Hello, I have 5 amplifier which consume 1A under Vds=50V so the biasingnetwork sees 10 ohms as load.
My mosfet whose source is connected to 10ohm load ,the load is connected to gnd.I was reccomend to connect UCC5304DWVR to it in order to open it.
But UCC5304DWVR is using a floating ground ,so how exactly we connect a floating ground component to real ground component?
Its like connecting oil and water
Thanks.

 

.I was reccomend to connect UCC5304DWVR to it in order to open it.

I assume you mean to turn it on or activate it.
In electronics "open" is no connection.

So why was that isolated driver recommended?
Do you need isolation for some reason?

 

I was told that isolated driver connected purely to the vgs opens the mosfet.
so my question is mainly why using UCC5304DWVR which is a floating ground , can be used in a device with real ground?
Its like connecting oil and water

 

UPDATE:

Floated means driver output ground is not the sameas input output ground.
so for example we have 5V potential difference between the to ports of the output.
what are the limits can this vice handle 0-5V 10-15V as well as 800-805V?

 

Still not clear what you want/need.
Are the two grounds at a different potential or not connected to each other?
How fast (frequency) do you need to switch the power?

 

hi yef,
Please post a simple diagram showing the project problem section.
E

 

UCC5304DWV

The UCC5304 needs a floating power supply.
Can you give us a schematic of what you are doing. A picture is worth 1000 +/-10% words.

 

Beasickly I have a 15V voltage supply and a 3.3V pulse.

I want the DCP021515U to supply the VDD for UCC5304DWVR .
The UCC5304DWVR will be connected to VGS of the mosfet .
However as you can see below my mosfet source is not connected to ground but it has ressistive load and I need current to be supplied to that load.
So how connecting a resistor between floating ground and a real one will make the mosfet open and drive the current threw the resistor?
Thanks.

https://www.ti.com/lit/ds/symlink/d...ps%3A%2F%2Fwww.ti.com%2Flit%2Fgpn%2Fdcp020503

 

A MOSFET is controlled by the voltage between its source and gate. The arbitrary voltage called “ground” has nothing to do with it. The MOSFET has no idea of what ground is.

 

This my Idea described below, I have DCP021515U which inputs are 15V and GND of a power supply.
The output of the DCP021515U goes to VDD and VSS_1/2 of UCC5304DWVR.
UCC5304DWVR VCCI_1/2 is 5V made from UA78L05ACLP.
Pin 1 of UCC5304DWVR gets a pulse 0-3.3V.
The out of UCC5304DWVR in such case will be 15V which is connected to the Vgs of IRF7854PbF.
Given that the drain is 50V and Vgs is 15V according to the Vgs .
Will the mosfet will open with no problem ?
Thanks.

https://www.infineon.com/assets/row...n.pdf?fileId=5546d462533600a40153560c559a1d29

 

You can use an opto-isolator when you have no common ground.

 

Hello,What about my idea in post 10, will it work?
Thanks.

 

I cut one wire and connected one with a red line.
The 15V isolated power supply powers the gate driver.
The gate driver "out" goes to Gate of MOSFET.
The Gate driver "VSS" goes to Source of MOSFET. (that in a N-MOSFET I hope. I did not check)

 

Pin 1 of UCC5304DWVR gets a pulse 0-3.3V.

Why does the gate driver have to be isolated and what is the frequency of this 0-3.3V pulse?
The IRF7854 is only rated to 30VDC max.

EDIT: the IRF7854 is rated to 80Vdss

 

Why does the gate driver have to be isolated

Because the Gate needs to be at 0V for off and 65V for on, as measured from ground.

The IRF7854 is only rated to 30VDC max.

what is the frequency of this 0-3.3V pulse?

Good question. Also what is the min and mix duty cycle. 0%? 100%?

 

Because the Gate needs to be at 0V for off and 65V for on, as measured from ground.

Correct the IRF7854 is rated to 80Vdss.
How to get 65 volts output with a 15 volt supply from the UCC5304?

 

How to get 65 volts output with a 15 volt supply from the UCC5304?

All we need to do is get 0 or +15V from Gate to Source on the MOSFET. That is done by making a floating 15V supply and connecting it to the Source of the MOSFET. The floating supply moves with the load. (0V or 50V)

 

This my Idea described below, I have DCP021515U which inputs are 15V and GND of a power supply.

That supply costs too much. Try this one. MEV1S1515SC

 

Hello Ronsimpson, Yes I forgot to connect the virtual gnd Vss_2/1 to the S_1 S_2 S_3.
Its intresting how exactly UCC5304DWVRwith isolated ground which means he have potential difference between VSS and OUT can just open a mosfet and not affect the load resistor.I am not used to work with floating ground components.
I am used to thimk at circuits as every potential must have one reference point so here I am not sure how to think of the the logic of connecting such device to a mosfet.

Another question is :
Where do you reccomend to put decouling capacitors in this system?
Thanks.

I cut one wire and connected one with a red line.
The 15V isolated power supply powers the gate driver.
The gate driver "out" goes to Gate of MOSFET.
The Gate driver "VSS" goes to Source of MOSFET. (that in a N-MOSFET I hope. I did not check)
View attachment 360114

 

.I am not used to work with floating ground components.

You can think of a floating supply as a battery. The (-) end of the battery can connect anywhere, not just to Ground.

Where do you reccomend to put decouling capacitors in this system?

You might look in data sheet to see if they want a certain size or range of capacitors.
1- put a small cap very near the supply pins of the ICs. Maybe a 0.01uF or 0.1uF. Do not put the cap from +15V to GND just anywhere. Put the small cap from Vs+ to Vs- at the IC. You can add a 10uF at a distance from +15V to GND.
2-UCC5304 VCC to VSS_1,2 (not connected to GND!)
3-7805 voltage regulator needs an input cap and an output cap.
4-The 50V needs a cap.
Arrow is where a cap needs to live.

At low frequencies energy comes from the 50V supply, through the MOSFET, load and into GND. At high frequencies (edges of turn on/off) the energy does not have time to come from 50V. It comes from the capacitor and back to the capacitor. In all these examples the fast-changing energy comes from the cap not the trace or wire all the way back to the source of power.