simulate this circuit – Schematic created using CircuitLab
Note 1: The input voltages are only \$V_{cc}\$ and \$V_\text{High Voltage}\$. You don't apply anything at the \$V_{BS}\$ node. It is only for representation.
Note 2: Notice that there are two different type of grounds. Those grounds must not be directly connected to each other.
You must drive the MOSFET between its gate and source terminals. Since the source terminal voltage of a high side MOSFET will be floating, you need a separate voltage supply (VBS: \$V_\text{Boot Strap}\$) for the gate drive circuit.
In the schematic below, VCC is the voltage source of the rest of the circuit. When the MOSFET is off, ground of the boot strap circuit is connected to the circuit ground, thus C1 and C2 charge up to the level of Vcc. When the input signal arrives to turn the MOSFET on, ground of the gate drive circuit rises up to the drain voltage of the MOSFET. The D1 diode will block this high voltage, so the C1 and C2 will supply the driving circuit during the on-time. Once the MOSFET is off again, C1 and C2 replenish their lost charges from VCC.
Design criteria:
- RB must be chosen as low as possible that will not damage D1.
- Capacity of C2 must be chosen enough to supply the driving circuit during the longest on-time.
- Reverse voltage rating of D1 must be above \$V_\text{High Voltage} - V_\text{CC}\$.
The input signal must be isolated from the boot-strap circuit. Some possible isolaters are:
Optocoupler
Optocoupler is the most basic method for isolation. They are very cheap compared to other methods. The cheap ones have propagation delay times down to 3\$\mu\$s. The ones with less than 1\$\mu\$s propagation delay are as expensive as isolated gate drivers though.
Pulse Transformer
Pulse transformer is a spacial type of transformer for transferring rectangular pulses. They have less number of turns in order to avoid parasitic capacitance and inductance and larger cores for compensating loss of inductance due to reduced number of turns. They are much faster than optocouplers. Delay times are less than 100ns in general. The image above is for illustration only. In practice, the current they can provide is not enough for driving a MOSFET fast; so they need additional circuitry in practice.
Isolated Gate Driver
Isolated gate driving is a relatively new technology. All the complexity of gate driving is encapsulated in one single chip. They are as fast as pulse transformers, yet they can provide a few amperes of peak gate current. Some products also contain on-chip isolated DC-DC converters, so they don't even need boot-strapping. However, all these super features come with a cost.
First of all why do you short the output pin via R1 resistor?
In LTspice \$10m\$ is interpreted as \$10m \Omega = 0.01\Omega \$.
Also we can find in data-sheet this information:
In your circuit \$V_{DD} = 12\$ therefore \$Vin > 0.67*12V = 8.04V \$
LM5134B is a TTL version
After I fix this the simulation look like this:
Best Answer
IN pin should not be pulled high continuously , this driver is a bootstrap driver , which means it needs to recharge the bootstrap capacitor through the low side mosfet when the high side mosfet is off.
buttom line , IN pin should be PWM from 0 to around 95% but never to 100%.
if you want 100% , ( for some applications you do ) there is solutions like adding a charge pump ... etc
check out this app note from Texas instruments Providing Continuous Gate Drive Using a Charge
EDIT:
After checking the datasheet it seems you have another mistake , that is powering the IR2104 from 5v , the IR2104 accepts 3.3v and 5v and 12v logic inputs at IN and SD , but its supply (aka VCC ) should be above between 10v to 20v. the IGBT voltage can be 5v no problem .