Have 5v regulator need 9 volts

On Thu, 24 Jan 2013 14:09:15 -0600, amdx <amdx@knologynotthis.net>
wrote:

On 1/24/2013 12:43 PM, John Fields wrote:
On Thu, 24 Jan 2013 11:24:09 -0600, amdx <amdx@knologynotthis.net
wrote:

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
Click to expand...
---
Yes, precisely.

The regulator input needs to be at least 1.7V higher than the output,
at all times, for its own internal circuitry to work properly.

--
JF
 
On Jan 24, 11:54 am, "Michael A. Terrell" <mike.terr...@earthlink.net>
wrote:
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.- Hide quoted text -

- Show quoted text -
Click to expand...
Well the pinout on my wall is for the 79XX so that's what I quoted.
(I can remember the 78xx pinout. :^) Others have pin ups on their
wall, I've got pin outs, wire gauges, screw sizes and Drill tables
(And some drawings done by my kids when they were younger.)

George H.
 
George Herold wrote:
On Jan 24, 11:54 am, "Michael A. Terrell" <mike.terr...@earthlink.net
wrote:
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.- Hide quoted text -

- Show quoted text -

Well the pinout on my wall is for the 79XX so that's what I quoted.
(I can remember the 78xx pinout. :^) Others have pin ups on their
wall, I've got pin outs, wire gauges, screw sizes and Drill tables
Click to expand...

I have all those on my hard drive(s), and can print them out if
needed. Some data sheets that were made from catalog pages, where the
OEM never prnted a real datasheet. :)


(And some drawings done by my kids when they were younger.)
Click to expand...

As long as they aren't trying to design electonics with crayon &
newsprint. ;-)
 
"amdx"

** 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.
Click to expand...

** Your stupid problem - fuckhead.
 
On 1/24/2013 5:16 PM, Phil Allison wrote:
"amdx"


** 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.


** Your stupid problem - fuckhead.



Is that you Phil? It's seems a little weak.
Click to expand...
Imposter!

Mikek
 
On Thu, 24 Jan 2013 17:38:53 -0600, amdx <amdx@knologynotthis.net>
wrote:

On 1/24/2013 5:16 PM, Phil Allison wrote:
"amdx"


** 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.


** Your stupid problem - fuckhead.



Is that you Phil? It's seems a little weak.
Imposter!
Click to expand...
No, Phyllis is normally weak. ...or were you commenting on the
"little"?
 
On 1/24/2013 2:09 PM, amdx wrote:
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


When testing, my line voltage was 123.3vac. When I got to my boat and
Click to expand...
checked the line voltage it was 112.9vac with my heater running.
I decided to use a 12v wallwart instead of the 9v.
The 9v Wallwart only had ~ 0.4 of headroom at full load.
 
On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net>
wrote:

an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested
Click to expand...
What's a walkie talkie? That sounds like some sort of gay sex device!
 
On 1/29/2013 2:04 AM, kaweeziur@criminal_investigation_services.com wrote:
On Wed, 23 Jan 2013 20:27:50 -0600, amdx <amdx@knologynotthis.net
wrote:

an infrared receiver (4.5v) and a walkie talkie (4.5v). I have tested

What's a walkie talkie? That sounds like some sort of gay sex device!
Click to expand...

Your persuasion has jumped to a conclusion.

In case you're really confounded, here's a link.



http://www.bhphotovideo.com/bnh/controller/home?O=&sku=604274&is=REG&Q=&A=details

Mikek
 
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