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I want to use a 5V 500mA to my circuit, also this will have a current spike for 1A. The input will be from 9V - 36V. I have reviewed the difference between the linear and switching regulator and obviously I should be using the switching regulator. How suitable is this switching regulator for my application with a real 36V input?

https://www.digikey.com/products/en?keywords=OKI-78SR-5%2F1.5-W36-C

This is on its datasheet. enter image description here

I know that switching has less power dissipation than linear, but how do you compute the power required, or power dissipation, power loss considering the input, ouput voltage and the output current on a switching regulator?

Will I able to use it on 36V input to have a really 5V, 1A output?

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At 36V input, 5V 0.5A output, efficiency is rated at ~80%.

So to output 2.5W of power (0.5A*5V), you draw 3.125W (2.5W/0.8). Losses are 0.625W.

enter image description here

Will I able to use it on 36V input to have a really 5V, 1A output?

Yes, the output at 1.5A with 36Vin is represented below:

enter image description here

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  • \$\begingroup\$ Would that mean I am heating 625mW? Is that good or bad? Would I exhaust it? How much power am I allowed? \$\endgroup\$ – Nikko Jul 11 '18 at 9:30
  • \$\begingroup\$ Basically heat, yes. 625mW is quite ok, especially for a device that size. Take a 1/2W resistor as reference for example. I looked in the datasheet and couldn't find the thermal resistance of the device, but you can calculate a ballpark of the expected temperature rise if you had it. Anyway, 625mW isnt an issue unless you are actively insulating the device. \$\endgroup\$ – Wesley Lee Jul 11 '18 at 10:18
  • \$\begingroup\$ Okay man! Thanks. Btw, another question. Whats a safety factor to use for a specified voltage? For example, I needed 36v. Should I buy a max voltage higher than that i.e. 42V? Is this for LDO only or can be applied to Switching Regulators? \$\endgroup\$ – Nikko Jul 12 '18 at 7:12
  • \$\begingroup\$ I would personally not use a device rated at 36V at 36V for critical situations. That said the datasheet itself mentions Vin=36V so I guess the internal device is rated at a higher voltage. If its a personal project then I'd go for it and watch it a bit closely in the beginning. You could add a series diode or two to drop the voltage a bit too - a bit of waste of energy but then you get reverse polarity protection too. In datasheets of specific devices you often have recommended range and abs max ratings. On this module I couldn't find it. \$\endgroup\$ – Wesley Lee Jul 12 '18 at 10:16
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You need to use the manufacturer's quoted efficiency to calculate how much power will be lost in the conversion. To be on the safe side, use the worst case value they give.

If Vo is the output voltage, Io is the output current, and E is the efficiency in percent, then to a reasonable approximation, the power lost is given by:

P = ((100/E) - 1) Vo Io

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