LED in series with power rail? - Page 1
Forum section: electronics / beginners

author: elimenohpee
author_rank: Contributor
timestamp: April 02, 2022, 12:45:24 pm
content: I was looking at the schematic on this demo board from ADS:https://www.analog.com/en/products/ltc6090.html#product-overviewOn the negative rail, there is an LED in series. How does this LED not blow up?? It looks like a standard LED according to the BOM (P/N PANASONIC, LNJ951C4BRA)

author: Benta
author_rank: Super Contributor
timestamp: April 02, 2022, 05:51:46 pm
content: I don't see a big problem.The boards draws a few mA, so doing this is OK.A clue is, that the positive output is +65 V, the negative -62 V corresponding to the 3 V across the blue LED.

author: golden_labels
author_rank: Super Contributor
timestamp: April 02, 2022, 11:56:19 pm
content: If you still don’t see how Benta’s answer explains the situation: diodes have a steep I-V curve. The current through it is the primary factor determining the power dissipated. Not the voltage that could potentially, no pun intended, be applied to it.

author: elimenohpee
author_rank: Contributor
timestamp: April 03, 2022, 10:48:45 am
content: That is what is the confusing bit for me. This demo board is for a power opamp that can source / sink upwards of 50 mA @ 140V. What if the user puts a load on that exceeds the LED's current rating?

author: wraper
author_rank: Unknown
timestamp: April 03, 2022, 11:06:14 am
content: I don't think that SMPS can provide anywhere close to 50mA @ 65V. If it does not exceed 10mA, it's fine.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 03, 2022, 11:44:39 am
content: Still, a cheapskate design.

author: wraper
author_rank: Unknown
timestamp: April 03, 2022, 11:59:21 am
content: Quote from: Terry Bites on April 03, 2022, 11:44:39 amOverengineering without any practical result is useless. IMHO smart cheapskate with the same end result is the best.

author: golden_labels
author_rank: Super Contributor
timestamp: April 04, 2022, 09:38:14 am
content: Quote from: elimenohpee on April 03, 2022, 10:48:45 amThe important word: can. Doesn’t mean it does. And even if it would hit 50mA for a dozen milliseconds, the LED will not mind.

author: SeanB
author_rank: Super Contributor
timestamp: April 04, 2022, 09:52:58 am
content: They want a power indicator on the secondary side, and a LED likely will double the current draw, simply to lose a lot of power in a series resistor. With a 10mA current draw, limited more by the transformer impedance and limited power capability, they used the supply rail current instead, and D2 cleverly also regulates the voltage on both rails, but only dissipates half the power, running the converter at a fixed duty cycle on the primary side as well.

author: tooki
author_rank: Super Contributor
timestamp: April 04, 2022, 04:37:59 pm
content: Quote from: elimenohpee on April 03, 2022, 10:48:45 amBzzzzzt!1. “Absolute maximum ratings” are NOT the same as the design specs!From the datasheet: “The LTC6090 is specified to source and sink 10mA at 140V. If the total supply voltage is dropped across the device, 1.4W of power will need to be dissipated. If the quiescent power is included (140V • 2.8mA = 0.4W), the total power dissipated is 1.8W.”2. Specified maxima often cannot be reachedall at the same time.This is a power op-amp, but only because it’s a high-voltage device, not because it’s a high-current device!! You most certainly cannot run it at 140V/50mA for anything beyond tiny peaks, because that’d be 7W of power (plus the 0.4W quiescent, so 7.4W). The best-case thermal resistance is 33°C/W, meaning that 7.4W would cause a temperature rise of 244°C, far,farexceeding the maximum junction temp of 125°C in the highest grade (only 85° or 70° in lower grades). The worst-case thermal resistance (when no thermal pad is present) is stated as up to 120°C/W, meaning a theoretical temperature rise of 888°, which obviously isn’t gonna happen. Either the internal thermal protection will disable the chip, or it’ll fry itself.The maximum power is the maximum allowable temperature rise (e.g. 125° max Tj - 25° ambient temp), divided by the thermal resistance (e.g. 33°/W). So if your maximum temp rise is the 100°, divided by 33°/W, then you can dissipate up to 3W. If you’re running it at 140V total, then that’s 21.4mA, of which 2.8mA is the quiescent current, leaving you a maximum output current of just 18.6mA — and that’s still higher than the 10mA design spec.On the other hand, if it’s running at 60V, then you can consume 50mA total, less the 2.8mA quiescent, so 47.2mA output.But again, well beyond the design spec, and something you wouldn’t want to do long term.

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 04, 2022, 05:56:11 pm
content: Is the asymmetric +65V/-62V a "feature" here? Why power the opamp with an asymmetric dual voltage? Is there a reason as far as the opamp is concerned, or is it just *because* they use an LED as a power indicator (which will then drop about 3V)?The benefit of this arrangement is not just to show that the power supply is on, it does show that the load (I suppose, the opamp here) is drawing power from it. But I fail to see a real plus if you lose 3V in the negative rail along the way. Just a thought. I wouldn't be worried about the max current as has been said already above, but I still think it's a relative poor arrangement for a questionable reason, unless again the asymmetry is actually a requirement. Which I doubt, but if it is and I missed the reason, then it's all good.

author: Benta
author_rank: Super Contributor
timestamp: April 04, 2022, 06:35:56 pm
content: Just add a blue LED in the positive rail as well, then it's +/-62 V and you can stop hyperventilating.

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 04, 2022, 07:20:08 pm
content: We could just a third blue LED, if it can help you stop playing smartass.

author: Benta
author_rank: Super Contributor
timestamp: April 04, 2022, 10:47:20 pm
content: Quote from: SiliconWizard on April 04, 2022, 07:20:08 pm

author: thm_w
author_rank: Super Contributor
timestamp: April 05, 2022, 12:37:59 am
content: Please read the datasheet of the eval kit for explanation:Quotehttps://www.analog.com/media/en/technical-documentation/user-guides/dc1979af.pdfWhat they don't specifically tell you (although it is implied) is that this circuit is designed to drive high impedance measurement equipment, so 1M oscilloscope or 10M DMM's. Where the current consumption will be very low. The LED becoming bright would indicate an overload, although, I suppose it still could burn up before the opamp shuts down.