What is happening here? - Page 1
Forum section: electronics / beginners

author: Mario87
author_rank: Frequent Contributor
timestamp: September 29, 2019, 08:35:55 pm
content: Hi all, I have a circuit as shown below using an IRF520 MOSFET and the pulse is controlled by an arduino.Now when I use a 1K resistor, or even a 220 ohm resistor in place of R1 I get the following output, which is exactly what I am looking for....Basically, the 5v rail is pulsed down to GND for 4038us, allowed to go high for 577us, then pulsed back down and repeats. Perfect.....However, the circuit I will be using will eventually be drawing circa 2A on every pulse, so for a test I hooked up a 2.2ohm, 50w power resistor (V=IR -> 5=2.2I -> I=2.27A).Problem is, when I do that I get the following output where it is only pulling the 5v rail down to circa 3.3v before being allowed to go high again, there are also very large overshoots when going high and going low. Only for a few ns, but overshoots none-the-less. The transistor also becomes very hot, pretty quick. Can anyone explain this to me, as I am a bit confused by it. Screenshot below....MOSFET datasheet here...https://www.vishay.com/docs/91017/91017.pdf

author: isometrik
author_rank: Regular Contributor
timestamp: September 29, 2019, 08:44:28 pm
content: If you look at the Vds vs Id vs Vgs characteristics curves, you will notice that the resistance of the IRF520 is about 0.4 ohm at Vgs = 5V.

author: iMo
author_rank: Super Contributor
timestamp: September 29, 2019, 08:48:51 pm
content: You need a driver which creates more than 5V at the gate..When the mosfet is not fully opened, it works in a linear region and heats up.

author: madires
author_rank: Super Contributor
timestamp: September 29, 2019, 08:56:26 pm
content: Or use a logic level MOSFET. The spikes are generated by the power resistor since it's wirewound and therefore inductive.

author: Mario87
author_rank: Frequent Contributor
timestamp: September 30, 2019, 09:20:13 am
content: Quote from: isometrik on September 29, 2019, 08:44:28 pmOk, so I take it that because the resistance is approx 0.4 ohm then I am effectively creating a voltage divider with the 2.2ohm resistor, so I will never be able to pull the voltage down to 0v, as some of it will always be across Vgs (approx 0.6v by my calcs). However, this obviously becomes a negligible voltage at higher resistances of R. Is that right?Quote from: imo on September 29, 2019, 08:48:51 pmOk, I have tested and you appear to be correct. When looking at the graphs on the datasheet (I hate logarithmic scales), it appears that at 5v Vgs & 5v Vds the drain current is circa 1.4-1.58AHowever I cannot find anything in the datasheet that states in order to drive a Vds of 5v I need to have a Vgs geater than 5v...so what is happening here?? I am still a bit confused due to lack of information in the datasheetQuote from: madires on September 29, 2019, 08:56:26 pmI think you are correct, logic level MOSFET would be best. I did actually suspect inductance in the wirewound resistor causing the spikes, so I added a flyback diode to try and eliminate them, but it didn't actually do anything.Adding a 1k resistor to the gate however, did remove the spikes when the gate powered off (voltage output went to 5v).Any further explanations on the above?It is clear you guys are correct, but due to lack of info on the datasheet I am still a bit perplexed as to why it reacts as it does.

author: magic
author_rank: Super Contributor
timestamp: September 30, 2019, 11:19:06 am
content: Quote from: Mario87 on September 30, 2019, 09:20:13 amEverything is on figure 1 (typical output characteristics).At some value of Vds, the FET saturates and drain current can't go any higher. You need to jump to the next curve (increase Vgs), then higher current is possible.Quote from: Mario87 on September 30, 2019, 09:20:13 amAlso figure 1Gate resistor reduces current which charges gate capacitance. Vgs raises slowly from 0 to 5V and therefore drain current ramps up gradually. This reduces the inductive effects. Notice that the rising/falling edge is much slower with a gate resistor. The FET heats up a bit in the process, this may become a problem at high switchng frequency.

author: mikerj
author_rank: Super Contributor
timestamp: September 30, 2019, 11:43:57 am
content: Quote from: Mario87 on September 30, 2019, 09:20:13 amYou are not attempting to drive a Vds of 5v, your goal is to get Vds as low as possible so use figure 1 in the datasheet as magic suggests. Vds with the MOSFET switched off is pretty much irrelevant.

author: Mario87
author_rank: Frequent Contributor
timestamp: September 30, 2019, 11:55:34 am
content: OK, that seems to make more sense. Mikerj, you are correct, I am not driving Vds of 5v, I am trying to drop it as low as possible. Just confusing myselfThanks guys, so seems like my best bet going forward to stick with the arduino control is a logic level MOSFET.

author: Kilrah
author_rank: Unknown
timestamp: September 30, 2019, 12:00:13 pm
content: If you look at the Rdson and other current-related ratings they're specced at Vgs = 10V, so that's a good value to consider as what you have to use to get the MOSFET fully conductive.