Multiple LL N-Channel MOSFET's in Series

[TheDood]

Just that the op-amp powered from a "sky hook" is the electronic version of the mechanical beginner's shaft passer.

The crux of it is this: it's easy to forget that the op-amp is itself made of transistors.  It cannot generate more voltage than supplied with; the full load current (plus a little extra!) is drawn from those supplies; and the full voltage difference between supply and output is burned as power dissipation.

Op-amp input pin current is pretty low, right, but the output current has to come from somewhere.

So, making a synchronous or precision rectifier, with op-amps, and expecting any efficiency out of it, is a fairly early and simple mistake to make.

Once this is realized, it's easy to see why this can never work, but also that one must explore circuits a bit more clever to actually implement it.


So... not to beat you to death about it, just to be perfectly clear what it is, and that we've all been there at some point.

Using transistors is indeed one such case; they must be controlled to turn on and off at the appropriate voltages/currents, but one should also be careful not to slam them on and off (e.g. using a comparator and gate driver), because that will lead to oscillation and ugly recovery* effects!  A modest gain factor, between Vds and Vgs, is probably best.

*Recovery is effectively the time a diode needs to "decide" when being turned on or off.  There is forward and reverse recovery.  Forward recovery is a relatively high forward voltage drop, for the first however many, usually nanoseconds, as the forward current rises to reach its nominal value; this manifests effectively as a series inductance (higher dI/dt --> shorter t_fr, higher peak V_fr).  Forward recovery usually isn't very substantial.

Reverse recovery is the time taken for the current to go through zero, to some negative peak I_rr, during which time the voltage drop remains near Vf; this manifests effectively as a parallel capacitance (higher dV/dt --> shorter t_rr, higher peak I_rr).

Diagrams of reverse recovery and more discussion can be found here: [url]https://www.allaboutcircuits.com/technical-articles/switching-losses-effects-on-semiconductors/[/url]

In an emulated or active diode like we're discussing here, the recovery is determined by the speed of the op-amp and transistors.

Here's a fairly simple example as a single diode, although, I guess I don't know how explanatory it will be since it uses a handful of BJTs -- I may be inviting something more complicated than you're prepared for..?  Explanation below:



The leftmost transistor is used as an inverted* diode (base strapped to emitter).  This is done for two reasons: one, the forward voltage drop matches that of the next transistor; two, the breakdown voltage is higher (about Vceo, say 60V for 2N3904).  The second transistor is also inverted, in a common-emitter configuration (except the collector is common, because inverted; but again, it works the same way!), so if the "Cathode" voltage is high, the first diode is reverse biased and the 470 ohm resistor turns on this transistor, pulling its output (the schematic emitter) down towards ground.

*BJTs are always introduced like a Tetris piece, a stack of N-P-N or P-N-P blocks.  But this is really pretty disingenuous as to how they work, or are made.  Well, most BJTs aren't symmetrical, one end is made stronger than the other, giving that side has a lower voltage rating (typically 2-10V) than the other (typically 20-100V+), but also better "emission" of electrons/holes into the transistor, hence it's called the "emitter" and is typically used as the common (grounded / reference) terminal.  It still works just transistorly if we swap E and C -- but, the performance is very different, hFE(inv) typically being quite low, under 5 say.  So that will be one drawback that applies here.

The PNP and three other resistors is a simple current source, about 0.6mA.  This biases the second transistor, giving it voltage gain and a working range of about 0-5V.

Finally the two rightmost transistors are a complementary emitter follower; they boost the gain node's output current from ~0.6mA to >60mA.  This is able to drive the MOSFET gate reasonably quickly.  The follower is not biased, which is kind of unfortunate (this introduces another nonlinearity, probably making this circuit awkward to use on arbitrary AC waveforms, say), but probably not worth addressing when used with most power circuits (like mains or switching supply rectification).

So, conversely, when the "Cathode" voltage dips negative, the first transistor turns on, shunting current away from the second one, turning it off, and allowing the gate voltage to rise.  This limits the FET voltage drop (Vds) to perhaps -60mV at low currents (the "decision range" is driven by the BJT's exponential gain, so a few multiples of 26mV goes from reasonably-fully-on to reasonably-fully-off), and I_load * Rds(on) at higher currents.

Two or three major downsides to this circuit are: the supply requirement (5V referenced to "Anode"); the relatively high bias requirement (mainly the 470 ohm resistor, or ~11mA; this is due to the low hFE(inv)); and probably the switching speed, say for switching supply application (this will take some 100s of ns, which is worse than a high speed PN diode, and despite the lower voltage drop, it may not turn out to perform better than a schottky diode).  For mains rectification it would be okay, but the first two issues are still unfortunate.

Note another characteristic of this: the transistors act as an amplifier, with a limited output voltage range -- 0-5V.  The input might be 40V.  It's not enough to have a gain function here: there also must be a limiting function.  If we used an ideal op-amp (one with a true skyhook for supplies!) to set Vgs = Vds * -100 say, we'd end up applying a tremendous voltage to the gate, blowing it in no time.  We need a limited output, so Vgs remains in a reasonable range.

Also, since it's a 5V supply, we would want to use a logic-level FET, which is fine, they're plentiful.   The supply can of course be increased to use others, but the bias current will be even more annoying then, and a better solution would be attractive.


There are of course ICs for this; LT for instance makes one, if you don't mind paying for it of course: [url]https://www.analog.com/media/en/technical-documentation/data-sheets/4320fb.pdf[/url]
Which, despite the multi-dollar price tag, is really a pretty good deal when you're tight on space and can't afford the heatsinks or the temp rise (let alone the raw efficiency points) of a passive (diode) solution.  The knock-on savings are heatsinks, mounting hardware, fans, all those sorts of things.

Tim

 
That was phenomenal! Its posts like this that allow people to really begin to understand and comprehend. Thanks. A. Ton! I've given it a once over but it will honestly take me a min to fully digest & comprehend. Appreciate the insight and methodical explanation. No offense taken. (I'm only trying to acquire knowledge so to me it would be well worth it regardless, but also, thanks for going easy lol)

[T3sl4co1l]

Cheers!

Tim

[mzzj]

Can eBay be trusted on MOSFETS? Any components?

No.

Ebay is riddled with fake mosfets.
Some other components are actually worth the gambling but high performance mosfets from china/ebay are not.

[Ice-Tea]

Can eBay be trusted on MOSFETS? Any components?

No.

You could put the mosfets in parallel to reduce power in each  but together they would still generate same amount of heat.

In case someone stumbles on this thread later: that's not true. Provided current is perfectly balanced, power loss would be half.

[Circlotron]

Can eBay be trusted on MOSFETS? Any components?

No.

You could put the mosfets in parallel to reduce power in each  but together they would still generate same amount of heat.

In case someone stumbles on this thread later: that's not true. Provided current is perfectly balanced, power loss would be half.

I'll second that.
Each mosfet would have half the current and the voltage drop would be  halved, so each mosfet would have one quarter the dissipation. Times 2 = half.

[T3sl4co1l]

Total Rds(on) goes down by half, so for the same load current, voltage drop is half, and power is half.

Tim

[TheDood]

 @Tim (and others)

So I've tried to comprehend what you're telling me but still not quite there.

What I've got so far:

Op-Amp output is limited to OP-Amp PS.

OP-Amp dissipation = deltaV between input & output

Current flow through a diode when turning off is somewhat like an inelastic collision, its almost like the current bounces off the diode, these elastic type of reactions are described as forward and reverse recovery characteristics?

The duration of the recovery times is dependent on the amount of charge stored in the depletion zone and/or N-P boundary?

Some more questions:

If the op amp is in an essentially open loop gain configuration, the only power loss is the Vf of the diode between output pin and inverting input pin, right(?), but at what current? Is the current dependent on the diode or the op-amp? When you said the current has to come from somewhere were you saying that a decent portion of the load current flows through the OP-Amp output pin, or is it only the corresponding current of the diode at its Vf (if op-amp PS were limited to diode Vf)?

Trying to understand where the greatest power losses come from, attached are (2) MOSFET recovery charts, 1 on, 1 off. Are we mainly concerned with the time duration it takes to complete the switching action (refer to chart)?

Also, it looks like the efficiency plateau's around 92%? This is a generalized graph? The efficiency would be dependent on RDSon and the speed that Vds is changed? Not all FETs would plateau here? At 1kHz there'd be an even greater efficiency, albeit not much greater?

As far as the schematic, I get the inverted diode config, but that's about it. Trying to understand it more but having difficulty visualizing the current path when cathode is +V and when cathode is -V.
Also, why not just use 1 transistor with a 1kΩ on the base instead of the cascading triplet?
When cathode is +5V, the MOSFET is turned on with ~5V @ gate, but when cathode <5V, the inverted diodes begin to turn on and bypass of the MOSFET gate which begins to close MOSFET. It seems that this cct doesn't slam MOSFET on & off? It almost seems like the inverted diodes or load line feedback is buffering the speed of the switching? Gate V goes from 0V to 5V but dependent on how fast load V is transitioning between 0V and 5V?

[TheDood]

Actually after further review, I don't understand the inverted diode configuration.

When does T1 ever conduct?

Is T2 emitter current = to its base current?

Is there significance in T3 being PNP? Could you use an NPN there instead, and get same type of action?

Is T5 redundant? Ie, does it ever flow current?

[T3sl4co1l]

If you need help visualizing the operation you can put it in SPICE.  The inverted hFE of SPICE models is not usually very realistic, but it should still be adequate to show operation.

For a testing circuit I would recommend a SINE voltage source and series load resistor, with the "anode" staying common ground.  Try it without the active circuit (just the MOSFET, gate grounded), and with.

Tim

[TheDood]

Whats a good SPICE simulator to use? Tried EAGLE and partsim. EAGLE required more know-how than I was anticipating, are there better options?

[Ice-Tea]

LTSpice

[TheDood]

Thanks @Ice-Tea I'll check it out

Can you or someone help me find the ESR for the 2.2μF capacitor on this data sheet?

[mzzj]

Thanks @Ice-Tea I'll check it out

Can you or someone help me find the ESR for the 2.2μF capacitor on this data sheet?

15-20 milliohms guessed from impedance graphs.

[TheDood]

Thanks @Ice-Tea I'll check it out

Can you or someone help me find the ESR for the 2.2μF capacitor on this data sheet?

15-20 milliohms guessed from impedance graphs.

Thanks, could you help me understand how you are calculating/guesstimating?

I've been finding the Xc per Hz but when I plug into a 305V cct I get pretty much the exact figures that are plotted on the Irms charts.

I was trying to note the perceived minor differences in the calculated ideal current and the plotted current, then solving for R in each scenario, and chalking up any differences in R to ESR, but I'm nowhere near precise enough. It seems I'm getting whole R values not 10s of mΩ values and not really sure this is what you were meaning?

Example:
2.2μF, @ 800Hz

Xc = 1 / ( 2×3.14×800×0.0000022)
Xc = 90.42Ω

I = 305 / Xc
I = 3.37A

Chart looks maybe closer to only 3.3A???

(305V / 3.3A) - (305V / 3.37A)
=
2Ω

[mzzj]

Thanks @Ice-Tea I'll check it out

Can you or someone help me find the ESR for the 2.2μF capacitor on this data sheet?

15-20 milliohms guessed from impedance graphs.

Thanks, could you help me understand how you are calculating/guesstimating?

Lowest dip in the impedance graph is indication of the ESR, ignore frequency and ignore the Irms charts. (page 12 Impedance Z versus frequency f)
Not by any means 100% accurate method but better than nothing. 

[TheDood]

I've been working with LT Spice, but it takes really long to run only a 5s simulation (about an hr). Any ways of making it faster? The task manager says it's only under a ~25% CPU load and it's an old laptop. I've changed the tolerances to 0.01 hoping it would speed it up, but if it did I sure couldn't tell lol. I see a "Simulation Speed" at the bottom, is there a way to bump it up?

[T3sl4co1l]

Five seconds?  What are you trying to find that can't be learned in 1ms?

Tim

[Zero999]

To answer the original question: yes, MOSFETs can be connected in series, in an arrangement known as a cascode, but it's a bit more tricky, than simply connecting them in parallel.
https://en.wikipedia.org/wiki/Cascode

For help, regarding LTSpice: please post the .asc file.