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I have an application with multiple components and multiple voltages in use and current calculations are giving me a headache.

A simplified version of my application would include:

  • 1x device drawing 300mA @5V
  • 1x device drawing 100mA @6V
  • 1x device drawing 500mA @12V
  • 1x 13.7V SLA battery being trickle charged

In order to trickle charge the battery I need a 15V PSU with current limiting resistor fed from the 15V PSU.

Done.

In order to supply 12V, I am using a voltage regulator MC7812 fed from the 15v PSU.

Done.

6V device gets power from a step down converter 62180 fed from the 15V PSU.

Done.

The 5V device feeds off the 62180 6V supply and uses current limiting resistors to reduce down to 5V.

Done.

But what wattage PSU do I need?

Using P=IV suggests P=0.3 x 5 = 1.5w but as my voltage supply is ultimately 15V and not 5V, would I need to multiply this by 3 (to get current draw at 15V) before adding to the other calculations or can I just calculate the wattage for each device (@ each devices voltage level) and then add them up?

using method 1 (excluding the SLA battery):
(0.3*5) * 3.0  = 4.5w +
(0.1*6) * 2.5  = 1.5w +
(0.5*12)* 1.25 = 7.5w
               =13.5w

However using method 2 (excluding the SLA battery):
0.3*5  = 1.5w +
0.1*6  = 0.6w +
0.5*12 = 6.0w
       = 8.1w
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  • \$\begingroup\$ if you are using switching converters, then you add the powers, with a 10/20% extra allowance for efficiency. If you are using linear regulators, then you add the currents, and expect the regulators to get hot. 7812 is linear, 62180 is switching, so you'll need to do both. If you just add currents, you'll get a safe figure. \$\endgroup\$
    – Neil_UK
    Commented Feb 7, 2020 at 11:12
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    \$\begingroup\$ Using a "current limiting resistor" to reduce 6V to 5V is probably not the best solution. A linear low drop out regulator or a buck converter would probably be better. \$\endgroup\$
    – JRE
    Commented Feb 7, 2020 at 12:04

1 Answer 1

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device drawing 500mA @12v
The device drawing 500mA @ 12V uses a linear regulator. For a linear current regulator applies: the current drawn at the output is also drawn at the input of the regulator. So, it will draw 500 mA (plus a small additional current for the regulator itself) from the 15V PSU.
Note the difference in input power and output power is dissipated in the regulator: (15V-12V) * 500 mA = 1.5W. This requires using a heatsink for the MC7812.

device drawing 300mA @5v
The device drawing 300 mA @ 5V draws its current directly from the 6V supply through the is current limiting resistor. So, again, output current is input current, and 0.3W is dissipated in the resistor.
So, current drawn from the 6V supply is 300 mA.

device drawing 100mA @6v
The 6V is generated by a step down converter 62180 fed from the 15V PSU.
Note to take in account the 300 mA from the 5V as well. So, the 6V supply needs to provide 400 mA.
From the datasheet of the 62180 you can estimate from Figure 11 that the efficiency is about 90%. So, 6V * 400 mA = 15V * I * 90%. It will draw 178 mA from the 15V PSU.

In total, 500 mA + 178 mA + trickle charge current is drawn from the 15 V PSU, so, somewhat more than 10W.

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  • \$\begingroup\$ So total drawn from the 62180 = 400mA + 178mA, and total drawn from the PSU= 578mA+500ma = 1.078A + trickle charge current? \$\endgroup\$
    – Sam Youé
    Commented Feb 7, 2020 at 11:54
  • \$\begingroup\$ @SamYoué No. Please carefully reread my answer. the 6V supply needs to provide 400 mA This is the 100mA from the 6V plus the 300 mA from the 5V device \$\endgroup\$
    – Huisman
    Commented Feb 7, 2020 at 11:59
  • \$\begingroup\$ @SamYoué The input current from a switched mode converter does not equal the output current. For such converters applies: power out = power in * efficiency. Therefore the 400 mA drawn at 6V corresponds with 178 mA at 15V. 500 mA (from the 7812) + 178 mA * (from the 62180)*+ trickle charge current is drawn from the 15 V PSU \$\endgroup\$
    – Huisman
    Commented Feb 7, 2020 at 12:01
  • \$\begingroup\$ THats what I was missing! So if I'm now drawing 5.4A @6v from the 62180: 6V * 5400 = 15V * I * 0.9. So current drawn from the PSU by the 62180 = 2.4A? Then add the other 12v devices and the trickle charge current? \$\endgroup\$
    – Sam Youé
    Commented Feb 7, 2020 at 12:12
  • \$\begingroup\$ @SamYoué Yes. Your conclusion and equation are correct, but … check Figure 11 again … efficiency improves to 94% (no big change, but sometimes a different load may change efficiency with 10%.) \$\endgroup\$
    – Huisman
    Commented Feb 7, 2020 at 12:31

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