MOSFET DS/Transistor CE voltage drop - Looking for a solution to switch DC line. - Page 1
Forum section: electronics / beginners

author: tifkat
author_rank: Regular Contributor
timestamp: October 22, 2020, 02:42:52 pm
content: Hi All,I'm brainstorming a project which mashes together 2 devices which run from USB, and have hit a wall with handling power switching. Basic block diagram follows.So, the USB Multiplexer handles switching the data lines between the 2 circuits, but not power. The 'S' input is low for one output, and high for the other. There is an OE (normally low) input to disable/enable the outputs.I'd like to have a push button for the user to toggle between the two outputs, a selectable default state at power on (maybe via jumper pins/pads on the PCB). I'd also like to drop 5v power to the 2 circuits for maybe 1 second before making the switch, and then 0.5 secs after, the 5v power is applied to the desired circuit. The MUX and other 'front end' stuff always stays powered while the USB/DC input is connected.I'm feeling that the best way to handle the 'smart' side of the switching is with a small MCU, which can signal H/L on the S line, handle the user button presses, and control which circuit receives 5v power.Firstly, I'm not sure the MCU can supply enough current, and then there's the 3.3v / 5v nature of chosen MCU. I have considered Transistor/Mosfet but don't know if the associated voltage drop would have a dramatic effect. I like the thought of a pull up resistor and driving the line low, but this would probably also have a voltage loss. A relay would be ideal, but size is an issue. I'd love to know if there is an IC equivalent of a relay. Something small (like SOIC/TSSOP or SOT??) but I'm having troubles finding anything like this. As usual, I'm finding that a lot of pretty obvious stuff has been flooded onto the web and youtube, using keywords which are related, but this seems maybe to be more esoteric in nature, and not really covered. Of course it may just be that I don't know what to search for.The switching capability doesn't need to be high voltage nor high speed. I wouldn't have thought 500mA is high current either, but maybe that is a bit of an ask for a small SMD IC?Closest thing I could find that seemed to fit is the Vishay VOR family.Any suggestions?EDIT: That 'Power disable switch (NC)' was from a previous iteration of the dawing, where I only had one switch, cutting/providing power to both circuits. I felt it may be better to be able to apply power only to one specific circuit at a time. Please ignore that artifact. The 500mA remains true though.Collin

author: tifkat
author_rank: Regular Contributor
timestamp: October 22, 2020, 11:50:27 pm
content: I've decided I need to ask a more pertinent question:Does a MOSFET or Transistor have a (significant) voltage drop across the DS or CE junction?

author: Peabody
author_rank: Super Contributor
timestamp: October 23, 2020, 03:34:46 am
content: Well first, since USB is already 5V, you don't need a 5V regulator. And your ATtiny 402 will run at 5V. I don't know anything about the USB multiplexer. If there's a 5V version of that, you wouldn't need a 3.3V regulator either.If you're switching the positive voltage rail, you would use a P-channel mosfet, and one with good specs would have a fairly low RDSon, and a very low voltage drop at 500mA. That's the equivalent resistance of the mosfet when it's conducting, and you can calculate what the voltage drop would be. Take a look at the DMP1045U. Look at the diagrams showing Rds(on). It looks like RDSon would be about 0.03 ohms, which at 500mA would be a drop of 0.015V, which is essentially zero.Lucky for you, P-channel mosfets are basically Australian - upside down and backwards. :-)

author: tifkat
author_rank: Regular Contributor
timestamp: October 23, 2020, 11:42:46 am
content: So if I was to use (for example) the Diodes Inc DMP2035UTShttps://www.digikey.com.au/product-detail/en/diodes-incorporated/DMP2035UTS-13/DMP2035UTS-13DIDKR-ND/2355965I would hook each of the S lines to my separate circuits, the common D to the 5v rail and the G1 and G1 to the ATTiny MCU digital out pins. I'd hold those pins high to stop current flow or low to allow current flow. Does this sound correct?I'm also guessing that each S has two pins to allow enough current to pass through, and the two D pins allow the same. Two D and two S when either channel is active.The USB MUX I found is the TI TS3USB30. It's capable of up to 7v supply voltage, so 5v should be good.

author: Peabody
author_rank: Super Contributor
timestamp: October 23, 2020, 02:54:55 pm
content: I don't think that mosfet pair will work. You would need to connect the sources to the 5V rail, and the drains to your circuits. But that pair has common drains. If you'll look at the diagram at the beginning of the datasheet, you'll see that current needs to flow from the source to the drain because of the body diode. The other way, current would still flow through the diode even when the mosfet is off. Something like this should work:https://www.digikey.com/en/products/detail/diodes-incorporated/DMG9933USD-13/4249923The datasheet for the new multiplexer has 7V as the absolute maximum, but under recommended operating conditions it has Vcc at 4.3V maximum. But you could probably just power it through a diode (1N4007 or similar) which would give you a 0.65V drop. But I'll say again that I have no experience with USB multiplexers, and don't know how they work.

author: tifkat
author_rank: Regular Contributor
timestamp: October 24, 2020, 04:02:48 am
content: Ok, so is it correct to say in FETs, it is true that:1. Source, is where the conducted charge comes in2. Drain is where the conducted charge goes out3. Gate is where an appropriate charge is supplied to active/deactivate the FET (change the state of the depletion zone) - to open or close the gate, so to speak4. A P channel FET conducts positive charges5. An N channel FET conducts negative charges6. Negative charge is associated with presence of electrons (flow with electrons)7. Positive charge is associated with an absence of electrons (flows against electrons)So for an N channel FET, the powered circuit acts as a negative charge pool for the FET and is thus is connected to the Source. Source of FET connects to +voltage rail of the powered circuit. And FET Drain is connected to the +voltage supply rail. The powered circuit -voltage rail connects to -voltage supply rail.For N channel FETs, the +voltage supply rail is the consumer of the charge. The powered circuit is the producer of the charge.For N channel FET, appropriate Gate charge is greater than Source. Gate should be pulled low (towards/to Source) for normal open/off.While to a P channel FET, the +voltage supply rail acts as a positive charge pool for the FET and is thus connected to the Source. Drain of FET is connected to the +voltage rail of the powered circuit, and Source of the FET is connected to the +voltage supply rail. The powered circuit -voltage rail connects to the -voltage supply rail.For P channel FETs the powered circuit is the consumer of the charge and the +voltage supply rail is the producer of the charge.For P channel FET an appropriate Gate charge is less than Source. Gate should be pulled high (towards/to source) for normally open/off

author: Peabody
author_rank: Super Contributor
timestamp: October 24, 2020, 04:27:10 am
content: I don't really understand electrons and holes.An N-channel mosfet is usually placed between the load and ground, with the drain connected to the load and the source connected to ground. So it's a "low-side switch", and is turned on by bringing the gate voltage above the source voltage. Conventional current flows from drain to source.A P-channel mosfet is placed between the high-side power rail and the load, with the source connected to the rail and the drain connected to the load. It's a high-side switch, and is turned on by bringing the gate voltage below the source voltage. Conventional current flows from source to drain.You might search on Youtube for mosfet tutorials.

author: tifkat
author_rank: Regular Contributor
timestamp: October 24, 2020, 04:40:01 am
content: Thanks Peabody. I was still editing my post to make sure I communicated it correctly, and I think your reply pretty much backs up my understanding.I think both of my hookup ideas are High-side though.If anyone knows, are there any caveats with using N channel FETs as High-side switch, or P channel FET as low-side switch?

author: tifkat
author_rank: Regular Contributor
timestamp: October 24, 2020, 04:53:38 am
content: My guess (for circuits using +voltage and 0v ground)Low side switch has 'direct' connection to 0v or ground, so it will be the lowest it can be.High side switch has 'direct' connection to supply voltage (+voltage) and thus it will be the highest it can beSo for N channel, it's easier to create Vgs on low side, and for P channel it's easier to create Vgs on high side.

author: tifkat
author_rank: Regular Contributor
timestamp: October 24, 2020, 05:45:29 am
content: Looks like I'm S.o.L. On finding a hand solderable P channel MOSFET array with common source. Looks like I'll have to go with discrete MOSFETs.Seems everyone makes their common source arrays to be used in phones and they're microscopic ball grid array chips.

author: Peabody
author_rank: Super Contributor
timestamp: October 24, 2020, 04:34:36 pm
content: You don't have to have a common source. They could be independent, and you would just connect both to the upper rail.