+1 for the Clear question, associated data and response for queries in the comment.
Can I power a Raspberry Pi with \$24V\$ through a TLE4271?
Let us look at some tables in the Datasheet.
The TLE chip comes with several features. Do you need all of them? Do you have many TLE in stock? anyways, let us see whether...
- The TLE can supply sufficient current or not ?
The estimate from google result (consider with a grain of salt) estimates to be around \$300 mA\$ (I believe you aren't watching a video) any other wireless application will not demand such huge continuous current (Bluetooth, Sub GHz etc).
So, let us assume a current about \$200\$ to \$300 mA\$. The TLE can definitely support it given the input voltage is \$24 V\$. When the input voltage is higher than \$36 V\$, the current is limited to \$300 mA\$.
- The TLE temperature rise** is within limits or not?
The voltage difference \$24 V - 5V = 19 V\$ will be present across the regulator. When the current flow is about \$300 mA\$, the power dissipated across the Regulator is $$V * I = 19 V * 300 mA = 5.7 W$$
From the second table above, the junction to ambient temperature resistance is \$65^o C/W\$.
What it means is: for every \$1 W\$ of power dissipation in the regulator, a temperature rise of \$65^o C\$ can be expected. For our assumed case it will be about \$5.7\$ times \$65^o W\$ that is more than \$300^o C\$. This is way beyond the temperature the IC can function (\$150^o C\$ is the absolute Max). The IC has internal temperature safety feature, so it will cut off the power anyway.
Let us see how we can still make it work. One way is to reduce the current to about 50 mA or so. Which may not be in our hand or the application can't function with such a low current. The same table also mentions about ambient to case thermal resistance of \$3^o C/W\$. Sounds good. For our \$5.7 W\$ dissipation, it would mean only about \$18^o C\$ rise in temperature (above Ambient). For this to workout we have to find a ideal heatsink to mount it on the TLE regulator.
If we assume reasonably, we can find one with \$4^o C/W\$ thermal resistance. This could mean that over all thermal resistance will be \$4^o C/W + 3^o C/W = 7^o C/W\$. This implies, the final temperature rise will be about \$ 5.7 * 7 = 40^o C\$ (5.7 Watt because of 200 mA current). Considering ambient as \$20^o C\$, the final temperature of the regulator will be about \$60^o C\$. It will be running hot but still within specification.
- We can improve the conditions
We can reduce the input voltage which greatly reduces power dissipation across the regulator. Or we can also reduce the current consumption in the Pi as much as possible. We can also have a switching regulator who can switch the input voltage to a value of 7 or 8 V and TLE regulator can then convert the voltage to 5 V. Benefits: low noise for the Pi device because of the LDO. Low power dissipation across the Regulator due to small voltage drop (3 V compared to 19 V). Depending on your complete system and actual condition of the design, you can optimise the design.