess* is that unless
the wall wart is rated at more than 9v @1Amp you have no chance of making
it work as you have suggested.

There are a few easy solutions to this: get a 9v regulated wall wart or
get a 12v unregulated wall wart but I suspect as much as possible you want
to roll-your-own. This regulator might solve your problem and still use
your wall wart:

http://www.ebay.com/itm/Boost-Buck-Voltage-Module-3-35V-to-2-2-30V-Step-Up-Down-Converter-Regulator-/290833291375?pt=LH_DefaultDomain_0&hash=item43b703d86f

http://www.ebay.com/itm/290799116764 $1.54

 

[Jasen Betts]

On 2013-01-24, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek

http://pdf1.alldatasheet.com/datasheet-pdf/view/89091/NSC/LM140.html

There it is right on the front page.

adding a capacitor between pin2 and ground will slow the voltage rise
at power-up but and may improve regulation under variable load
conditions,




--
⚂⚃ 100% natural

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[David Eather]

On Thu, 24 Jan 2013 16:42:49 +1000, Michael A. Terrell
<mike.terrell@earthlink.net> wrote:

David Eather wrote:

pOn Thu, 24 Jan 2013 12:27:50 +1000, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider
on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

Stacking regulators in your case may not work. For the top regulator to
work its quiescent current must go through the load on the 5v rail. If
there is no load on the 5v output then it won't regulate. You can fix
this
situation by putting a 470 ohm resistor permanently across the lower
regulator (this provides a continuous load of 11ma which is greater
than
the top regulators 8ma maximum quiescent current so it will regulate
properly - see data sheet link below). This may be the best solution if
the wall wart is up to it.

But the wall wart could be a problem since it is unregulated and is
rated
at 9v at some specified current which you haven't provided. All
regulators
need some voltage 'overhead' to work. From the LM7805 datasheet

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

The TYPICAL drop out voltage is 2 volts so if you want 9 volts out of
the
regulator you need 11 volts going into it TYPICALLY - some 7805's may
need
more. When you start putting a load on that wall wart the 14 volts
output
is going to fall and at 350ma it may be too low. My *guess* is that
unless
the wall wart is rated at more than 9v @1Amp you have no chance of
making
it work as you have suggested.

There are a few easy solutions to this: get a 9v regulated wall wart or
get a 12v unregulated wall wart but I suspect as much as possible you
want
to roll-your-own. This regulator might solve your problem and still use
your wall wart:

http://www.ebay.com/itm/Boost-Buck-Voltage-Module-3-35V-to-2-2-30V-Step-Up-Down-Converter-Regulator-/290833291375?pt=LH_DefaultDomain_0&hash=item43b703d86f


http://www.ebay.com/itm/290799116764 $1.54

Thats not going to solve a problem where he wants regulation unless he can
get 10.5v input at whatever his needed current is (unusual to see a choice
of colors though)

 

[amdx]

On 1/23/2013 10:18 PM, Phil Allison wrote:

"amdx"

Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my divider
to bring it up to 4 volts.


** All you need is a 3.9V zener in series with the ground lead.



... Phil

Yes, but I don't have one locally and I want to get done today.

Thanks, Mikek

 

[John Fields]

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net>
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

---
View using a fixed-pitch font:

Here's how I'd do it, but there are a couple of gotchas, the main one
being that in order for the 78L09 to be happy, its input must see at
least 10.7V with the supply fully loaded.

That means that the ripple valleys must be at 10.7V or higher in order
for the regulator's output to stay at 9V.

It also means that if the supply's peaks fall below 10.7V, then the
output will drop accordingly.

C1 can be used to build up the valleys, but there must be enough
headroom from the supply to never let the peaks fall below 10.7V

Can you post what the output of the supply looks like when it's fully
loaded?




.. 9VSUP 78L09 IRTX
.. +-----+ +---+ +-----+
..MAINS>-----|~ +|-----+----| |-+-----|+ |
.. | | | +-+-+ | | |
..MAINS>-----|~ -|-+ | | [C2] +-|- |
.. +-----+ | | | | | +-----+
.. GND | GND GND GND
.. |
.. | 7805 IRRX
.. | +---+ +-----+
.. +----| |-+-----|+ |
.. | +-+-+ | | |
.. [C1] | +-|- |
.. | | | +-----+
.. GND | GND
.. | XCVR
.. | +-----+
.. +-----|+ |
.. | | |
.. [C3] +-|- |
.. | | +-----+
.. GND GND

Choose C1 so that at full load the output ripple never falls below
10.7V.

For example, assuming that the wall-wart puts out 12V with 2V of
ripple at full load,

I dt
C = ------
dv

Where C is the capacitance,
I is the load current,
dt is the period of the ripple, and
dv is the allowable ripple voltage.

Assuming that the load current is 150mA at 9V, the ripple frequency is
120Hz, and the allowable ripple is 12V - 10.7V = 1.3V, then


0.15A * 0.0083s
C = ----------------- ~ 1000ľF
1.3V


--
JF

 

[amdx]

On 1/23/2013 9:50 PM, John Larkin wrote:

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

You can stack regulators: hang the ref pin of one 5V reg on the output of
another.

See U9:U10 here:

https://dl.dropbox.com/u/53724080/Circuits/ESM/ESM_power.pdf

Dang, I could have done that with U11 and U12, too.


John, I had thought about that but I had a concern that 10 volts

would be to high for my 9v device. Guess I could stick a diode in series
with the output giving me 9.4 Volts, pretty close to a fully charged 9
volt battery.
I think that might be an easy solution, need to reread thread, someone
mentioned quiescent current is a problem.

Thanks, Mikek

 

[amdx]

On 1/23/2013 11:36 PM, David Eather wrote:

pOn Thu, 24 Jan 2013 12:27:50 +1000, amdx <amdx@knologynotthis.net> wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider
on the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

Stacking regulators in your case may not work. For the top regulator to
work its quiescent current must go through the load on the 5v rail. If
there is no load on the 5v output then it won't regulate. You can fix
this situation by putting a 470 ohm resistor permanently across the
lower regulator (this provides a continuous load of 11ma which is
greater than the top regulators 8ma maximum quiescent current so it will
regulate properly - see data sheet link below). This may be the best
solution if the wall wart is up to it.

But the wall wart could be a problem since it is unregulated and is
rated at 9v at some specified current which you haven't provided. All
regulators need some voltage 'overhead' to work. From the LM7805 datasheet

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

The TYPICAL drop out voltage is 2 volts so if you want 9 volts out of
the regulator you need 11 volts going into it TYPICALLY - some 7805's
may need more. When you start putting a load on that wall wart the 14
volts output is going to fall and at 350ma it may be too low. My *guess*
is that unless the wall wart is rated at more than 9v @1Amp you have no
chance of making it work as you have suggested.

There are a few easy solutions to this: get a 9v regulated wall wart or
get a 12v unregulated wall wart but I suspect as much as possible you
want to roll-your-own. This regulator might solve your problem and still
use your wall wart:

http://www.ebay.com/itm/Boost-Buck-Voltage-Module-3-35V-to-2-2-30V-Step-Up-Down-Converter-Regulator-/290833291375?pt=LH_DefaultDomain_0&hash=item43b703d86f


I haven't dealt with this supplier or used this circuit but it seems
reasonable and is from a top rated supplier on ebay so chances are good,
and if you are a hobbyist getting into electronics you will want to get
used to using ebay.

The voltage divider question:

The 78xx series is not a great candidate for this type of modification
because it's quiescent current is so high. It's not impossible but the
LM317 is much better for this (search http://www.fairchildsemi.com for
the data sheet).

The simplest way to jack up the voltage output is to put a zener diode
in series with the gnd pin - for your example a 3.9v diode will do the
job nicely. The quiescent current is typically about 5ma and may be as
high as 8ma which is fine for the zener diode. Since the quiescent
current only changes by a maximum of .5ma this will have almost no
effect on the zener voltage and hence the output voltage

A brute force approach for the 7805 with voltage divider would go like
this:

We want to be able to plug any old 7805 regulator into the circuit and
be within 10% of the 9v value so we set the current through the voltage
divider to be much greater than the maximum quiescent current of the
regulator (8ma). To achieve this We make the design decision that the
divider current will be at least 80ma. Therefor the total resistance of
the divider must be no more than 112 ohms. The regulator gnd pin must be
raised by 4 volts so the bottom resistor (Rb) of the divider must be
4v/9v * 112 and the top resistor (Rt) equals the remainder. Rb = 49.8
ohm Rt= 62.2. The closest E24 values are 51 and 62 ohms (47 and 56 for
E12 values). Because of the amount of current involved you should check
the power dissipation in these resistors which is I*I*R which comes to
approximately 0.4 of a watt. In this example you need to use at least
1/2 watt resistors for Rt and Rb (if the circuit is to sit around
powered up for years and be reliable than 1 watt resistors would be a
better choice).

This design continually wastes 0.8 watt of power but has the advantage
that you could use it on a production line and get the right result
without adjustment but it is wasteful and inefficient.

An approach more suitable to a designer who will adjust his work is like
this:

The maximum change of the quiescent current in the regulator is 0.5ma
(DqI)so wee want to make that small (less than 10%) relative to the
current in the voltage divider. The top resistor (Rt) will always have 5
volts across it and for 5ma (DqI * 10) has a value of 1k ohm. The value
for Rb has to be selected by testing as quiescent current of the
regulator can vary from about 4ma to 8ma and that plus the current from
Rt must cause a 4v drop in Rb. So the value for Rb will be between about
360 to 440 ohms. In this case the power dissipated in Rt is just 25mW
and Rb is under 50mW. Common 1/4 watt resistors will be perfect and the
wasted power is under 0.1 watt. Just grabbing a 390 ohm resistor would
put the circuit within about a maximum of 11% of the desired output but
most people would trim that to be closer to the 9v.

The reason for not using Rb alone is because of how much the quiescent
current can change between devices (4 to 8ma) and how much it can change
in one device (0.5ma) depending on how it is used.

The zener diode works well with the 7805 but the resistor divider not so
well. The resistor divider works well with the LM317 because of its low
current through the adj pin but also for that reason the simple zener
approach does not work so well with the 317

HTH

For now, I'll try the stacking with a series diode on the output to
drop .6 volts.
The wall wart is not a problem, I have a box full of various voltages
and currents.
The wallwart is where my problem started. I used a 6v 700 ma, with a
no load output of 9.6 volts. All worked good until the walkie talkie
went into transmit. Then the wallwart voltage dropped to about 7 volts.
It still worked but I could hear hum on the receive walkie talkie.
If I went to a higher voltage wallwart then the voltage was to high for
the 9 v device. So, the need for a second regulator.
Thanks, Mikek

with s

 

[amdx]

On 1/24/2013 7:32 AM, John Fields wrote:

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

---
View using a fixed-pitch font:

Here's how I'd do it, but there are a couple of gotchas, the main one
being that in order for the 78L09 to be happy, its input must see at
least 10.7V with the supply fully loaded.

That means that the ripple valleys must be at 10.7V or higher in order
for the regulator's output to stay at 9V.

It also means that if the supply's peaks fall below 10.7V, then the
output will drop accordingly.

C1 can be used to build up the valleys, but there must be enough
headroom from the supply to never let the peaks fall below 10.7V

Can you post what the output of the supply looks like when it's fully
loaded?




. 9VSUP 78L09 IRTX
. +-----+ +---+ +-----+
.MAINS>-----|~ +|-----+----| |-+-----|+ |
. | | | +-+-+ | | |
.MAINS>-----|~ -|-+ | | [C2] +-|- |
. +-----+ | | | | | +-----+
. GND | GND GND GND
. |
. | 7805 IRRX
. | +---+ +-----+
. +----| |-+-----|+ |
. | +-+-+ | | |
. [C1] | +-|- |
. | | | +-----+
. GND | GND
. | XCVR
. | +-----+
. +-----|+ |
. | | |
. [C3] +-|- |
. | | +-----+
. GND GND

Choose C1 so that at full load the output ripple never falls below
10.7V.

For example, assuming that the wall-wart puts out 12V with 2V of
ripple at full load,

I dt
C = ------
dv

Where C is the capacitance,
I is the load current,
dt is the period of the ripple, and
dv is the allowable ripple voltage.

Assuming that the load current is 150mA at 9V, the ripple frequency is
120Hz, and the allowable ripple is 12V - 10.7V = 1.3V, then


0.15A * 0.0083s
C = ----------------- ~ 1000ľF
1.3V


All that looks great but the objective of this exercise is to make it

work with the parts I have on hand today. The parts I have are two-5v
regulators. I have many wallwarts, so I have one that won't fall to low.
Hmm... I just recalled, there is an audio repair shop in town and the
guy sells parts. it's worth waiting till he opens and checking if he has
a 9v regulator in stock.
Thanks, Mikek
Thanks, Mikek

 

[George Herold]

On Jan 23, 9:27 pm, amdx <a...@knologynotthis.net> wrote:

Hi all,
  Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
  If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
  So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
  Should I put a capacitor across the lower leg of my divider?
How big?
                            Thanks, Mikek

I think that's pretty standard... did you look at any 5 volt regualtor
spec sheets? I bet they have a circuit.
I've got a LM79L05 spec sheet tacked to my wall. (I can never
remember pinoputs.)
It's got what you want on the back.
R1 from out to 'adjust', R2 from 'adjust' to ground. (Adjust is the
old ground)

Vout = -5V -(5V/R1 +IsubQ)*R2
5V/R1 > 3*IsubQ
0.1uF cap across R2.

All you need is the quesient current for your regualtor.

George H.

*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
  I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

 

[Michael A. Terrell]

George Herold wrote:

On Jan 23, 9:27 pm, amdx <a...@knologynotthis.net> wrote:
Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek

I think that's pretty standard... did you look at any 5 volt regualtor
spec sheets? I bet they have a circuit.
I've got a LM79L05 spec sheet tacked to my wall. (I can never
remember pinoputs.)

Did you mean the LM78L09? The 79xxx series are negative regulators,
the 78xxx are positive regulators.

 

[amdx]

On 1/24/2013 7:32 AM, John Fields wrote:

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

---
View using a fixed-pitch font:

Here's how I'd do it, but there are a couple of gotchas, the main one
being that in order for the 78L09 to be happy, its input must see at
least 10.7V with the supply fully loaded.

That means that the ripple valleys must be at 10.7V or higher in order
for the regulator's output to stay at 9V.

It also means that if the supply's peaks fall below 10.7V, then the
output will drop accordingly.

C1 can be used to build up the valleys, but there must be enough
headroom from the supply to never let the peaks fall below 10.7V

Probably meant troughs rather than peaks.

Can you post what the output of the supply looks like when it's fully
loaded?

Ok, project complete, boxed and working. Picked up a 9v regulator at
the audio repair shop. $3.02 :-(
I used a 1000uf cap to filter the wallwart and .1uf at the output of
each regulator. There are input filter caps in the devices.
At idle, with every device powered the input is 13 volts with 100mv
of ripple. When the walkie talkie transmits the input voltage trough is
11.1 volts and 800mv of ripple. This is fine, there is no sign of any
ripple or droop in the output of either 9v or 5v regulator.
I used a 9v 1 amp wallwart.

Thanks for everyone's input.
Mikek

 

[Jim Thompson]

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net>
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

How about something like this...

http://www.analog-innovations.com/SED/amdx_9V_%28SEB%29.pdf

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[amdx]

On 1/24/2013 11:35 AM, Jim Thompson wrote:

On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

How about something like this...

http://www.analog-innovations.com/SED/amdx_9V_%28SEB%29.pdf

...Jim Thompson

Hi Jim,

I think I like it because it looks like it solves the current flow
problem from the reference pin.
Is that correct?

Thanks, Mikek

 

[Jim Thompson]

On Thu, 24 Jan 2013 11:59:28 -0600, amdx <amdx@knologynotthis.net>
wrote:

On 1/24/2013 11:35 AM, Jim Thompson wrote:
On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

How about something like this...

http://www.analog-innovations.com/SED/amdx_9V_%28SEB%29.pdf

...Jim Thompson

Hi Jim,
I think I like it because it looks like it solves the current flow
problem from the reference pin.
Is that correct?

Thanks, Mikek

Yes. That's why I posted it >:-}

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[John Fields]

On Thu, 24 Jan 2013 11:24:09 -0600, amdx <amdx@knologynotthis.net>
wrote:

On 1/24/2013 7:32 AM, John Fields wrote:
On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

---
View using a fixed-pitch font:

Here's how I'd do it, but there are a couple of gotchas, the main one
being that in order for the 78L09 to be happy, its input must see at
least 10.7V with the supply fully loaded.

That means that the ripple valleys must be at 10.7V or higher in order
for the regulator's output to stay at 9V.

It also means that if the supply's peaks fall below 10.7V, then the
output will drop accordingly.

C1 can be used to build up the valleys, but there must be enough
headroom from the supply to never let the peaks fall below 10.7V

Probably meant troughs rather than peaks.

---
No, I meant peaks, since if they go below 10.7V the regulator output
will start to droop regardless of where the valleys are.
---

Can you post what the output of the supply looks like when it's fully
loaded?


Ok, project complete, boxed and working. Picked up a 9v regulator at
the audio repair shop. $3.02 :-(
I used a 1000uf cap to filter the wallwart and .1uf at the output of
each regulator. There are input filter caps in the devices.
At idle, with every device powered the input is 13 volts with 100mv
of ripple. When the walkie talkie transmits the input voltage trough is
11.1 volts and 800mv of ripple. This is fine, there is no sign of any
ripple or droop in the output of either 9v or 5v regulator.

---
Excellent!
---

I used a 9v 1 amp wallwart.

Thanks for everyone's input.
Mikek

--
JF

 

[amdx]

On 1/24/2013 12:43 PM, John Fields wrote:

On Thu, 24 Jan 2013 11:24:09 -0600, amdx <amdx@knologynotthis.net
wrote:

On 1/24/2013 7:32 AM, John Fields wrote:
On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

Hi all,
Tomorrow I'm working on a power supply to replace a 9v battery.
All I have is a *5 volt regulator. I think can use a voltage divider on
the output, then tie the normally grounded leg to the center of my
divider to bring it up to 4 volts. So I have 4 + 5 = 9 volts.
If I used 5.1K and 4.3k in series to ground. that would give me 9.1
volts. However there is some current from the reference pin.
So, I don't know how to calculate the proper ratio and how much
current do I need to flow through my divider.
Should I put a capacitor across the lower leg of my divider?
How big?
Thanks, Mikek


*My local Radio Shack only carries 5V and 12 V.

PS. In case this isn't feasible,
Here's the scene, I have three devices, an infrared transmitter (9v)
an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
both of the 4.5v units on 5 volts and I'm comfortable they will be OK.
I'll be running this all from a 9v wall wart, we'll say it puts out
14v at no load. The big load is the walkie talkie during transmit
~350ma. At idle it is less then 10 ma for all devices combined.

---
View using a fixed-pitch font:

Here's how I'd do it, but there are a couple of gotchas, the main one
being that in order for the 78L09 to be happy, its input must see at
least 10.7V with the supply fully loaded.

That means that the ripple valleys must be at 10.7V or higher in order
for the regulator's output to stay at 9V.

It also means that if the supply's peaks fall below 10.7V, then the
output will drop accordingly.

C1 can be used to build up the valleys, but there must be enough
headroom from the supply to never let the peaks fall below 10.7V

Probably meant troughs rather than peaks.

---
No, I meant peaks, since if they go below 10.7V the regulator output
will start to droop regardless of where the valleys are.
-

Hmm... I guess you just get hum if the troughs go below 10.7v.
Mikek