Not all FETs work as switches? - Page 1
Forum section: electronics / projects

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 26, 2018, 02:13:13 am
content: Are Mosfets the only type of FET that can be used as a switch?I am using the attached circuit as a high side switch which is driven by a PIC IO pin. I first tried a FDV304P. However, when I turn the FET on by driving the PIC pin high, the FET gets too hot to touch. It doesn't matter if a load is connected or not.So I ordered some DMP3099L-7 parts to replace the FDV304P. The DMP3099L-7 part runs cool as a cucumber with or without a load connected.That indicates to me that a standard FET cannot be used as a switch. But I could be wrong. The FDV304P is described as a DMOS digital FET by the manufacturer.

author: BrianHG
author_rank: Super Contributor
timestamp: January 26, 2018, 02:19:39 am
content: The FDV304P does work as a switch, you are just over driving the gate. The maximum gate voltage of the FDV304P is -8v and it has an internal 8v zener diode between the gate and source protecting the gate. You are driving it with -12v, damaging the internal zener diode, heating it up whether or not you have a load on the drain...

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 26, 2018, 04:13:59 am
content: Thanks Brian. I missed that about the zener and am still learning FETs. I will keep using the DMP3099L-7 since the board is already designed.But, for the sake of knowledge, if I put another 10K resistor between the collector of the 2N3904 and the FDV304P gate it would work? I believe that would drive the gate with 6 volts.

author: floobydust
author_rank: Super Contributor
timestamp: January 26, 2018, 04:52:52 am
content: Note the on-resistance is much greater for theFDV304P0.87R at 0.5A ; could also be why it ran hot.DMP3099L-7is 0.065R at 3.8A much better for bigger loads but slower.Adding a 10k gate resistor doesn't do much for the DMP3099L as it has no gate protection zener. For the FDV304P though, it would limit the current through that 8V zener when you are giving 12V drive.If your power supply or load could ever spike above 20V, then I'd include a gate protection zener, for reliability.

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 26, 2018, 06:19:29 am
content: QuoteThe FDV304P ran hot without a load so it wasn't a question of high on-resistance. I think it was just the 8V zener creating the heat. The circuit I am driving with 12V will only need about 5 mA so the high on-resistance would not be a problem.I would not add the 10K resistor for the DMP3099L, I was just curious if that solves the issue for the FDV304P. Out of curiosity I will test it tomorrow to see. The DMP3099L is actually 1 cent less than the FDV304 from Mouser and has a higher voltage rating so that is the part I will use. If the power supply ever spikes above about 15 volts it will damage a lot more circuitry than just the FET.The main question I had was whether a standard FET could be used as a switch just like any bipolar transistor can be used as a switch. Apparently they can as long as you understand the operating parameters.

author: Zero999
author_rank: Super Contributor
timestamp: January 26, 2018, 09:03:38 am
content: Quote from: rcbuck on January 26, 2018, 04:13:59 amBetter still, move R1 to between Q1's base and 0V.

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 26, 2018, 06:05:16 pm
content: QuoteI don't see how that would help anything. In fact, that would make it impossible for the PIC to drive Q1's base.This morning I did confirm that putting a 10K resistor between Q1's collector and Q2's base does allow Q2 to turn on and remain cold.

author: Zero999
author_rank: Super Contributor
timestamp: January 27, 2018, 03:00:01 pm
content: Quote from: rcbuck on January 26, 2018, 06:05:16 pmWhy do you think that?Moving R1 to Q1's emitter will limit the voltage across Q2's gate to the PIC's output voltage, minus Q1's base drop. Assuming the output voltage from the PIC is 5V and Q1's base-emitter voltage is 0.6V, the gate drive on Q2 will be 4.4V. If that's too low, reduce the value of R1.QuoteThat will work, but the advantage of having an emitter resistor, is the gate voltage depends on the PIC's output voltage, which will be fairly well regulated. The power supply, powering the load, can then be varied over a wide range, without affecting the gate voltage.

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 28, 2018, 05:48:01 pm
content: QuoteBecause you said move R1 to between Q1's base and 0V. You didn't say anything about Q1's emitter.The circuit will work the way you show it. However, I don't like the idea of connecting the PIC pin to Q1's base directly with no current limiting. Since the DMP3099L-7 doesn't have a zener in it, I will leave R1 where it is.After reading floobydust's post, I have decided to add a 15 volt crowbar circuit right where the 12 volt supply comes into the box. The power supply "should" never go above about 12.5 volts. But you never know for sure. I would rather take out a fuse rather than damage other circuitry. Hopefully the fast blow fuse would go before anything else did.

author: floobydust
author_rank: Super Contributor
timestamp: January 28, 2018, 06:59:34 pm
content: This is a high-side switch scheme I use without trouble in environments with transients and harsh conditions.Q1 can be a small MOSFET like BSS138, but add a gate pull down resistor so no drama if the MCU O/P floats by accident (stuck in bootloader etc.). BJT is better for low voltage MCU's and HV rating, but more current used for the base drive.D3 protects the MOSFET switch from -ve spikes dues to any load inductance.I've seen designs with no series resistors and the MOSFET failed short, took out the driver MOSFET and then the MCU exploded.

author: Zero999
author_rank: Super Contributor
timestamp: January 28, 2018, 09:36:48 pm
content: Quote from: rcbuck on January 28, 2018, 05:48:01 pmYou're correct. I apologise. I meant to say between the emitter and base.QuoteYour concern is unfounded. The emitter resistor will limit the base current to a much lower level, than it would be if the resistor was in series with the base. The base current will be equal to the current through R2, divided by Q1's Hfe, which will be around 4.6µA, assuming an Hfe of 100. It will also speed up the turn-off time, as Q1 will never go into saturation, although I appreciate speed isn't a concern here and it will never be fast with such high resistor values.QuoteI think that's a good idea.If you experience nuisance tripping, you ca add a slow down capacitor to the crowbar and another, slightly higher voltage zener to catch short, high voltage spikes.

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 29, 2018, 04:11:15 pm
content: QuoteYes, that was my concern. If the Mosfet shorted and took out the driver, even with a 10K resistor you are still applying 12 volts at 1 mA to the MCU output pin. Hopefully the MCU would survive. With no resistor it would be destroyed along with everything else tied to the same Vdd bus.

author: Zero999
author_rank: Super Contributor
timestamp: January 29, 2018, 07:08:10 pm
content: Quote from: rcbuck on January 29, 2018, 04:11:15 pmIn this case, Q1 is rated to over 12V, so even if the MOSFET failed short it would survive and nothing bad would happen to the MCU.EDIT:Have you considered a proper MOSFET drive IC, such as the MCP1401 or MC34152?https://www.onsemi.com/pub/Collateral/MC34152-D.PDFhttp://ww1.microchip.com/downloads/en/DeviceDoc/20002052D.pdf

author: rcbuck
author_rank: Frequent Contributor
timestamp: January 30, 2018, 03:39:00 am
content: QuoteNo, but the MCP1402 looks interesting. It would certainly work nicely with a higher powered Mosfet.