Hi James, There's capacitance in the MOSFET between gate and drain. When you hit the gate with voltage, some current couples to the drain befor= e the gate reaches it's turnon threshold. With the LED in place and not on, the LED is essentially an open circuit. So that coupling from the gate causes a little bump in voltage at the drain= .. With the LED shorted, the resistor puts enough load on the drain that there= is very little bump. Also there's lots of other coupling going on everywhere because of the brea= dboard, but that's not what's causing the effect you are seeing. Cheerful regards, Bob ________________________________________ From: piclist-bounces@mit.edu on behalf of James = Cameron Sent: Friday, February 3, 2017 8:42 PM To: piclist@mit.edu Subject: [EE] why drain rises as N-channel MOSFET opens with LED not 220R? Slightly puzzled. A microcontroller output pin is wired through a 4k7 resistor to the gate of an N-channel MOSFET (BUK477 or MTP3055). The MOSFET source is grounded. The MOSFET drain is connected to a 220R resistor, then a red LED, then to 5V. The micro is running PWM, at 1kHz, with a very small duty cycle. My question; why does the drain voltage rise briefly as the gate rises? Why does it not rise if the LED is shorted? http://dev.laptop.org/~quozl/DS1Z_QuickPrint105.png (with LED) http://dev.laptop.org/~quozl/DS1Z_QuickPrint104.png (with LED shorted) Legend; - yellow is micro output, - cyan is gate, - purple is drain, - blue is 5V nearby. Circuit is on socketed breadboard with 70mm hookups. --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .