Negative Current - Page 1
Forum section: electronics / projects

author: sambran
author_rank: Contributor
timestamp: December 28, 2015, 07:04:16 pm
content: My analog skills are a bit rusty and I need some help.I am trying to build a wireless charging circuit based on the following inscrutablehttp://www.instructables.com/id/Low-Power-Wireless-Charging/I built the following circuit for simulation:The following is the simulation result of the current from the DC Source:As you can see the current is negative at times so I was wondering what it means with regard to a DC source. I eventually plan to power this circuit from a USB power brick, I'd imagine that most dc bricks would have circuitry preventing the reverse current.Any advice would be appreciated.

author: Fungus
author_rank: Super Contributor
timestamp: December 28, 2015, 07:56:39 pm
content: It means your inductor L1 is generating a voltage which is higher than the DC source.I'd put in a diode to block reverse currents. Many DC sources won't be happy otherwise.

author: sambran
author_rank: Contributor
timestamp: December 28, 2015, 08:05:22 pm
content: Quote from: Fungus on December 28, 2015, 07:56:39 pmRight, but shouldn't the current flow through R1 and R2?Quote from: Fungus on December 28, 2015, 07:56:39 pmI tried that, in that case the circuit won't oscillate.

author: Zero999
author_rank: Super Contributor
timestamp: December 28, 2015, 09:06:43 pm
content: LTSpice always shows the current taken from a positive voltage source as negative. It's where the current goes positive it's being returned to the power supply.There's no need for diodes. In real life, just connect a capacitor across the power supply which will simply charge up a little when the current goes positive.

author: Fungus
author_rank: Super Contributor
timestamp: December 29, 2015, 05:32:14 am
content: Quote from: sambran on December 28, 2015, 08:05:22 pmSome of it will, yes, but there's 20k of resistance there and the source is in parallel.Quote from: sambran on December 28, 2015, 08:05:22 pmWhere did you put it? It should be on the output of the DC source, V1.

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2015, 06:12:29 am
content: Be suspicious of any circuit that parallels bipolar transistors without individual emitter degeneration resistors to aid even current sharing.

author: TimFox
author_rank: Super Contributor
timestamp: December 29, 2015, 09:58:16 pm
content: In general, avoid connecting multiple diodes in parallel (as the B-E diodes in this circuit) without current-sharing resistors.The actual forward voltage for a given current varies somewhat from device to device (of the same part number). However, since the current is an exponential function of the voltage, the current at a given voltage (all parallel devices have the same voltage) will vary dramatically.

author: c4757p
author_rank: Super Contributor
timestamp: December 29, 2015, 10:14:47 pm
content: Quote from: Fungus on December 29, 2015, 05:32:14 amAnd just where exactly do you think the current is going to go at the moments when the polarity changes? Ithas to, that's how this oscillatorworks.sambran, ditch the diode, it's a bloody stupid idea. Just put a nice big capacitor across your power supply in the real-life circuit, it'll take the reverse current. (It won't have any effect in the simulation.) You could use a capacitoranda diode, if you really wanted to be sure, but the lilies are probably gilded enough.

author: Fungus
author_rank: Super Contributor
timestamp: December 30, 2015, 01:07:54 am
content: Quote from: c4757p on December 29, 2015, 10:14:47 pmI haven't analyzed the circuit but R2 looks like a candidate.

author: c4757p
author_rank: Super Contributor
timestamp: December 30, 2015, 01:40:40 am
content: R2, at 10k, looks like a candidate to sink that negative current, which was 1A?

author: Fungus
author_rank: Super Contributor
timestamp: December 30, 2015, 02:17:41 am
content: Quote from: c4757p on December 30, 2015, 01:40:40 amHuh? The energy in an inductor isn't measured in amps, it's measured in joules.But ... if you say a capacitor works then who am I to argue? You obviously know so much. I'm just not very sure that applying high voltages to the output of a power supply is a good idea.

author: c4757p
author_rank: Super Contributor
timestamp: December 30, 2015, 06:06:47 am
content: Quote from: Fungus on December 30, 2015, 02:17:41 am....and the energy is a direct function of the current, as E = 1/2 L I2.QuoteIf you decouple the supply, youwon'tapply high voltages to anything. That current will flow into the capacitor with only a very small voltage rise, if the capacitance is sufficient and the ESR sufficiently low.Yourmethod, on the other hand, is likely to generate a quite high voltage and shove it right up the diode. I hope you use a nice high-voltage one.Though you'll probably get a whole one or two good spikes out of it before it settles and just stops oscillating...