Have 5v regulator need 9 volts

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

See U9:U10 here:

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

--

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

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"amdx"

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

... Phil

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

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:

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

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

$1.54

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,

--
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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)

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

Thanks, Mikek

--- 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 = ----------------- ~ 1000µF 1.3V

-- JF

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

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

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

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.

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

Probably meant troughs rather than peaks.

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

How about something like this...

...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  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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I love to cook with wine.     Sometimes I even put it in the food.

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  |             | 
| Voice:(480)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.

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

on

.1

ors,

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.