Low side switch puts negative voltage on load - Page 1
Forum section: electronics / beginners

author: knotlogic
author_rank: Regular Contributor
timestamp: April 11, 2016, 08:53:12 am
content: Well I've been trying to figure this one out for a while, but I'm stumped. I have NMOSFET as a low side switch, driven by an ATtiny. The circuit is meant to cycle the power on a load on a regular basis. I can cycle power fine, but I've noticed that when I switch off the MOSFET I see a negative voltage across the load.I've tested it with a green LED driven through a resistor. The LED has a forward voltage of about 2.0 V, and when the MOSFET is off, the low side of the LED load floats at about 11 V above supply ground (The positive voltage rail sits at 12V). The bit I can't figure out is that I'm measuring -2.0 V across the LED when the MOSFET is off.That's confounding enough. When I hook up my actual load to the circuit, the low side of the load floats at 19.3V when the MOSFET is off, and the voltage I measure across is is approximately -8.4 V. Not only can I not figure out why I'm getting a negative voltage across the loads, but the numbers don't seem to add up. Unless I'm missing something incredibly obvious.... Any thoughts?

author: amspire
author_rank: Super Contributor
timestamp: April 11, 2016, 09:02:50 am
content: Sounds like your 12V DC is actually AC. When the MOSFET is off, it acts as a diode that conducts when the 12V is negative.

author: danadak
author_rank: Super Contributor
timestamp: April 11, 2016, 11:02:50 am
content: Is you meter on AC ? If so when MOSFET is off you could see higher thanpower supply voltages due to AC coupling from mains wiring in your lab.Regards, Dana.

author: knotlogic
author_rank: Regular Contributor
timestamp: April 12, 2016, 02:22:23 am
content: Quote from: danadak on April 11, 2016, 11:02:50 amYou mean plugged into mains? No it was a handheld meter. (In DC mode, just to cover that too...)Quote from: amspire on April 11, 2016, 09:02:50 amWell the brick I'm using has a slew of safety approvals on it, so I'm a little surprised it's that bad. But I stuck a USB scope (which is all I have on hand at the moment) across the green LED, and yep, 50 Hz stuff clamped by the LED on the positive, and the input range of the scope on the negative. That matches local mains frequency.I'm now wondering what I can do about that. Series diode for reverse protection on the supply line? I'm not sure if it's even needed, but for the sake of robustness I figure I can suffer through a small board respin.

author: knotlogic
author_rank: Regular Contributor
timestamp: April 12, 2016, 06:19:53 am
content: So thinking about it some more, I'm not sure how much a single diode inline with the supply would work (D1 in the attached image). It would prevent reverse current* but not really help with the reverse voltage across the load.So how about I use diode D2 to clamp the voltage across the load under reverse conditions? I figure whatever's in the load should be able to withstand a reverse diode drop's worth of voltage**. By itself I would be worried that D2 and the MOSFET body diode would short out the supply, but together with D1 I should be safe.Thoughts and criticism welcome.* Um, at least down to the reverse current leakage of the diode? When my load is a LED and series resistor I couldn't measure any voltage drop across the resistor with the MOSFET off so I don't know if that matters anyway.** Would that be a safe assumption? Or put another way, I haven't fried my loadyet, so -0.7 V should be fine. Right?

author: danadak
author_rank: Super Contributor
timestamp: April 12, 2016, 11:27:31 am
content: When yous say reverse voltage across load, MOSFET off, you are actually seeing MOSFETdrain more + than the rail side of the load ?Ofetn in MOSFET switched loads, a diode (or back body diode) is used for transientsuppresion when switching L loads, as L loads will generate large reverse polaritytransients.Regards, Dana.

author: knotlogic
author_rank: Regular Contributor
timestamp: April 12, 2016, 02:16:16 pm
content: Quote from: danadak on April 12, 2016, 11:27:31 amYes, exactly that. The drain is above my 12V supply load. How much more seems to depend on the load I have attached.And yes, I'm aware of the use of a recirculating diode together with inductive loads. In this case I'd only be using it as a voltage clamp, the load isn't inductive as far as I'm aware.

author: danadak
author_rank: Super Contributor
timestamp: April 12, 2016, 03:51:54 pm
content: Man this is crazy. If the only potential in the system is 12 V, there areno other energy storage or sources in system than can deliver energy,you have a miracle on your hands....Regards, Dana.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: April 13, 2016, 01:38:37 am
content: Quote from: danadak on April 12, 2016, 03:51:54 pmEither that... or one or more of the assumptions about the actual circuit and its behaviour are wrong.(And I don't believe in miracles.)

author: knotlogic
author_rank: Regular Contributor
timestamp: April 13, 2016, 02:50:23 am
content: Heh, no. It looks like the problem is that the AC-DC adapter acting as the source is "leaking" an AC signal through to the load. I don't have the gear on hand to investigate this fully, so I'm stuck with the limited measurements I can make. amspire called it:Quote from: amspire on April 11, 2016, 09:02:50 amAnd yes the adapter issupposedto have a DC output. At least that's what it says on the label.