Forums

Have 5v regulator need 9 volts

Started by amdx January 23, 2013
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.
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
"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
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
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
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, -- &#9858;&#9859; 100% natural --- news://freenews.netfront.net/ - complaints: news@netfront.net ---
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)
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
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 = ----------------- ~ 1000&#2013266101;F 1.3V -- JF
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