Back to back MOSFET Load Switch- Common Source or Common Drain? - Page 1
Forum section: electronics / projects

author: petersanch
author_rank: Regular Contributor
timestamp: June 22, 2021, 07:27:33 am
content: There are conflicting explanations for bidirectional load switches that have the P channel MOSFETs in common source setup and others in common drain. Goal is to control low voltage AC from either end. Whats the "right" way?Cheers!

author: Zero999
author_rank: Super Contributor
timestamp: June 22, 2021, 09:33:23 am
content: It's not clear what you're asking.Only two MOSFETs, connected back-to-back are requred for AC power control. They work in common source mode.Here are a couple examples of a simple solid state relays. The left showing a transformer and right a couple of Y-rated safety capacitors to couple the signal, to the MOSFETs. The gate capacitance keeps the transistors on, through the short length of time, when the signal voltage is zero. When the current through the MOSFET is reversed, the drain and source positions exchange. MOSFETs are bidirectional devices, but the body diode bypasses the device in one direction.

author: Wolfram
author_rank: Frequent Contributor
timestamp: June 22, 2021, 10:17:13 am
content: This looks more like a CD4066-style bilateral signal switch than an AC load switch. In this case, the second pair of MOSFETs (of opposite polarity) allow the control voltage to be referenced to the supply rail rather than to the source. Or rather, it allows full signal swing within the supply rails while minimizing the effective on-resistance, provided that Vcc is significantly above 2* Vgsth. Note that the PMOS and NMOS gates are driven from complementary signals. When the switch is turned on, the PMOS pair gets biased on when the input signal drops towards ground, and the NMOS pair gets biased on when the input signal is close to VCC. Series resistance is at a maximum when the input signal sits just between the NMOS and PMOS threshold voltages, around VCC/2 depending on the relative scaling of PMOS and NMOS transistors.If you need to switch real power, say over a few tens of milliamps, the schemes posted by Zero999 are better. They only require half the number of power devices, Vgs (and therefore Rdson) doesn't change with the signal level, and the input signal (and VCC) can be higher than Vgsmax (20 - 30 V typically). Another way to do the gate drive is to use a photovoltaic optocoupler like the VOM1271.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 22, 2021, 01:31:44 pm
content: Have a look at the first datasheet of a MOSFET you can get your hands on.Voltage between gate and source is a very important design parameter.When it's below around 4V, most MOSfets are usually closed, and when it gets above 20V they usally let the magic smoke escape.So draw the schematic in whatever way you want to connect your Fet's, then start entering some numbers for voltages in the schematic in different situations.If you have a basic understanding of how MOSFET's work, this will be pretty easy. If you have trouble with this, you have to think harder and can learn a bit from it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 22, 2021, 02:22:54 pm
content: Gate voltage is referenced to source. Drain-to-drain makes little sense as you need two isolated gate drivers. The top-left diagram is erroneous.If by "low voltage" you mean just a few volts, so that Vgs(on) dominates and Vgs(max) is in no danger of being violated, it doesn't make much difference.Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: June 22, 2021, 05:06:48 pm
content: The case of very low voltage may happen when switching current shunts. In this case one may even get away with only MOSFET if the voltage never gets higher than some 100 mV.The CMOS switch chips (e.g. 4066, DG201 and similar) use fets with a separate substrate contact and no reverse diode. So no need for 2 fets in series in this case.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 22, 2021, 05:09:54 pm
content: Quote from: Kleinstein on June 22, 2021, 05:06:48 pmHeh, in fact they use two inparallelinstead (complementary P and N), to extend the voltage range and get a flatter R(on) vs. Vcm.Only works within the supply rails, so it's good for signals; or, done with power transistors, yup, can do that directly on shunts.Power transistors are very rare with substrate connections (there are a few!), but if your signal isthatlow (i.e. low 100s mV), you might not even care about the body diode -- only needing the one transistor, no series or parallel connections.Tim

author: AndersJ
author_rank: Frequent Contributor
timestamp: June 22, 2021, 07:09:25 pm
content: What is the purpose of the discussed circuitry?One poster mentioned ”AC power control”.That suggest applying varible power to a load, like a triac can do.Or is the purpose to make a on/off switch for a AC load?What functionality is wanted?

author: petersanch
author_rank: Regular Contributor
timestamp: June 23, 2021, 10:57:52 am
content: Quote from: Zero999 on June 22, 2021, 09:33:23 amIs that a charge pump at the gate for isolation? Can it be driven by microcontroller PWM on one pin and ground on the other?Quote from: T3sl4co1l on June 22, 2021, 02:22:54 pmIf I understand this right, if Vgs is being driven as described in my previous post (between -10 V and +10V) then connecting common source or common drain will not make a difference?Quote from: Doctorandus_P on June 22, 2021, 01:31:44 pmSI1539CDL is a MOSFET pair I had in mind. It has Vgs +/-20 V, Vds +/- 30 V. Dont see it to be a problem for this circuit.Cheers

author: petersanch
author_rank: Regular Contributor
timestamp: June 23, 2021, 11:05:36 am
content: Thanks for all the feed back!I meant in the range of +/- 10 Volts for low voltage AC. I wanted to distinguish from AC mains voltage or similar when i wrote low voltage.Quote from: AndersJ on June 22, 2021, 07:09:25 pmIn a previous thread I asked how to switch from 2 audio amplifier outputs to digitally controll which amplifier is driving the speaker. Kleinstein and David Hess suggested using a photo voltaic circuit like a solid state relay but also gave the idea of using 4 MOSFETs.https://www.eevblog.com/forum/projects/analog-switch-to-control-low-voltage-ac/To learn more about MOSFETs I will still like to do this experiment with discrete MOSFETs instead of photo voltaics if I can.I want to make certain the switch can both source and sink current to drive a small speaker and be able to pass frequencies in the KHz.Will 2 series N channel MOSFETs be enough to for this job? I plan to drive the gates to -10 V to turn them off and +10 V to turn them fully on but I don't know if N channels can sinkand sourcecurrent while connected like this.Thanks

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 23, 2021, 02:28:58 pm
content: Quote from: petersanch on June 23, 2021, 10:57:52 amIt is; note that common mode between driver and power path will turn it on just the same. And that intermediate drive levels will partially bias the transistors, leading to high dissipation. It's reasonable for limited applications, but most of those will have the driver isolated, in which case, err, it doesn't need to be cap-coupled at all and can be tied common source with the switches...Or the caps could be dimensioned small enough that AC mains for example doesn't trip it, in the mean case -- but can't be made small enough that the exceptional case (EMI or transients) won't. So it's not very safe for that reason.Also, speaking of transients, there's nothing limiting gate voltage, which a transient could easily overdrive, and then your switch is always-on which is rather inconvenient... (Semiconductors tend to fail shorted.) Easy solution there, use 12V zener diodes for the charge pump.A more refined version could use an LC resonant tank, to filter out most of that junk (mind, it's still susceptible at the resonant frequency); in which case the coupling capacitors can be made quite small thanks to the enhanced power coupling of a double-tuned resonant filter. Drive of course also needs to be fixed frequency, but that's rather easy to obtain from an MCU with crystal (or even the internal RC oscillators are usually precise enough for this) and a timer-counter.Still better would be a balanced LC resonant tank, using a centertapped choke to couple +/- balanced into the network, so that common mode interference is cancelled out. Now you're using a transformer anyway, so I don't know that it matters much at this point... pulse transformers aren't terrifically expensive and the left circuit is simple and easy.The one advantage to the resonant method is, if you don't mind the frequency being a LOT higher (alas, out of the range of an MCU + timer), you could use coreless planar PCB structures at zero parts cost (just design time and board area). Run it at say 433MHz from a gated crystal oscillator (use small schottky diodes for the low capacitance and no recovery loss). The isolation can also be extremely good (thickness of FR-4 is good for many thousands of volts). Another advantage is, the keying can be quite fast now (a modest fraction of Fo), better than most optoisolators and maybe even competitive with digital isolators (some of which use the same construction internally!).But I digress; there's a fair amount of network theory and poking around to prove out something like that, and even in an ISM band, it may not be easy to meet emissions limits. More just as a flavor of what's possible, not necessarily feasible.QuoteYes, for Vg(on) - Vgs(max) < Vcm < Vg(on) - Vgs(on)(min).For Vgs(max) = 20V and Vg(on) = 20V, the CM range is only 0...15V say (for a minimum Vgs(on) of 5V).Like I said, it's practical for small voltages, a few volts.QuoteQuote from: petersanch on June 23, 2021, 11:05:36 amYeah that's not gonna happen, the span is already as much as Vgs(max) and that's already distorting as it runs out of Vgs(on) at the peaks.BTW, another way to use the common-source switch, is to bias the gates from an external supply. Say the source and drive are common ground, and drive is +/-30V current limited to say 0.1mA, something small like that. (Easily enough done with a handful of transistors and resistors, as level shifting and CCS.) Tie this to the gatesand add a S-G zenerso that Vgs(on) and (off) are limited, regardless of Vcm. Now you can run Vcm up to +25/-30V peak and have reliable operation. Note that the drive injects some current into the source or load, so this won't be an acceptable solution for every situation; for something like an audio amp, it's fine.BTW, AC/DC (MOS type) SSRs are just the same thing over again, using a small photovoltaic (solar panel) stack powered by the LED itself. This delivers minuscule current (some uA), hence they switch quite slowly (some ms). They're quite handy -- just look at how much circuitry they save!Tim

author: Zero999
author_rank: Super Contributor
timestamp: June 23, 2021, 04:18:32 pm
content: Quote from: petersanch on June 23, 2021, 10:57:52 amThe capacitors are for isolation. They pass high frequency signals, but block DC, or mains frequencies. They need to be suitably safety rated and the PCB needs too have sufficient clearance for double insulation. Yes, a microcontroller can be used to drive it. If one pin is 0V and the other 5V, the gate voltage will be a couple of diode drops lower. Schottky diodes can be used to reduce the voltage drop. Note that most MOSFETs are specified with a gate-source voltage of 10V, so it's better to drive the capacitors with opposite phases.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 23, 2021, 05:49:17 pm
content: https://www.wikiwand.com/en/Transmission_gate

author: petersanch
author_rank: Regular Contributor
timestamp: June 24, 2021, 10:32:58 am
content: Very helpful @T3sl4co1l and @Zero999 thanks for the explanations!Quote from: T3sl4co1l on June 23, 2021, 02:28:58 pmParticularly interestingQuote from: Zero999 on June 23, 2021, 04:18:32 pmThanks! Sorry for so many questions as I understand your circuit.Is R1 there to discharge the gate?Will the control signal for the charge pump couple through R1 or C2 to the common sources and into "Phase" or "Neutral"? Is there a way to avoid coupling?Can the extra capacitance from C2 make this circuit unsuitable as a switch for high frequency signals?A 200 KHz square wave turns on the switch but how do you fully turn off the switch? By stopping the square wave and letting R1 discharg the gate?Quote from: Terry Bites on June 23, 2021, 05:49:17 pmThanks. This is the concept I was thinking but the examples for transmission gates show the body of the MOSFETs accessible when commonly found components dont have pins to the body.Cheers!

author: Zero999
author_rank: Super Contributor
timestamp: June 24, 2021, 04:33:11 pm
content: Quote from: petersanch on June 24, 2021, 10:32:58 amR1 discharges the MOSFET gates, allowing them to turn off.Good point about the signal leaking through. The amount of leakage will depend on how well isolated the high frequency AC source is from the mains. I would ditch the idea of driving C1 & C2 antiphase from a 5V MCU output and MOSFET driver, or some transistors to drive it with a >10V square wave.This circuit is only designed for switching mains frequencies.The MOSFETs will turn off, when the 200kHz squarewave is turned off.