If switching the Ground-side is not practical for your application,
use a Photo-Voltaic-Gate-Driver as shown below.
This works with High-Side, or Low-Side, Control equally well, no P-Channel required.
But, the Photo-Voltaic-type-Gate-Drivers can not be used for "High-Speed-Switching"
as they are somewhat "slow", ( several Milli-Seconds, depending upon the specs of the FET used ).
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View attachment 324158.

So, this circuit diagram shows a resistor to protect the input LED. Does it also need a resistor to protect the output LED?

 

Does it also need a resistor to protect the output LED?

The input LED needs a resistor like any LED.
On the output side (Gate) needs no resistor.
Many opto-isolators have an IR LED and a photo transistor. In this isolator the receiver is not a photo transistor but more like a solar panel that turns light into voltage.

 

The input LED needs a resistor like any LED.
On the output side (Gate) needs no resistor.
Many opto-isolators have an IR LED and a photo transistor. In this isolator the receiver is not a photo transistor but more like a solar panel that turns light into voltage.

The bench-tests went well ... circuit worked well at all Voltages, +12, 24, 48 and 20 Amps. Vgs stayed a consistent 7 Vdc. Such a very good simple solution. Thank you kindly for all of the detailed assistance )

 

You can lead a horse to water, but You can't make him drink.
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The bench-tests went well ... circuit worked well at all Voltages, +12, 24, 48 and 20 Amps. Vgs stayed a consistent 7 Vdc. Such a very good simple solution. Thank you kindly for all of the detailed assistance ). NAH NAH

 

Schematic Updated with more Part-Numbers and Component-Prices.

Limiting the Gate Turn-On-Current with a Resistor will make the Switching-Time excessively long,
this will generate tremendous amounts of Heat in an extremely short period of time.

Remember that a FET-Gate has Capacitance,
and if You install a Resistor going into that Capacitance, You have created a "Delay-Timer".
The more "TIME" that a FET remains in it's Linear-Region, the more HEAT will be generated.
HEAT is the Enemy.
It is advantageous to "Switch" the FET as fast as possible to prevent this Heat generation.

The FDA215 Gate-Driver is a "Dual" Device,
this means that You can wire both Outputs in parallel to double the Output-Current and
cut the Switching-Time in half.
This may allow using a much smaller Heat-Sink for the FET,
since almost all the Heat that will be generated, will be generated during the delay in
getting the Gate-Voltage ( Gate-Capacitance ) charged-up from zero, to around ~8-plus-Volts.
At ~8-Volts and above, the FET will look like an extremely low-value Resistor, ( see the Rds-On Specs ),
and therefore, it will generate very little Heat when continuously and fully turned "On".
But, during the Time that the Gate-Voltage is rising,
the FET will have a substantial amount of Resistance,
and this will generate a lot of Heat for a very short period of Time,
this "Switching-Time" needs to be kept as short as possible to prevent smoking the FET.

Putting the Outputs in parallel will mean that the Input-LEDs must both be driven.
The Input-LEDs must must be driven hard enough to create at least ~5mA each.
Since You have ~12-Volts, or more, to work with, this is easy to accomplish.

The Switching-Time for the below Schematic, with the Components specified,
will be somewhere in the range of ~1.5 Milli-Seconds, which is very reasonable,
and will not generate a lot of Heat when Switching "On".
Switch-Off-Time is roughly one tenth of that Time, so it is not a concern.

However, care should be taken to insure that the Switching-Input-Voltage is always "Clean and Fast",
with no chance of the Input-Voltage having a "Slow-Rise-Time".
If the Input-Voltage spends any significant amount of Time around the ~2 to ~3-Volts-Level,
the FET could be smoked within Seconds.
The Input-Control-Voltage must always be either,
below ~1-Volt, or in excess of ~4-Volts,
any wandering Voltages between those above could cause the FET to be smoked within Seconds.
This applies to virtually all FETs controlling a Heavy-Load.
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Limiting the Gate Turn-On-Current with a Resistor will make the Switching-Time excessively long,
this will generate tremendous amounts of Heat in an extremely short period of time.

Remember that a FET-Gate has Capacitance,
and if You install a Resistor going into that Capacitance, You have created a "Delay-Timer".
The more "TIME" that a FET remains in it's Linear-Region, the more HEAT will be generated.
HEAT is the Enemy.
It is advantageous to "Switch" the FET as fast as possible to prevent this Heat generation.

The FDA215 Gate-Driver is a "Dual" Device,
this means that You can wire both Outputs in parallel to double the Output-Current and
cut the Switching-Time in half.
This may allow using a much smaller Heat-Sink for the FET,
since almost all the Heat that will be generated, will be generated during the delay in
getting the Gate-Voltage ( Gate-Capacitance ) charged-up from zero, to around ~8-plus-Volts.
At ~8-Volts and above, the FET will look like an extremely low-value Resistor, ( see the Rds-On Specs ),
and therefore, it will generate very little Heat when continuously and fully turned "On".
But, during the Time that the Gate-Voltage is rising,
the FET will have a substantial amount of Resistance,
and this will generate a lot of Heat for a very short period of Time,
this "Switching-Time" needs to be kept as short as possible to prevent smoking the FET.

Putting the Outputs in parallel will mean that the Input-LEDs must both be driven.
The Input-LEDs must must be driven hard enough to create at least ~5mA each.
Since You have ~12-Volts, or more, to work with, this is easy to accomplish.

The Switching-Time for the below Schematic, with the Components specified,
will be somewhere in the range of ~1.5 Milli-Seconds, which is very reasonable,
and will not generate a lot of Heat when Switching "On".
Switch-Off-Time is roughly one tenth of that Time, so it is not a concern.

However, care should be taken to insure that the Switching-Input-Voltage is always "Clean and Fast",
with no chance of the Input-Voltage having a "Slow-Rise-Time".
If the Input-Voltage spends any significant amount of Time around the ~2 to ~3-Volts-Level,
the FET could be smoked within Seconds.
The Input-Control-Voltage must always be either,
below ~1-Volt, or in excess of ~4-Volts,
any wandering Voltages between those above could cause the FET to be smoked within Seconds.
This applies to virtually all FETs controlling a Heavy-Load.
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View attachment 324545
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Thank you kindly for excellent content … glad that you signed your work ). When my dad made me wood bowls, I insisted that he sign his art )

 

Schematic Updated with more Part-Numbers and Component-Prices.
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