I'm powering a router, two webcams and a GPS unit, all working off of 5V for a total of 1.3Amps (input is 12V to the regulators). I have two L7805C-V voltage regulators . The L7805C-V regulators are rated up to 1.5Amps each with proper heat sinking. I have no heat sinking. I tried powering the GPS unit with a single regulator (0.1Amp @ 5V = 0.5W) and the regulator got too hot to touch. I tried putting an additional regulator in parallel with the first in the hope that I would now be distributing the current load between them. I put in a temp sensor and measured a temperature of 170F/77C on one regulator while the other was fairly cool.
Does putting identical regulators in parallel pose any problems? If one is trying to put out precisely 5.01V and the other precisely 4.97V, does that essentially put more of a burden on one regulator than if there were only one regulator? Is it false to assume that putting two regulators in parallel simply doubles the amount of current I can use?
How should I go about taking a 12V supply (I'm looking at burglar batteries on Amazon, about 5Ah @ 12V, which would give my robot some time to roam around) and turning it into 5V? I'm using the 12V line to power my motors (0.3A) and the Ardunio, so I haven't looked at 5V batteries. I've just completed some tests and it seems that current into a the regulator is equal to the current out of it, so I'm looking at ( 7V ) * ( current used ) as the heat dissipated by the regulator. At 1.3A, that's 9W I need to get rid of, which doesn't seem feasible (at least not when using a single regulator) since the heat sinks on Digikey only go up to 7.5W, and require a significant airflow to keep them cool. Should I be looking at PWM (at 42% duty cycle) on the 12V signal with a large capacitor to generate the 5V signal? The webcams and router are off-the-self, so I'm not sure how much power fluctuation they're willing to take...
If you cannot buy a heat sink then make one by bolting the regulator to a LARGE lump of aluminium
If your 5 volt load is reasonably constant, you can dump some of the excess 7 volts across a power resistor before the regulator. Say the load is 1 amp, then dump 4 of the 7 volts across a 3.9ohm, 10watt resistor.
Install a small PC fan to keep the regulator cool
You can boost regulator output by using its output to drive a power transistor, but this is somewhat overcomplicating the issue
1.5A at 7V difference across the regulator is just over 10W - a heatsink recycled from an old computer complete with fan would be one solution, or find some aluminium sheet about 10cm by 10cm to bolt the regulator to...
A 12V->5V DC-DC converter might be a better choice, these waste less power since they are switch-mode.
Does putting identical regulators in parallel pose any problems? If one is trying to put out precisely 5.01V and the other precisely 4.97V, does that essentially put more of a burden on one regulator than if there were only one regulator? Is it false to assume that putting two regulators in parallel simply doubles the amount of current I can use?
You've identified the root of the problem in paralleling, which is that one regulator will always be slightly lower in voltage than the other, and no matter what it does it can't lower the output voltage since the other regulator is keeping it up. Thus, it essentially shuts down and passes no current, leaving the other regulator (with the higher voltage) to carry all the current.
You can balance out the situation to an extent by adding "ballast resistors" in series to the output of each regulator, small values like 0.1ohm or so, and then tying the far ends of the resistors together to form the "5V" node. Each resistor will have a voltage drop RI1 for regulator 1 and RI2 for regulator two, where I1 is the current carried by regulator 1 and I2 the current carried by regulator 2. The distribution of currents has to satisfy:
5.01 - RI1 = 4.97 - RI2 = Vout
I1+I2 = 1.3A
assuming the 5.01 and 4.97 numbers from your example, and 1.3A of total current. Just to finish this example with R=0.1 ohms you will find that I1 = 0.85A and I2=0.45 A. So there will be slightly more equal current sharing but still not exact. The bigger you make the resistor R, the more equal the current sharing, but the lower the final output voltage Vout becomes, and the more power (heat) is dissipated in these ballast resistors. Nothing comes for free.
--
The Ruggeduino: compatible with Arduino UNO, 24V operation, all I/O's fused and protected
Is it false to assume that putting two regulators in parallel simply doubles the amount of current I can use?
Yes, false without using something to help them 'share' the total current load. An old application datasheet for a LM7805 regulator showed one method of paralleling their regulators by wiring isolation diodes from the outputs of each regulator ( the anode lead) and wiring all the cathodes together for the load connection. This allows some sharing as if one regulator is passing too much current it's output voltage will start to decrease then other regulators will start to pick up the load, so a rough balancing act is in place. However the disadvantage is that the output voltage seen by the load will be around .7vdc less then the nominal regulator output voltage due to the forward voltage drop across the diodes.
If you want the heat shared you can put them in series, 12V down to 9V and then that into the regulator down to 5V. In that way the heat you have to burn off is split. You will probably still need a heat sink though.
Thank you all for the in-depth replies. I dug through my supplies and took apart an old PC, got the heat sink off the CPU and took the power block fan (0.2A @ 12V), so I'm using those to keep the regulators cool. The fan really makes a difference, between 35C and 75C with and without it on (room temp is 27C). I split up the regulators so one powers the router (0.6A) and the other powers the webcams (0.6A) and GPS (0.1A).
Using RuggedCircuit's method, I have the equation:
5.04V - I1* R1 = 5.08 - I2 * R2 = 5.09 - I3 * R3.
I wanted to optimize for about 27 amps, so I1 + I2 + I3 = 27A. With:
R1 = 0.005 ohms, R2 = 0.01, R3 = 0.01, I get:
I1 = 9A, I2 = 8.5A, I3 = 9.5A. Awesome.
Then, I checked out what would happen if my load only pulled 5A (the draw will vary from 0A, possibly all the way up to 27A). With those same resistances, when:
I1 + I2 + I3 = 5A,
I1 = -2A, I2 = 3A, I3 = 4A....
What is the significance of the negative amperage here? Will the current flow backwards through the regulator and back into the battery?
I've seen LDO regulators in parallel burn out a whole board - never parallel voltage regulators,
they are high gain devices and if you get instability you potentially fry everything. LDO's are
particular risky as they are non-linear and marginally stable in the first place (which is why you
have to use caps with the right ESR rating).
Hm, ok, thanks. That is unfortunate news. I would like to convert 12V down to 5V, with up to 30A coming out at 5V. How would you recommend I go about doing this?
EddieCarlson:
Hm, ok, thanks. That is unfortunate news. I would like to convert 12V down to 5V, with up to 30A coming out at 5V. How would you recommend I go about doing this?
. The L7805C-V regulators are rated up to 1.5Amps each with proper heat sinking. I have no heat sinking. I tried powering the GPS unit with a single regulator (0.1Amp @ 5V = 0.5W) and the regulator got too hot to touch. I tried putting an additional regulator in parallel with the first in the hope that I would now be distributing the current load between them. I put in a temp sensor and measured a temperature of 170F/77C on one regulator while the other was fairly cool.