How are Power Mosfets driven by Gate Drivers if the Vgs needs to be so high?

I'm working on the circuit in the attached image. I've quickly realized there is not any good way to activate the Vgs of my power mosfet with my gate driver. I've looked online at some other gate drivers, and all of them only output low voltages around 20-30V MAX.
Every single datasheet for gate drivers intended to be used with power mosfets describes their output in terms of current. Sink or source, up to 4A/6A, etc.
These are 2 of the gate drivers that I have, one of them is also an isolated gate driver specifically used for power mosfets.

http://www.ti.com/lit/ds/symlink/ucc20520.pdf

https://www.mouser.com/datasheet/2/698/isl55110-11-1302115.pdf
I understand that the mosfet gates activate upon charging to a certain Qg, but if the gate voltage needs to be 10V > than Vs, and Vs is 200V+, how is it possible to turn the gate on and let current flow from Vs to Vd? I tried simply switching the gate and drain so that the 10V output of my gate driver would be Vgs = 10V, and I realize now that would not work. Is this because Vgd is the parameter for activating the gate when higher voltage is applied to the drain?

Hi Michael-Vs (voltage on the Mosfet source pin)is not 200 volts.  I would expect it to be around 4 or 5 volts. the voltage on the drain pin will be over 200 volts.  The circuit appears to be a constant current sink circuit. Look at it this way: Start with the gate  voltage of zero.  The MOSFET is off, there is no current being drawn, so the voltage on the source pin is also zero. Now assume that you begin raising the voltage on the gate.  The MOSFET begins to turn on somewhere between 3 and 5 volts, and now current is beginning to flow. Now you will be dropping voltage across  to 50 ohm resistor, so the voltage difference between source and gate is a little less that what the voltage source feeding the gate is putting out. As you increase the gate voltage source, the current through the 50 ohm resistor increases, also the voltage drop across it increases, further reducing the voltage difference between source and gate. .By the way- don't be confused by the current rating of the gate driver chips. The current rating only comes into play when you are turning the MOSFET on and off very quickly, which is not the case in the circuit example you gave.  MOSFETS have a relatively high gate capacitance so the high current is necessary only for a few microseconds to quickly charge or discharge the gate. In your case the gate current will be close to zero.. 

Michael, the configuration in your drawing is called a "source follower", because the source-to-ground voltage follows the gate-to-ground voltage as the gate-to-ground voltage varies. As Rick explains, the voltage between source and ground will be a few volts less than the voltage from gate to ground. You have built a unity-gain analog amplifier rather than a switch.
The switch-mode gate drivers are meant to be connected between gate and source, not gate and ground, so all the gate-driver voltage goes between gate and source. For your circuit, this means the gate driver must run from a floating power supply or battery.

Why not use a P channel device if you are simply load switching?You may need an N channel for level shifting, and perhaps a resistor or two, and a zener to limit Vgs, these options depend on the configuration chosen, see   High side switch, or   another PDF for example,cheers,Richard

Hi,
Take a look at the Infineon IR2125 device, or similar.These devices work as they have a floating channel arrangement. Once the FET start switching on, the source voltage is fed back to the device where a comparator type circuit, keeps the Vgs voltage 12-18V above the supply. So with a 200V supply, the gate of the FET will be at least 212V. If the Vgs voltage drops below the Vgs th value, typically 5-10V for high voltage MOSFETs, it will conduct poorly in the linear mode, dissipating a lot of power or it may stop conducting.Taken from the IR2125 datasheet:

The Linear Technology/Analog devices datasheets give better descriptions of how these devices work. The Infineon/International Rectifier datasheets are terse.
I've used this style of high side drivers and some from Linear Technology (LTC7001) , they do what they say in the datasheet. Your UCC20520 device is intended for a power converter or H-bridge arrangement, you just want to switch a high voltage on or off. If you try and use the UCC20520, or similar, you will fail as these devices drive a half bridge. In the 'power down' or failure state, they switch the low side MOSFET, not something you want in your application. 

 The design in your illustration is known as a "source supporter", in light of the fact that the source-to-ground voltage pursues the door to-ground voltage as the entryway to-ground voltage changes. As Rick clarifies, the voltage among source and ground will be a couple of volts not exactly the voltage from entryway to ground. You have constructed a solidarity increase simple speaker instead of a switch. By the way-don't be confounded by the ebb and flow rating of the door driver chips. The current rating just becomes an integral factor when you are killing the MOSFET on and rapidly, which isn't the situation in the circuit precedent you gave. MOSFETS have a moderately high entryway capacitance so the high current is vital just for a couple of microseconds to rapidly charge or release the door. For your situation the door current will be near zero..

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The max Vgs to fully turn off a
typical MOSFET is never more than 10V.
Remember, Vg to ground is
not always the same as Vgs.

In your circuit you need to ground
the transistor source and place the 200V in series with RL and the
transistor drain.
That way the Vgs of 10V will fully turn on the
MOSFET and apply 200V across RL.

If you need to have RL
grounded, then you would need to either use a P-MOSFET with a high
driver, or an N-MOSFET with a bootstrap driver.
Here's an example driver
circuit with its LTspice simulation, to generate a 200V, 100ns
pulse.Since the pulse has a low duty-cycle, it can be AC coupled
to the high-voltage P-MOSFET.(Note that the I only had a 200V
P-MOSFET model, so you would need find a P-MOSFET with at least a
250-300V rating for the actual circuit.