Floating MOSFET Gate Drive- how does it work?

Thread Starter

shortbus

Joined Sep 30, 2009
8,581
Modernator's note: This thread was created from a discussion in https://forum.allaboutcircuits.com/threads/choosing-full-bridge-driver.176093/page-3#post-1598192 Where the gate drive circuit that is the basis of this thread was discusssed, but the discussion has been moved to here so the thread starter can get his problem solved.

Oh yeah,
the price of the FETs is around ~$5.oo each
How exactly are the mosfet gates being turned on? How are the highsides staying on?
 
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LowQCab

Joined Nov 6, 2012
538
If you will look at the Spec Sheet for the FDA215's you will see that they are actually
"Photo-Voltaic-Cells" with a 5.5 Volt Output each, specifically designed to drive a MOSFET Gate .
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Thread Starter

shortbus

Joined Sep 30, 2009
8,581
If you will look at the Spec Sheet for the FDA215's you will see that they are actually
"Photo-Voltaic-Cells" with a 5.5 Volt Output each, specifically designed to drive a MOSFET Gate .
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You have a very different way of making schematics. That said, there is no information showing that the FDA215 is capable of making a highside mosfet switch.
 

LowQCab

Joined Nov 6, 2012
538
When using an Isolated Voltage/Current Source, "High-Side" or "Low-Side" is completely irrelevant .

These chips act as a completely "floating" Voltage Source,
just like a single battery cell which can be turned on or off by shining an LED on it .
They are NOT referenced to Ground,
they are referenced to the Source Pin of the FET that they are controlling .

The only thing that the FET cares about, is the Voltage difference between the Source and Gate Pins .

Like I said, read the Spec Sheet ........

"""The FDA215 is a dual photovoltaic MOSFET driver that uses a pair of
optically coupled LEDs to
drive two photodiode arrays.
When the input current is applied to the LED,
the light emitted activates the photodiode array, and generates a voltage at the output.
The photodiode arrays are capable of generating a

floating source voltage and current sufficient to drive high power MOSFET transistors."""

https://www.ixysic.com/home/pdfs.nsf/0/3BECA6A45A150E6985256BC3004F5DD7/$file/FDA215.pdf

Turn-On-Rates vs Gate Capacitance Charts are provided .

Always read the instructions, 95% of the time it's useless blather,
but 5% of the time you might just learn something !!!

As I noted before, these units are NOT FAST, because they have very little Current generating capability.
Although, they can be paralleled to increase their Current/Switching-Speed,
or wired in series for increased Voltage, or both .
Their turn-off speed is roughly ~100 times faster than their turn-on speed .
But in this low frequency application, high switching speeds are not necessary, or even desirable .
High Switching Speeds, in this application, would only be likely to cause serious Flyback/Ringing issues .
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Thread Starter

shortbus

Joined Sep 30, 2009
8,581
The only thing that the FET cares about, is the Voltage difference between the Source and Gate Pins
Exactly. This is why the bootstrap is needed. As that source pin voltage rises the gate voltage also needs to rise to keep the Gs voltage correct. If this voltage is 12V to turn it on there needs to be another 12V (24Vtotal) added to the gate. Since this is a highside switch. The component would work great for a lowside. Another thing form the data sheet is it is only a 5V device.

Then there is the problem of using the opt-interrupter for commutation. The design of the "project' only show it to have 2 poles on the rotor. So the interrupter only has a 50-50 chance of starting it to rotate. If the solenoids are not in the correct polarity. It would then need a push to get it to rotate.

A better way would be a ring counter to make a rotating field electronically instead of mechanically. Then even if the poles weren't correct they would be on the next step of the counter.
 

LowQCab

Joined Nov 6, 2012
538
Sorry man, you are mistaken on "Bootstrapping",
this arrangement bears no relationship to the usual Bootstrapping principles and rules,
this is a "FLOATING VOLTAGE SOURCE" not referenced to ground, or any other Voltage Source .

The Floating Voltage Source used here is "referenced" to whatever Voltage is present at
the "Source Terminal" of the particular MOSFET that it is attached to.

The Device has a 5.5V maximum output Voltage,
which can be wired in series with any number of additional, similar devices,
to create output voltages of 5.5V, 11V, 16.5V, 22V, etc,
They may also be wired in parallel for additional Current capability.
They can be thought of as 5.5V "Batteries", which can be turned on and off with an LED.

I already addressed the question of Commutation, and advised using a 3-Phase Rotating Field scheme,
but James has other, rather strange, ideas ........
He will find out that,
the Motor will NOT start by its self .
But, after the Commutator is "clocked" correctly in relation to the Windings and Magnets,
the position of which must be found by a "Trial-and-Error" method,
because of the bizarre Motor configuration,
the Motor should continue to run after manually spinning the Shaft by hand.
The "clocking" of the Commutator will also determine the RPM and Efficiency of the Motors operation.

Hey James .........
Nikola Tesla invented 3-Phase Power, and 3-Phase Motors and Generators, for a lot of very good reasons .......
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DickCappels

Joined Aug 21, 2008
7,432
@shortbus , you probably missed the fact that each channel is made of an LED and the photodiode array that the LED shines upon. When illuminated the photodiodes develope enough voltage to turn on a MOSFET.
 

Thread Starter

shortbus

Joined Sep 30, 2009
8,581
@shortbus , you probably missed the fact that each channel is made of an LED and the photodiode array that the LED shines upon. When illuminated the photodiodes develope enough voltage to turn on a MOSFET.
So all of that means that a high side mosfet switch works differently when using one of these? That the high side no longer needs to have a gate voltage higher than the source out put voltage?
 

Thread Starter

shortbus

Joined Sep 30, 2009
8,581
The Floating Voltage Source used here is "referenced" to whatever Voltage is present at
the "Source Terminal" of the particular MOSFET that it is attached to.
So then where does the return to the Floating Voltage Source go? I'm not trying to troll but to understand a very new concept, at least a new concept to me. I can see it working when it is a low side, since the low side fet is referenced to ground, but high side? And the data sheet is not a very good one shows no "common use circuit". And even though it says for solid state relay use, many of them are ground/common referenced or in other words a low side switch.

Quote LowQCab, " but James has other, rather strange, ideas". That is one thing we can both agree on.
 
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wayneh

Joined Sep 9, 2010
17,152
So then where does the return to the Floating Voltage Source go?
This is new to me too, but I believe you tie the photovoltaic to the MOSFET source pin. Shine a light on it and the PV generates the gate voltage. Clever. $3.12 at Mouser though. I guess that's a bargain in applications where getting the high-side voltage is challenging otherwise.
 

Thread Starter

shortbus

Joined Sep 30, 2009
8,581
This is new to me too, but I believe you tie the photovoltaic to the MOSFET source pin. Shine a light on it and the PV generates the gate voltage. Clever.
Hard to get my head around. It flies in the face of everything I've ever read or been told about high side mosfets. Have always been told the gate needs to be the Gs voltage referenced TO the source voltage when it is turned on. And this doesn't, as far as I can see, do it.

I understand the photo part of the new driver, how that makes a voltage, But how that gets around the need for gate to be higher than source voltage?????

If this device can be used to drive a high side mosfet, it would make life much easier and more than one company would be doing it, but no one else is.
 

wayneh

Joined Sep 9, 2010
17,152
I understand the photo part of the new driver, how that makes a voltage, But how that gets around the need for gate to be higher than source voltage????
Picture placing a battery between the source pin and gate pin. Boom, the MOSFET conducts. No connection whatsoever to the rest of the circuit. That's all this thing does. It's slow as noted, so only for certain applications.
 

DickCappels

Joined Aug 21, 2008
7,432
The photodiode array has one end connected to the source and the other to the gate. It makes volts and it is, as you indicated, volts between the source and the gate that make MOSFETs come on. The circuit does not care where ground is as long as it connects to the source and gate.

Think of it like a relay - the relay coil usually does not care what its voltage to ground is but it still manages to close the relay contacts as long as there is enough voltage across the coil.
 

james31207

Joined Sep 19, 2018
72
Sorry man, you are mistaken on "Bootstrapping",
this arrangement bears no relationship to the usual Bootstrapping principles and rules,
this is a "FLOATING VOLTAGE SOURCE" not referenced to ground, or any other Voltage Source .

The Floating Voltage Source used here is "referenced" to whatever Voltage is present at
the "Source Terminal" of the particular MOSFET that it is attached to.

The Device has a 5.5V maximum output Voltage,
which can be wired in series with any number of additional, similar devices,
to create output voltages of 5.5V, 11V, 16.5V, 22V, etc,
They may also be wired in parallel for additional Current capability.
They can be thought of as 5.5V "Batteries", which can be turned on and off with an LED.

I already addressed the question of Commutation, and advised using a 3-Phase Rotating Field scheme,
but James has other, rather strange, ideas ........
He will find out that,
the Motor will NOT start by its self .
But, after the Commutator is "clocked" correctly in relation to the Windings and Magnets,
the position of which must be found by a "Trial-and-Error" method,
because of the bizarre Motor configuration,
the Motor should continue to run after manually spinning the Shaft by hand.
The "clocking" of the Commutator will also determine the RPM and Efficiency of the Motors operation.

Hey James .........
Nikola Tesla invented 3-Phase Power, and 3-Phase Motors and Generators, for a lot of very good reasons .......
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This all makes quite good sense, even to me; particularly if starting from scratch with MOSFETs and your FDA215, as I may end up doing. To reiterate however, this MCPG apparatus though is NOT a motor of any sort, nor, as a generator, is it intended to drive a motor, or anything else. The rotor already spins easily by itself over 135 degrees, and once the H-Bridge switch 'tells' the electromagnets when to reverse polarities, thus to begin to repel the core which has just been attracted to it, that core/rotor will certainly spin continuously; and we can start investigating what's happening in the winding.

Also, although there are 12 field coils (geometrically at the midpoints of the 12 edges of a cube), these are are just two poles divided 3 ways, twice: one pole component in the horizontal or equatorial plane, and the other two at 45 degrees elevated and declined, and also rotated; repeated at 90 degrees around the equatorial plane. So, with respect to the drive aspect, there are 4 'phases', but these will arise naturally through the switching mechanism.

As you point out, the old Serbian took care of power generators and their motor destinations through the convenience of polyphase windings (so introducing the whole planetary system of mass-produced centralised electricity required in the 20th century). Nonetheless, having visited his museum in Belgrade and seen first hand his earliest experiments on rotating magnetic field -- there's one where a brass egg is made to spin continuously under the influence of 4 end-to-end electromagnets connected to a battery --, I'm pretty sure he'd have gotten round to this sort of 'integrated single phase AC output (or infinite phase if you like)' sooner or later. Of course, by 1943 he was becoming infirm, mentally and physically, and may not have recognised such an idea even if it had come knocking on his incomparable door.
 

LowQCab

Joined Nov 6, 2012
538
So all of that means that a high side mosfet switch works differently when using one of these? That the high side no longer needs to have a gate voltage higher than the source out put voltage?
Evidently you have a fundamental misunderstanding about how a FET works ........
We have to clear-up some terminology first,

1) The Source doesn't "have" an "Output Voltage".
When the FET is "On", there is low Resistance between the Drain and Source .

2) The FET can be thought of as a variable resistance between the Drain and Source Pins.

3) That Variable Resistance changes based on the Voltage between the Source Pin and the Gate Pin, ONLY.

4) When the voltage between the Source Pin, and the Gate Pin, is less than the "Gate Threshold Voltage",
then the RESISTANCE between the Drain Pin and the Source Pin is considered to be infinite.

5) When the voltage between the Source Pin, and the Gate Pin, is more than ~15V,
then the RESISTANCE between the Drain Pin and the Source Pin will be the value of "RDS-ON",
( Resistance, Drain to Source ) .

6) When the voltage between the Source Pin, and the Gate Pin, is more than the Gate Threshold Voltage,
then current can flow equally well from the Drain to Source, or, from the Source to Drain .

7) The FET does not care about any other outside factor,
except for the Voltage-Level between the Source and Gate Pins .

( Technically these statements are not absolutely true, and these are not the "only" factors,
but these are the basic rules that always apply,
and they are all you need to know for a basic understanding of how FETs work,
and how to build a useful circuit using them )
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.Motor Control Flat 2 .pngFET Graph .PNGFET Graph 2 .PNG
 

Thread Starter

shortbus

Joined Sep 30, 2009
8,581
Evidently you have a fundamental misunderstanding about how a FET works ........
We have to clear-up some terminology first,
I understand all of that, and agree when it is used as a low side it is true, because the source is a ground/common level. But your number 1 is where I've been hung up. And over my few years of learning and doing it has been beaten into me on the forums that when you say,
The Source doesn't "have" an "Output Voltage".
When the FET is "On", there is low Resistance between the Drain and Source
That statement, I've been told so many times is wrong when it comes to a high side switch. Since in that high side situation there is no longer a direct ground/common connection. The voltage needs to go through the motor , resistor, etc. So now I'm being told that isn't true? Blowing my mind here.
 

LowQCab

Joined Nov 6, 2012
538
The first thing to do is to forget about the concept of "GROUND".
You "might" think of it as a "Common" Voltage Rail, but it is usually not connected to "Ground", (Earth).
And it could be a voltage potential that is Positive or Negative in relationship to Earth-Ground.

Think about the FET as being an isolated, floating, switch,
or a Relay with one end of its Coil attached to one side of the Switch, ( which is labeled "Source" ),
and the other end of the Coil attached to a terminal called "Gate",
the end of the Switch that is not connected to the Coil is labeled "Drain",
now attach a Battery between the Source and the Gate, and the Switch will close.
Now, you can put any crazy kind of Voltage/Current though the Switch Contacts, ( Drain & Source ),
and the Battery will not care at all, the Battery will be completely unaffected .

You can have 2000 Volts to Ground attached to the Drain Pin,
and 1950 Volts to Ground, attached through a Resistor, to the Source Pin,
and the FET will happily function just fine,
as long as THE DIFFERENCE between the Drain and Source Pins
does not exceed the Drain to Source Voltage Rating .

The Photovoltaic Cell used to turn on the FET Gate,
will produce 1960 Volts ( measured to Ground ), at the Gate Pin,
if the Source Voltage goes to 1950V ( measured to Ground ),
because its Negative connection is ONLY tied directly to the Source Pin, NOT TO GROUND .
So, when turned "on", the Photovoltaic Cell's ~10v Positive Output,
WILL ALWAYS BE ~10V higher than whatever the Voltage at the Source Pin might be, at any particular moment .
Whether this is measured relative to Ground,
or relative to some other point in the circuit,
IS COMPLETELY IRRELEVANT .

The Voltage DIFFERENCE between the SOURCE AND THE GATE, IS ALL THAT MATTERS, ever, end of story.

This G-to-S Voltage Difference IS NEVER MEASURED FROM GROUND,
it is measured ONLY from the Source Pin to the Gate Pin .

If the Voltage on the Source Pin goes up by ~5V, ( when measured from Ground ),
then the Voltage on the Gate Pin will also go up by an additional ~5V, automatically, you can't stop it .

If the Voltage on the Source Pin goes ~100V NEGATIVE, ( with respect to Ground ),
the Gate Voltage WILL REMAIN AT ~10V Positive, WITH RESPECT TO THE SOURCE PIN,
so the FET will remain turned "ON" .

When something is labeled as a "Floating" Voltage Source,
that means that it is NOT tied to ANY Voltage-Rail, including "Ground",
and therefore, its Voltage Potential must be measured across the device output pins, NOT TO GROUND .
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LowQCab

Joined Nov 6, 2012
538
Fascinating. A solid-state relay without a triac. Are there applications where you would choose this approach over a conventional SSR?
Yes,
A Triac is just like a Diode or a Bi-Polar Transistor, in that, it will always have a "Diode-Voltage-Drop".

So, let's say that you have a 10V circuit which draws 50 Amps .........
With a Triac in the circuit, there will always be at least a ~0.7 Volt Voltage-Drop,
so you only get a maximum of ~9.3 Volts at the Output of the Circuit.
Where did that Voltage go ????
It was wasted as HEAT.
Using Ohms Law, 0.7V X 50A = 35 Watts of Heat that must be dissipated by the Triac .

Now, take an example of a very stout N-Channel MOSFET .......
Lets say that the FET has a "Rds-(on)" rating of only 0.001 Ohms, or, 1-mOhm .
Now do the Voltage-Drop and Wasted-Heat calculations again ......
50A X .001 Ohms = 0.05V of Voltage-Drop through the FET,
now multiply Volts X Amps to get Watts,
0.05V X 50A = 2.5 Watts of Heat that must be dissipated by the FET, that's a massive difference .
AND, you only lose 0.05V, instead of 0.7V from your 10V circuit, (~9.95V vs ~9.3V ).

Large Heat-Sinks can be expensive .

At higher Voltages, and lower Currents, the differences become somewhat less extreme,
but are still quite significant,
and must be taken into consideration in the selection of components for your circuit .

And BTW,
FET-Based SSS (Solid-State-Relays) are readily available in a huge variety of sizes/ratings,
but they are quite often VERY expensive,
but the convenience and packaging may make the cost worthwhile in some applications .
Just search for "SSR".......

https://www.digikey.com/en/products/detail/sensata-crydom/84134020/1279975
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