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Old 09-29-2011, 01:34 PM #1
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Default Why a test-load IS still necessary with linear drivers... (another unique edge case)

Common wisdom has always been that you don't really need to use a test load, when setting a linear driver. The current measured without any load, should be the same as the current measured with an LD attached (presuming of course that you've got a sufficient supply voltage to cover your Vf + Vdropout).

Well, this may not be the case after all, at least not in certain very-unique situations. This isn't likely to change the way we approach testing linear circuits for LDs like a 445 or 635 - but it's an interesting point to be aware of as we start pushing some of the higher current diodes that are likely to start popping up (especially in the IR arena)


The story:

- I have a couple freak "7W" C-Mount 925nm diodes that I was testing last night.

- Without a bench supply, I turned to my flexible "DDL Box" to serve as a current source. It's basically a DDL driver built from an LM1084 chip (to handle high currents) with a heavy duty pot for current setting, a heatsink on the IC, a case for 18650s, and a tactile push-button for quickly discharging the caps. I've used this box to test virtually every diode that has come across my plate - from 445s, to the HL63133DG, 12Xs, high powered IRs, etc. It's a good box.

- For this 7W IR diode, I needed that 1084 chip to be delivering it's full "rated" 5A of regulated output. I used 2S2P high current 18650 cells that I knew could deliver on that demand.

- However, when I set the current as high as the pot would let me, my DDL Box would only register 3.9 A on a DMM in current mode. I thought this was odd initially, but chalked it up to either the fact that the 1084's "current rating" of 5A was really meant to apply to it's max current when used as a voltage regulator so perhaps didn't apply here, OR to the fact that I was setting current via a pot, and I didn't really remember what the reference resistance of that pot was when at its lowest (perhaps it was simply whatever resistance would have produced 3.9A - I supposed around 0.3 ohms)

- Ok I thought, 3.9A is likely to at least cause this diode to lase, even if it is rated at 7W.

- It certainly did cause the diode to lase, but with the DMM in current mode and in series with the LD, my read-out now indicated a bit over 6A of current.


What was going on?

- This had me stumped for a while. My DMM was fine, and this DDL Box has been rock solid for 6 months. I had used it on hundreds of LDs, and had never seen a different between the output current when measured with a test load, no test load, or with an LD. Regardless of the load (or lack thereof), this flexible driver circuit had been bang-on consistent. So why was the current now going UP, once I attached an LD?

- Then it occurred to me that I was asking the question backwards. In reality, the current was properly set at ~6A, and the readings I observed during operation with an LD attached, were correct.

- What wasn't as it should have been, was the 3.9A of current I was observing when running the circuit minus a diode or test load. This turns out to be one of those "edge cases" where some of our common wisdom deceives us. Without a laser diode in series with the driver's output, the only "load" (so to speak) on the DDL Box output, was the current-mode DMM. This meant that the entire ~7.5 Volts of input supply was being dissipated by the IC. That's almost 29.25W of power (7.5 x 3.9), which coincidently is almost exactly the 1084's maximum power dissipation as per the datasheet (30W).

- When the LD was attached to the DDL Box output, in series with the DMM, I had essentially offloaded 3V worth of power dissipation responsibility. Now, with less voltage to drop, the IC could manage delivering the entire 6A of current I had requested of it, while staying (just barely) under its 30W maximum power dissipation rating.

- We've always know that there were thermal considerations to pay attention to with linear drivers. However, it takes pushing the IC right to the edge of its power dissipation capabilities to run into a use-case where you risk diode damage if you neglect to use a test-load. You obviously need a test load if there is any concern over not having enough input voltage to cover Vf + Vdropout. I would now suggest that you also need a test load when there is concern that you have too much input voltage to be dropped by the IC alone, in the absence of a diode.


RHD, that's a lot of reading for a very small edge-case scenario!

- You're right. Sorry about that.


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- Federal Aviation Administration - Laser Safety Initiative (link)
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Old 09-29-2011, 01:49 PM #2
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Default Re: Why a test-load IS still necessary with linear drivers... (another unique edge ca

I read it all heh good lesson +1 for the warning
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Old 09-29-2011, 01:50 PM #3
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Default Re: Why a test-load IS still necessary with linear drivers... (another unique edge ca

I guess this is a feature common to many voltage regulators - they keep within their safe operating area at all costs and will reduce current if required. Often this only happens after they run hot and the limitation is thermal, but it doesnt have to be.

There is, however, a very simple doublecheck for this: When running the current source with a short circuit as the load, measure the voltage over the potmeter / current setting resistor. If this is lower than normal (1.25V i presume), the chip is out of regulation: either due to a lack of load, lack of voltage drop, or throttle back.

I guess it would be possible to add a circuit that detects this condition and lights a led warning you of nonregulation.
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Old 09-29-2011, 04:31 PM #4
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Default Re: Why a test-load IS still necessary with linear drivers... (another unique edge ca

A power resistor (value selected with ohm's law, ~0.3 ohms in this case) would still be simpler and cheaper. Besides, JUST the 1ohm resistor that you'd normally use for a test load would drop 6V in this case.

Or just set the current low and turn it up as needed.
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Old 09-29-2011, 11:35 PM #5
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Default Re: Why a test-load IS still necessary with linear drivers... (another unique edge ca

There are lots of better solutions than the lm317 style circuit, obviously. This example just points out one example how it could unexpectedly fry a diode when you just test the output current into a short.

But all drivers do have their limitations, and the fault-condition indicator could also be useful on a proper opamp based circuit... to indicate that the load cannot be driven at the requested current from the input voltage. On a circuit like Merghart.com - Opamp based adjustable current source this would only require a resistor and indicator led to be wired between the opamp output and ground.
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