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I'm trying to power a 12V 1A device with 1 cell lipo battery. At 12W, the lipo would need to provide 3.24A (assuming no power loss) to step up the voltage to 12V and providing 1A.

However the DC-DC Boost coverter (XL6009) is rated at 2A continuous input. So I've decided to connect 3 DC-DC Boost converter in parallel, so each of them share the load and keep everything cooled down. Did some googling and I'm not sure if this is a good idea, some say the small different in output would fry the converter with lower output over time.

My question is should I do this at all? If so do I need a diode to deal with the different in output voltage? And if so what diode should I use in my case?

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  • \$\begingroup\$ Here's the datasheet of XL6009: lib.chipdip.ru/164/DOC001164799.pdf . Datasheet says that minimum input voltage is 5V for full functionality. \$\endgroup\$ – Rohat Kılıç Oct 4 '16 at 13:05
  • \$\begingroup\$ Typically if a power device supports load sharing it is advertised as a feature. I'm not saying that parts which are not advertised can't be connected in parallel, but the others should do it more easily. \$\endgroup\$ – Arsenal Oct 4 '16 at 13:09
  • \$\begingroup\$ what is your load device exactly \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 4 '16 at 13:24
  • \$\begingroup\$ If you can copy a design and buy the PCB, TI.COM gives you the free tool to do this \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 4 '16 at 13:30
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In your post you said that you wanted 1A out at 12V, and that you calculated that you needed 3.24A in with no losses. You didn't state what input voltage you were using, but it can be calculated. 12V * 1 A / 3.24A = 3.7V. 3.7V is much lower than the minimum input voltage of 5.0V specified in the datasheet. So your supply probably won't work right.

With respect to the second problem of paralleling boost regulators. Yes it can be done.

The output of the XL6009 boost converter, like most boost converters includes a diode at the output stage. If multiple XL6009 are put in parallel, that diode will prevent the output current of one supply from flowing back into the others. Now in order to make sure that the supplies share the load you need to add some sort of feedback. The easiest way is to add a small series resistor at the output of each supply and then connect the other side of each resistor together to feed your load.

The XL6009 has a reference voltage tolerance of just under 3%, assuming a divider consisting of two 1% resistors is used to set the otput at 12V, then the total tolerance in the output is ±5%. So the output of each supply will be 12V ± 0.6V DC plus ripple.

Lets assume that you actually had a 5V input and that the supply was 92% efficient as stated in the datasheet. So if you want 1A at 12V out = 12W then you will need to put in 12V * 1A / 92% / 5V = 2.6A. In that case you probably only need to parallel two supplies.

You want the load sharing to be balanced such that neither supply needs more than 2A input at the worst case output voltage, which is 12.6V on the supply with the high output current, and 11.4V on the other supply. At 2A input the supply can output 2A * 5V * 92% / 12.6V = 730mA output. In that case the other supply would need to supply 1A - 730mA = 270mA.

To ensure that the load sharing is such that each supply is not worse than 270mA to 730mA on each output a small series resistor needs to be added to the output. There will be some drop in the load voltage through the series resistor. Lets call the load voltage Vout, and the output of each supply V1 and V2. Lets call the output current of each supply I1 and I2.

To solve for the resistor value we now have two equations and two unknowns.

Vout = V1 - I1 * R
Vout = V2 - I2 * R

Therefore

V1 - I1 * R = V2 - I2 * R

Solving for R gives...

R = (V1 - V2)/(I1 - I2)

We already know that in the worst case we want V1 = 12.6V, I1 = 730mA, V2 = 11.4V, I2 = 270mA.

Therefore R = (12.6V - 11.4V) / (730mA - 270mA) = 2.6 ohms.

The wattage in the resistor will be 2.6 ohms * (730mA)^2 = 1.38W worst case. Therefore you would probably need a 2W resistor.

Under that worst case sharing scenario the output voltage including the drop across the resistor would be 12.6V - 2.6 ohms * 730mA = 11.14V.

With both sharing equally outputting 11.4V at 500mA your output could be as low as 11.4V - 500mA * 2.6 ohms = 10.1V.

With both supplies sharing equally outputting 12.6V at no load your output could be as high as 12.6V.

If you want to back off from that 2A worst case input a bit you can increase the resistor value beyond 2.6 ohms to create more equal sharing at the expense of efficiency.

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So I've decided to connect 3 DC-DC Boost converter in parallel, so each of them share the load and keep everything cooled down

Scenario 1

The device that produces the highest output voltage (fractions of a few milli volts over the others) is the device that wins the race to supply all the current and will eventually burn or shut down. Then, the device that produces slightly more output voltage then the third device (but less than the first device) will win and it will continue to supply current until it burns or shut downs leaving the third device and that eventually burns or shuts down.

Scenario 2

However, as the first device warms it's output may reduce and gracefully start to share load duties with the 2nd device and this may extend to the third device but this is usually a bit of a fairy tale and won't happen / can't be relied on.

Scenario 3

Active synchronous power supplies will fight each other dumping as much current into each other's outputs in order to raise the voltage and win the race (as per scenario 1).

There are probably other scenarios but those involving output diodes fall into scenario 1 because the diode acts to prevent the current dumping mentioned in scenario 3. Diodes also drop anything up to 1 V and so energy is wasted and power is lost.

My recommendation - buy a single module designed for the job.

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You can use a current mirror to actively and accurately divide the power load between your three DC-DC boosters. All you need is just one NPN transistor for each booster. I have used three cheap TIP31s, but any medium power type would do, as long as they are all the exact same model. This is because a current mirror depends on all transistors having very simmilar properties regarding beta, thermal behaviour, etc. Building the current mirror is very easy: connect the collector od Q1 to the output od booster 1, collector od Q2 to output of booster 2, and collector of Q3 to to output of booster 3. The emittors of Q1,Q2,Q3 are to be tied together. This is your new combined power output, going to the input of your load. Next, connect all the bases together, and then - this is important - connect the base of Q1 also to the collector of Q1. Do this last step only with Q1. Q2 and Q3 (being the same model and sharing a common base with Q1, will therefore behave nearly identically and copy the the collector-emitter current of Q1, no matter how much higher the booster 1 and booster 2 voltages are. Lastly, you need to adjust your booster output voltage to compensate for one diode drop (because your transistors are now emitter followers). For 12v output, set booster 1 to 12.6v and booster 2 and 3 to 13v (they need to be equal or higher than booster 1 to be able to copy Q1's current, but this will not affect your output at all, because it is now actively controlled.) So, what have we gained from this approach? Same current through all parallel boosters times same output voltage (emitters are tied together) equals exactly same power delivered by each booster, no matter what the load, and no matter what the (higher) voltage of booster 2 and 3. Ohms law. And unlike just using resistors, the output voltage will remain nicely constant as regulated by Booster 1, no matter what the current draw of your load is. I hope this is helpful. Mike3301

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  • \$\begingroup\$ Can someone check if this really works? I've been reading about current mirrors since I read this and I am very inclined to try this with MOSFETs. \$\endgroup\$ – rmarques Sep 12 '17 at 10:26
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I'm reading that a current mirror made from discrete mosfets can't be used because you won't find more than 2 in box of 500 mosfets similar enough in voltage and gain - so the expenses, in purchase of 500 items, and then the time to find the matching pair, is large.. Then the matching Mosfets might be used only for low current. In practice, mosfets can be used in low current application in an IC design, because there close proximity in the IC ensures they match well enough .. but temperature variations prevent their use in high current even so.

For power (anything not small signal) use an extended /improved (4 transistor) wilson BJT current mirror..

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Yes, best option is to buy a more powerful module, but is almost always at least double the price of multiple modules in parallel of the same total current or more. Right now i'm using 6 modules in parallel to give me a boost of 28V 30amp from 13V 80amp each module has a output diode. What Andy says about one module that is set to a few milivolts more gives more current, is true, but only in few miliamps so really not a problem. Moreover, each module has a overcurrent protection so natturally if one gives more current, the protection automatically lowers the voltage thus the system is working flawlessly.

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