LED power IN-USE indicator (current indicator) - Page 1
Forum section: electronics / beginners

author: cprobertson1
author_rank: Frequent Contributor
timestamp: August 14, 2014, 11:58:29 am
content: I'm building a dev board for microprocessors - but ran into confusion when drawing up the circuitry for the power terminals and the associated LED indicators.I am in need of two LEDs - a power-good indicator (simple LED across the terminals): and a "power-in-use" indicator that, with any luck, wont interfere too much with the rest of the circuitry.The purpose of this is to indicate if there is any current being drawn by the header pins (which serve as a power source for anything on the dev board)What is the best way to do this?I had a couple of ideas on using transistors but I'm inexperienced with their usage.Attached is one circuit I thought might work - though the top "header connected indicator" made me wonder what would actually happen were it wired like this? I doubt it will work as expected - but at the same time I can't figure out what would actually happen! (I'm away from my workshop and can't make a mockup just now). Incidentally, are those two resistors in parallel when the jumper is connected?)My other problems was that of any resistors I put across the terminals would end up in parallel to any circuitry downstream of the headers - would I be right in saying I can reduce or isolate that problem using transistors as switches?Many thanks!

author: Refrigerator
author_rank: Super Contributor
timestamp: August 14, 2014, 02:04:20 pm
content: Maybe something like this could work ?

author: jakeypoo
author_rank: Regular Contributor
timestamp: August 14, 2014, 02:08:45 pm
content: You could put a shunt resistor in series with the v+ header pin. This would need to be a very small resistor, then amplify the voltage across the resistor to drive your "power-in-use" LED.Size the shunt so that voltage drop is at most 2-3% of your Vcc Voltage. ie: if you anticipate your downstream circuit of drawing 500mA max, pick a 0.2Ohm 1% resistor which will give you 100mV of voltage drop (which 5V circuits should be tolerant of).The problem exists that you must supply at least Vf to the LED to turn it on. So a competitor might be better over a simple non-inverting opamp setup.

author: David Hess
author_rank: Super Contributor
timestamp: August 15, 2014, 09:59:11 pm
content: Use a two transistor current mirror as a differential comparator with the inputs at the emitters. The voltage drop will only need to be millivolts then if the transistors are matched for Vbe and the circuit is simple.The new Raspberry Pi B+ uses this type of circuit to switch its p-channel MOSFET at its input but this is the best two transistor example I found online:http://www.beam-wiki.org/wiki/I-Mirror

author: cprobertson1
author_rank: Frequent Contributor
timestamp: August 16, 2014, 09:07:41 am
content: Thanks for the help folks!Im finally back at my workshop so I can start mocking these up - will probably do that tomorrow (sunday 17th) as I'm working back shift today and have quite a bit of soldering to do this morning (stripboard is great and all, but when it comes to more complex things it can be a bit of a time-sink! I wonder if PCB Etching is a viable alternative - will need to look into that)ANYWAY - will mock up power-sense circuirts tomorrow and let you know how I get on! Many thanks for the advice!

author: Zero999
author_rank: Super Contributor
timestamp: August 16, 2014, 10:23:39 pm
content: There are various ways to do this.What current do you need to turn the LED on at?A transistor can be used but it has the disadvantage of dropping 0.7V.

author: cprobertson1
author_rank: Frequent Contributor
timestamp: August 17, 2014, 11:11:46 am
content: Quote- any - its just to indicate that there is power in use somewhere in the circuit (i.e that I have something active connected to the power source - as opposed to merely the power source being connected to the board).Will mock up circuits today and let you know how I get on! (might have problems with that last one, I dont have a schottky diode handy - again, will let you know xD)Purely out of curiosity (as it doesnt apply for this project) - could I wrap the power-in wire round a toroidal ferrite and then wrap the sense wire round that same ferrite and amplify any signal I get through it to detect a given current flow (or even voltage)? Wait... have I just made a very inefficient transformer doing that?

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2014, 12:25:39 pm
content: A transformer will only work at AC, so if the current is continuously varying and isn't completely filtered out by decoupling capacitors it might work, otherwise a hall effect sensor will work at DC.

author: David Hess
author_rank: Super Contributor
timestamp: August 17, 2014, 12:31:03 pm
content: Quote from: cprobertson1 on August 17, 2014, 11:11:46 amThis would work but in order to sense DC current, a hall effect sensor would have to be used with the toroid. Signal conditioning would still be needed on the output of the sensor anyway to drive the LED.

author: cprobertson1
author_rank: Frequent Contributor
timestamp: August 18, 2014, 06:46:28 pm
content: QuoteAh! Good point! xDI've hacked together Hero999's circuit using a Schottky I salvaged from a... actually I don't know where that board came from? Er... Ithinkit may have been a printer? Orion? I'll figure that out later.Another quick question - would an optocoupler tied to a transistor been of any use? (not detecting current per say but I could use it to detect a high voltage on the wire leading to the jumpers for the board - excusing the voltage drop of the transistor of course)Many thanks for all your help by the way! It's been very much appreciated