Hello,In the diagram sown below I have a MOSFET called NDT3055L, the mosfet sends the current into the Qorvo TGA2590-CP drain.
When we test the circuit the source of the mosfet is basically floating although when I will connect the amplifier to source then the QORVO amplifier will introduce a load to the source of the mosfet.
Somehow the optocoupler helps with the transition of the state when we don't have QORVO amplifier load on the source of the MOSFET.
What is the role of the optocoupler in this situation?
I'll be glad to have some analog design logic behind putting this component in this position.
Thanks.

https://www.ti.com/lit/ds/symlink/ucc5304.pdf?ts=1713292114392

TGA2590-CP - Qorvo

 

The term 'biassing' when referring to a MOSFET switch is ambiguous/meaningless. Either the gate is +ve with respect to source and >>Vgs(th) so MOSFET is switched fully on, or the gate is <<Vgs(th) wrt source and the MOSFET is fully off. No other state should exist. The MOSFET is on or off purely based on the relationship of gate voltage wrt source. If Vgs is sufficient to turn the MOSFET fully on (typically >10v, or >3v for a logic-level MOSFET) the MOSFET behaves as a very low impedance resistor and the current flowing through it has little impact on its state.

In your circuit the gate driver (its not an opto-coupler), when sufficiently driven, drives OUT (MOSFET gate) close to VDD, otherwise it actively pulls OUT down to VSS. Since its output is isolated from the input it can float so when the MOSFET turns ON, so the gate driver's VSS line floats to the MOSFET source, which will now be very close to the 24v on its Drain. As long as the gate driver's supply input between VDD and VSS (note that VDD is not connected to a voltage source in your diagram) is isolated & floating, or if referenced to GND is suitably higher than the 24v rail the MOSFET stays ON. If not, as the source voltage rises the differential between gate and source reduces and the MOSFET begins to turn off. In the steady state the MOSFET will assume a intermediate state somewhere between ON and OFF - this is not a valid operating point for most MOSFETs and it means that the voltage to the amplifier will vary with current drawn.

 

Hello Irving , Yes you are correct , ucc5304dwvr is a gate driver.The mosfet is a N channel which has conditions for operating.

mosfet source is connected to VSS1 which is ground for secondary side by definition of the datasheet.
The gate of the mosfet is connected to the out of the driver.
what is the logic of the floating voltage here?
why Vss1 which is a ground in datasheet shown bellow is 24V?
Thanks.




https://www.onsemi.com/pdf/datasheet/ndt3055l-d.pdf
https://www.ti.com/lit/ds/symlink/u...939&ref_url=https%3A%2F%2Fwww.mouser.co.il%2F

 

Yes, VSS is 'ground' for the driver, but it's not ground for the circuit, it's connected to the MOSFET source. What is the driver's VDD connected to?

 

Hello Irving,extended schematics photos are attached the node is call VIZ and GIZ and its connected to DB02S2415A component as shown in another picture .
mosfter is opened and closed by the Vgs>Vt condition, however what i have inner logical conflict when on one hand we put "floating" Ground.
But on the other hand we the source is not connected to ground it could be connected to power amplifier and the draing of power amplifier is the load and not plain ground.
So how this virtual ground helps us here?
Thanks.

"Output of UCCxxx should be referenced again Source of MOSFET - this is the way how NOSFET works (it controlled by Gate-Source voltage)"

"BTW - there is no supply attached to isolated side of UCC5304 in schema. I think something was forgotten"

https://www.mouser.co.il/datasheet/2/632/DS_DB02S_D-3106392.pdf

 

Consider the MOSFET when turned on to be a very low resistance, approx 10mOhm, then you have a divider network with the load, which is around 60ohm (24v @ 400mA), so Vs will be around 23.9v. For the MOSFET to stay turned on, Vgs > 5v orr so, so gate voltage must be at approx 29 - 34v with respect to ground. In the absence of a 30+v supply the common option is to use an auxiliary supply above the supply rail. Where the MOSFET is continuously switched on & off. eg on the high side of an inverter H-bridge, a simple capacitive bootstrap is used, but here the MOSFET is permanently on so the alternative is an isolated DC-DC converter whose output can float with Vs.

There is an error in your circuit. The DC-DC converter as spec'd is 24v in, 15v out. You are powering it from, seemingly, +24 and -12v rails, ie 36v. While spec'd for 18 - 36v operation, there is no need to stress it unecessarily, you only need to power it from 24v and ground. There is no need to ground pin 8, its a NC on the single output device.

It is hard to read your schematics, they need some serious tidying up. Also, why so many decoupling capacitors, 1u, 100u ,1n, 100n on each device? 100n & 10u low ESR ceramics should be sufficient if you have a good PCB layout.

 

Other thoughts...

You don't need R4u & Du1 on the MOSFET gate. The UCC5304 is an active pull-down device, they wont turn the MOSFET off any quicker. What's the purpose of C4u? That will slow things down...

 

Hello Irving regarding what you said:

Point 1:
What is the exact voltage of -Vout(Viz) and +Vout(Giz) given the data sheet. In the photo shown below they say 15V but its not enough because i need to know the potential -Vout(Viz) and +Vout(Giz)?

"There is an error in your circuit. The DC-DC converter as spec'd is 24v in, 15v out. You are powering it from, seemingly, +24 and -12v rails, ie 36v. While spec'd for 18 - 36v operation, there is no need to stress it unecessarily, you only need to power it from 24v and ground. There is no need to ground pin 8, its a NC on the single output device."








point 2:
There is this boot strap here ,could you give an intuition regarding how this diode and capacitor play together?
Thanks.

 

Point 1:
What is the exact voltage of -Vout(Viz) and +Vout(Giz) given the data sheet. In the photo shown below they say 15V but its not enough because i need to know the potential -Vout(Viz) and +Vout(Giz)?

Why do you need to know this? The DC-DC converter output will be 15v within a few % unless you draw too much current from it.

point 2:
There is this boot strap here ,could you give an intuition regarding how this diode and capacitor play together?

They don't really. The capacitors provide local energy storage to counteract local voltage drop due to high current switching spikes and reduce noise from those spikes radiating from the power trace (which is inductive). Normally there wouldn't be an R4u, and the diode provides a lower impedance path for discharging the MOSFETs gate capacitance than the gate resistor R6u which is required to prevent the gate 'ringing' as the gate capacitance charges. R4u and Du1 are not really needed as the driver is an active pull-down. R4 isn't needed because the driver forces the gate voltage to zero when the driver is in startup and/or VDD is < 5v. C4u will just slow everything down and shouldn't be there IMHO.

 

Hello Irving, I have simulated an identical device gate driver shown below.
As you can see my output pulse moves from VDDA to VSSA .
If this is going to a mosfet then i need to know what voltage goes into the the MOSFET.
given that DB02s2415A is that source of my driver how do I know what voltage potential goes into the gate and what into the source.

I know to simulate the interactionof this driver and the mosfet and if i dont know what ecactly are VDDA and VSSA then i will not know what will go into my MOSFET and my simulation could the opposite ofthe voltage that will drive this mosfet in reality?
Thanks.

 

You're overthinking this. The auxiliary supply is floating relative to ground, and its -ve, and that of the driver, is tied to the MOSFET source. Therefore the driver VDD will always be 15v above the source, and the gate either close to the MOSFET source when the driver is off,, or close to source + VDD when it is on. Does that make sense now?

 

Hello Irving,according to the datasheet I have to put 20V on the drain of the TGA2590 power amplifier that means i need to put
20V on the source of the mosfet.
You said that its a floating voltage without reference ,how do i make sure i have 20V on the source of the mosfet (drain of TGA2590) as shown in the diagram below.
Thanks.

https://www.qorvo.com/products/p/TGA2590-CP

 

Either your 24v supply needs to be 20 - 21v or you need to post regulate between source and the amp.

Here's one option... a simple post-regulator. Q1 will dissipate around 8W so will need a small heatsink, though without knowing more about the environment it'll be working in I can't say what size yet.



A second option is to make the MOSFET do the regulation... This also does away with the driver and the aux power supply.

 

Hello Irving there is a basic confusion I have.
We have a mosfet with 24V drain its VGS i getting plugged with 15V floating from DB02S2415A floating difference.
But in my configuration as you see the drain of the power amplifier is the source of the mosfet.

So on one side we have the floating value of GIZ but on the other hand you said i should connect to the same node 24V to 20V converter that you reccomended in post 13.
So we are connecting two different voltage potentials on the same node.
Arent they going to conflict which other?
Thanks.

 

Update:
In my last photoi am manually fixing the Vds,you reccomended the following circuit.
Is there a way to impelement the methos you presented in the circuit below?
or other cool method so i could tune easily my Vds?
Thanks.

 

Do you want to tune Vds manually or electronically?

BTW yo are now showing 7.3A x 2 = 14.7A, not the 2A originally mentioned. The NDT3055L can't handle more than 2 - 3A continuous.

 

Hello Irving , I am working in pulses not continues.
Manually tuning the Vds will be great at first so I’ll understand the logic .
Thanks.

 

How long pulses?

 

Hello Irving 500ns on , period
1 micro seconds .
In the data sheet they don’t have microsecond periods , I am not sure how to know if my mosfet will get burned or not.
In the last page they show a plot but the r(t) values end at 0.0001

https://www.onsemi.com/pdf/datasheet/ndt3055l-d.pdf

 

Your duty cycle is 0.5 therefore r(t) = 0.5, Rthja(t) = 0.5 x 110 = 55degC/W with no heatsink



Your current simulation, with 6v on the gate shows a dissipation of 7.4W average across the cycle, but is not realistic because it draws a peak gate current of 1.6A from a 2W inverter - that's not going to happen, realistically you'll get around a 120mA, which won't even turn the MOSFET on fully in 500nS! Even if you provided more gate drive, at an average wattage per cycle is approx 7.4W, including switching transients, your junction temp will be 7.4 * 55 = 407 above ambient with no heatsink, and even on a decent lump of copper you're not going to get much better than 42degC/W or 7.4*42*.5 = 155 above ambient, as this is limitation of SMD package type. You need to limit your dissipation to < 5W. One option is to use two MOSFETs in parallel as shown below. Gate resistors R1, R2 and source resistors R3, R4 provide negative feedback and ensure M1 & M2 share the load equally. Each dissipates 1.98W giving a junction temp of around 113 on a reasonable heatsink. The idea can be extended to 3 or more if need be. Note: your chosen gate driver will provide drive for several MOSFETs, but your chosen DC-DC converter won't drive one, lt alone two or more - the simulation shows 500mA gate drive peak per MOSFET. You'll need an isolated AC/DC supply capable of 1A or more.