high side mosfet switch and circuit  - Page 1
Forum section: electronics / beginners

author: elektronJack
author_rank: Contributor
timestamp: March 20, 2023, 04:54:42 am
content: my understanding of this circuit is that it uses a npn mosfet high side switch.does that not mean then that the mosfet will not turn on and allow currentto pass unless the mosfet gate voltage is higher than the voltage at the drainby an amount of the gate threshhold eg if gate threshhold is between2 and 4 volts and the voltage at the drain is 5 volts, then the applied voltageat the gate must be higher than 5 + 2 volts ?********************************with the amended circuit and using a LED and a resistor as load, i. l. o. R1, the LED was on and there were no visible oscillations. no visible LED oscillations were expected.when i connected my oscope between GND and the blue mark of the schematic i was expecting to see oscillations.why didn't i see oscillations?i used mosfet K3569(tried to attach datasheet but couldn't)my DMM showed voltages :replaced 5v with 6vpin3 3.38vpin2 1.98vpin1 4.82v+Load 5.64vbetween blue mark on schematic and GND2.48v

author: Kim Christensen
author_rank: Super Contributor
timestamp: March 20, 2023, 05:17:13 am
content: Quote from: elektronJack on March 20, 2023, 04:54:42 amIt's called an N-Channel MOSFET... Not an npn.Your circuit wouldn't work as drawn: The load has no source of power.What is the purpose of this circuit?QuoteNo... As long as thegatevoltage is higher than thesourcevoltage (by an amount greater than the gate threshold voltage) then the FET will be on even if the drain voltage was lower than the gate voltage.

author: WattsThat
author_rank: Frequent Contributor
timestamp: March 20, 2023, 06:12:46 am
content: That is NOT a high side switch. It is an electronic load circuit - as the mosfet is being used as a variable resistor driven by the opamp. The gate voltage is controlled by the voltage drop across the 1 ohm resistor which is proportional to the current following through the mosfet.If you get rid of the opamp and the 1 ohm resistor, it’s the basic configuration of a LOW side switch.Sparkfun and Adafruit have decent tutorials on mosfet switches.

author: Psi
author_rank: Super Contributor
timestamp: March 20, 2023, 06:37:38 am
content: Put as simply as I can.If you want to use a N-channel mosfet as a high side switch you must have a power source to drive the mosfet gate that is at a higher voltage than the voltage the mosfet is switching.So, say your N-channel mosfet is controlling a 12V fan. And say the mosfet is rated to work for 5V gate drive.Because the mosfet is being used as a high side switch the 5V needed to turn it on is "on top of" the 12V going to the fan.So you need to get the mosfet gate up to 17V to make the 12V fan turn on.To understand why this is the case, understand that the voltage applied to the gate of a mosfet is not a voltage with respect to GND it's a voltage with respect to the source pin of the mosfet.If using the mosfet for low side switching the source pin IS gnd and it makes things easy.If using the mosfet for high side switching the source pin is at your output voltage (when on). So you need a gate voltage that is at your output voltage + the drive voltage.

author: BillyO
author_rank: Super Contributor
timestamp: March 20, 2023, 02:14:11 pm
content: What you have drawn is a variable current limit. As previously mentioned though, it wont' do anything because there is no power supply on the output. The "++ terminal should go the the load supply voltage and the "-" terminal should go to the drain of the MOSFET.

author: Terry Bites
author_rank: Super Contributor
timestamp: March 20, 2023, 05:54:31 pm
content: To clarify. This not a current source a variable resistor or an electronic load. Terminology matters. This circuit is a variable current sink.Once a voltage has been set with the potentiometer, the output will sink a constant or maximum current through the load from a positive supply .The circuit is incorrectly labelled. Current cannot normally flow from ground into the load terminal- it will only flow from positive supply into the sink. The terminal marked as GND Load should be connected to a positve potential.The current flowing in the sink (Isink) will be the voltage at the pot wiper (Vw) divided by R1. The Load terminal can't swing bellow (Vw+Vgs(th))+(Isink*R1)+(Isink*Rds). No one said it would be easy!Opamp charcateristics may add further restrictions on this. This may limit its usefulness when the load supply voltage is low. The MOSFET operates in its linear region and will generate heat.Isink*Vds watts. Something to watch out for.