Parallel voltage regulators and the detrimental affects of ballast resistors - Page 1
Forum section: electronics / projects

author: SharpEars
author_rank: Unknown
timestamp: August 21, 2015, 01:36:36 pm
content: Looking at the Linear LT1083 datasheet (http://cds.linear.com/docs/en/datasheet/108345fh.pdf) top of page 15, the show a sample circuit of how to connect multiple voltage regulators in parallel. This is a typical circuit where the outputs are connected together after a ballast resistor (2 feet of 18 gauge wire on each output) is connected in series with each output.Unfortunately, connecting series power resistors with the outputs hurts voltage regulation at the load, because of the voltage drop on the power resistors at high currents. This particular regulator is rated at 7.5 A of output, so 2 feet of 18 gauge wire can bring the voltage down by over a hundred millivolts at full load. For low voltages especially, a hundred or more millivolts can be a significant percentage of the output voltage.What can be done to alleviate this - keep the load regulation of the regulator while allowing for multiple regulators to be wired in parallel?

author: Kleinstein
author_rank: Super Contributor
timestamp: August 21, 2015, 04:03:20 pm
content: There are regulators with tighter specs, like the LT308x series. Here the load sharing resistor can be smaller, so you only loose something like 20-50 mV at full load.Still these LDOs are rather limited in power, so the max current is available only at a low drop.

author: digsys
author_rank: Unknown
timestamp: August 21, 2015, 04:18:17 pm
content: That's easy fixed - Use closely matches enamel type resistors, with one channel used as master. You inject the "loss" back into the regulator.

author: SharpEars
author_rank: Unknown
timestamp: August 21, 2015, 04:50:13 pm
content: Quote from: digsys on August 21, 2015, 04:18:17 pmI am not sure I understand how the precision of the resistor plays a role. The problem here is the voltage level of the regulators themselves being uneven. Even if I were to match them "exactly" at no load, before connecting them together, they may not stay exactly matched under varying load conditions.Also, two feet of 18 AWG cable is a pretty damn good match in resistance, if cut to the same length (0.01% territory there...). It's not the precision of the resistors that the issue, it's how to eliminate them or reduce their resistance to as close to zero as possible to preserve load regulation while still having good current sharing among the regulators.

author: SharpEars
author_rank: Unknown
timestamp: August 21, 2015, 05:02:01 pm
content: Quote from: Kleinstein on August 21, 2015, 04:03:20 pmI am not sure what you mean. The LT308x series regulator with the highest current capacity is the LT3083 (3 A), but it requires a control voltage in addition to Vinalthough that can be tied to Vinwith some caveats. Also, because it is only 3 A, I would need to parallelize five of them for say 15 A, vs. two 7.5 A LT1083 regulators. There is an application note for the LT3083 (Page 21 ofhttp://cds.linear.com/docs/en/datasheet/3083fa.pdf) that shows its use in parallel with 10 mOhm ballast resistors (compared to 13 mOhms for two feet of 18 AWG cable), so apparently Linear still thinks they are necessary.There is also the LT3086 that is single supply, but only provides 2.1 A of output and 0.1% load regulation just like the LT1083, so that's quite a step down.I chose the LT1083 for a reason - it's a very high current linear regulator, but if I need more current than 7.5 A, it's back to poor load regulation due to the ballast resistors.

author: digsys
author_rank: Unknown
timestamp: August 22, 2015, 03:32:48 am
content: Quote from: SharpEarsFair enough, I wasn't sure what level of analysis you were up to, so I gave you the basics, and left you to figure out the rest :-)Here's how it works :-)U2, R5, VR1, R6, R7 form your basic Linear regulator. R1 is a current sense resistor, and if you chose 5A per channel (giving you 2.5A headroom), it WILL get quite warm,at 5A, 50mR + 250mV, 1.25W or at 5A, 100mR = 500mV, 2.5W !! The LT1083 is a pretty efficient thermal package, but I still never push them >5A.U2 is my "Master" and drives all the other "slave" regulators. We have up to 20 in a product. There is an approx 0.5% - 1.0% variation between regulators, but that'snot even worth worrying about, after all we have plenty of headroom. There are no thermal runaway issues, as SHOULD a regulator try to run off, it's own internalthermal protection will shut it down. We've never had that problem though.The only other "channel matching error" are the equalisation resistors.Your normal "cheap" power resistors have 5-10% tolerance, and terrible tempco / stability and will affect sharing, so you need something stable.J2 and J3 are connected at the output (OR you may have separate output connections and use J3 as "lead loss correction" in the "outside" world.The op-amp sums the losses from the current sense resistor + wiring and ADDS it to the regulator output ! You need to play around with R9 to get the rightcorrection "curve". You can in fact have the Vout INCREASE with higher currents, if you wish.In other designs (usually HV), I run the op-amp at a logic 5V and drive the R6/R7 junction with the correction voltage. There's heaps of ways to do it.PS: R10 has an important role - Should the "correction input" J3, ever break, it ensures the output doesn't go flying up to Vin or some other uncontrolled voltage.You can use 10-100R etc, as long as it is 50x-100x total expected lead resistance.

author: SharpEars
author_rank: Unknown
timestamp: August 24, 2015, 02:01:46 pm
content: Quote from: digsys on August 22, 2015, 03:32:48 amThat's a bit over my head, but I will try to go through it and figure out what you saying.