Your first concern in selecting a gate driver is to find one that can drive enough current to switch your selected MOSFETs fast enough for your application. As a rough estimation, you can divide the total gate charge of your MOSFET by the current the driver can sink/supply.
$$ t_{on}=\frac{Q_g}{I_g}$$
Using the worst-case values for IRF1405 and the slower of your two gate drivers, IRS4253:
$$ \require{cancel} \begin{align}
t_{on} &= \frac{260 \cdot 10^{-9} C}{180 \cdot 10^{-3}A} \\
&= \frac{260 \cdot 10^{-6} \cancel{C}}{180 \cdot \cancel{C}/s} \\
&= 1.44 \mu s
\end{align} $$
Off is faster, because this driver (which is typical) can sink more current than it can source:
$$ \require{cancel} \begin{align}
t_{on} &= \frac{260 \cdot 10^{-9} C}{260 \cdot 10^{-3}A} \\
&= 1 \mu s
\end{align} $$
If your switching frequency is 10kHz, each switch period is \$1/10000 = 100\mu s\$ and you will spend \$ (1.44\mu s + 1\mu s) / 100\mu s = 2.44\%\$ of that time switching. Probably acceptable, but you should calculate your switching losses and check.
Also, keep in mind this calculation is an approximation. The current specified in the gate driver datasheet is current into a short circuit, but a MOSFET gate isn't that. Unlike a short circuit, the gate voltage rises as it is charged, which will reduce the current the driver can provide. Also, your layout may introduce more inductance and resistance than there was in the test circuit the manufacturer used, further reducing current. Consequently, your actual switching losses may be higher than this calculation suggests.
In selecting the bootstrap capacitor, you want to make sure it's significantly bigger than the gate capacitance it will be charging, so that the bootstrap voltage doesn't sag appreciably when you switch. It also needs to supply whatever leakage current there is as long as you keep the high-side switched on. You can calculate these leakage currents, or just make the bootstrap capacitor way bigger to be safe. 100 times bigger than the gate capacitance should be good, so at least \$26\mu F\$. Bigger doesn't hurt much, so round up to a standard value or whatever you already have in the BOM or stock.
Since this capacitor is the power supply for the high-side gate current, you also want it to be very low impedance. It wouldn't hurt to parallel your big capacitor with some smaller \$100nF\$ decoupling capacitors very near the gate driver(s).
Selecting a bootstrap diode isn't terribly difficult. It needs to be able to withstand the reverse voltage when the H-bridge is switched high. Also keep in mind that you will lose the diode's voltage drop from the gate voltage. A Schottky diode might be nice for this reason, but depending on your circuit, you may not find one that can take the reverse voltage. A simple 1N4148 can take reverse voltage up to \$100V\$.
The reverse recovery time of the diode can also be relevant if your are switching very fast; 1N4148 has a reverse recovery time of \$4ns\$, so you will have to have the H-bridge switched low for significantly longer than that for the bootstrap capacitor to have time to recharge between cycles.
While a schematic would have helped describe the problem statement better, one key concept might help in clarifying this matter:
- A voltage is the potential difference between two points in a circuit, it is not an absolute value of any physical characteristic at a single point in a circuit. Thus, there is no absolute potential involved, it is relative value, a difference.
How this applies:
The control side of the L293D is actually powered by the +5V from the Arduino only when the +5V has a reference ground available, that corresponds to that particular +5V supply, in other words, the ground of the Arduino board.
The L293D does not have a separate drive-side ground pin, just the "Heat sink and Ground" pins, which are also the ground reference for the \$V_{cc1}\$ pin.
If you note the schematic on page 3 of the datasheet, the ground references for \$V_{cc1}\$ and \$V_{cc2}\$ are one and the same, the GND pin(s). Thus, that reference needs to be connected to the Arduino's reference ground.
Best Answer
You do not need a full H bridge, just a couple of relays.
You can connect the purple wire to your positive power supply, while the other two wires should be connected to the negative via two separate relays. To hook a relay to an arduino just search the web, there literally are dozen of tutorials.
If for some reason you want to use the H bridge you can proceed as follows:
Hook the purple wire to your positive power supply, then power the L293D as usual. As Vcc you do not need a stiff power supply, as if you were powering the motors, you just need the digital Vcc.
Enable 1 and 2 drivers connecting pin 1 to Vcc, then hook the black and orange wires to the outputs (1 and 2). Now you can control your alarm and water pump via in1 and in2, please note that the device will turn on when the input is low, i.e. a digital zero.
When you turn off the devices there is a sim chance of damage since you are putting Vcc on a terminal that is probably meant for ground or high impedance. If you are concerned about this problem please add a protection diode in series, pointing towards the L293D.
See the schematic:
simulate this circuit – Schematic created using CircuitLab
Please note that I connected pin9 to gnd to disable the 3 and 4 drivers.