How to I drive video switch IC ?

Thread Starter

damyb

Joined Aug 27, 2021
2
I want to supply power 28V to 5 V for this specific IC (BH76330FVM, BH76331FVM, BH76360FV, BH76361F BH76332FVM, BH76333FVM, BH76362FV, BH76363FV V). I'm worried about if I used LDO is OK?
I choose 2 to regulator for this purpose:
1)Integrated 4V-60V, 150mA, High-Efficiency, Synchronous Step-Down DC-DC Converter with 50mA Linear Regulator/ MAX17670, MAX17671, MAX17672
2) XC6216/XE6216 / 28V Low Power Consumption 150mA Voltage Regulators
 

KeithWalker

Joined Jul 10, 2017
1,933
The information in your question is not very helpful It would take up a considerable amount of my precious time to study the data sheets of all the devices you have listed before I could give you a definite answer. Have you calculated from the datasheets what the total current of your circuit will be? Once you have done that, select a suitable step-down converter, allowing for at least a 20% extra current safety margin.
 

DickCappels

Joined Aug 21, 2008
7,836
The only definite question I was able to find was whether or not it is ok to use Low Dropout regulators. If you can find some with specifications that meet your needs, sure you can!
 

Papabravo

Joined Feb 24, 2006
16,938
You can evaluate the choices yourself with simple calculation(s).

Multiply the current required by the load by the voltage difference between the input and the output for the LDO. This will be quite high if the LDO is doing all the work. You might have to provide for a big heatsink in order for the LDO to handle this application.​
If you have a buck converter in front of an LDO it is just a bit more complicated. In the buck converter the power out will always be less than the power in. For a back-of-the- envelope calculation you can assume an efficiency of 80%. This assumes that 20% of the available input power is lost or wasted. Then do the same calculation as above for the LDO and you can estimate the total power loss of each scheme.​
Why is this approach valid? It is exact for the LDO alone. For the combined solution, if the efficiency of the buck converter is less than 80%, it is a poor design and you should seek an alternative. If it is better than 80% you will have a better situation then you anticipated and have some extra margin in the final design.

I can do a numerical example of both alternatives if that would be helpful to you.
 
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