Is my calculation correct?

I want to tranform 240Vac output of 5Vdc after voltage regulator.

The transformer I am interested in is VB 2,0/1/6: https://docs-emea.rs-online.com/webdocs/0f46/0900766b80f46e48.pdf

I decided to use NCP1117DT50G: https://uk.rs-online.com/web/p/products/5165871/

Now my calculations:

6 * 1.414 = 8.484 VacRMS No load factor 8.484 * 1.64 = 13.91 After rectifier 13.91 - 1.4 = 12.51 Vdc

Minimum voltage for 5v is 6.5V

This means if my circuit current is 300mA the power loss would be P = (12.51 - 6.5) * 0.3 = 1.8W.

Is this correct? It seems crazy how a 6V transformer can turn double with Vdc and have a high power loss into a lot of heat.

  • 1
    \$\begingroup\$ I highly recommend you reading this electro-tech-online.com/threads/… and this hammondmfg.com/pdf/5c007.pdf \$\endgroup\$
    – G36
    Aug 29, 2019 at 15:06
  • 2
    \$\begingroup\$ If you're drawing 300 mA, should you be applying the no-load voltage factor? The transformer is only 2.00 VA. Additionally, the transformer does have an efficiency of 52% so I'd expect some heat loss. \$\endgroup\$
    – user199402
    Aug 29, 2019 at 15:24

1 Answer 1


It was written: "6 * 1.414 = 8.484 VacRMS No load factor 8.484 * 1.64 = 13.91 After rectifier 13.91 - 1.4 = 12.51 Vdc"

Because it is a 'short-proof' transformer, it will be inefficient.

I interpreted the calculation this way and presumed to fix the apparent transposition of 1.414 and 1.64. I hope it makes sense:

6 * 1.64 No load factor = 9.84 VacRMS (no load) (means 6VAC at full load 2VA/333mA, and 9.84V at no load)

9.84 VacRMS * 1.414 = 13.91VAC peak (no load)

13.91 - 1.4 = 12.51VDC after rectifier and cap filter (no load).

So that is no-load. Now for 300mA load:

This is rough approximations because no curve is published for transformer: no load - full load = voltage drop 9.84VAC - 6VAC = 3.84VAC 0mA to 333mA = drop of 3.84VAC

volts drop per mA = 0.01153 (if linear)

0.01153V/mA * 300mA = 3.45945VAC drop from 9.84VAC

9.84V - 3.45945V - 6.380V For the 300mA load, VAC is 6.380

It will probably look like this: 6.380V * 1.414 = 9.0221V peak 9.02V - 1.2V bridge rectifier drop = 7.8221VDC peaks to charge filter. Filter charged to 7.8V LDO needs at least 6.2VDC at 800mA

So even with some mistakes, there should be enough voltage for your 1.2V dropout 5V regulator at 300mA.

If there is still too much doubt, suggest either ordering the transformer and just trying it on the workbench, or, contacting the manufacturer or distributor application engineer, sending your schematic over and asking them if it's OK.

The dissipation and very small size of the transformer also suggests using large copper tracks for each transformer lead to help carry away the heat.

  • \$\begingroup\$ Thank you so much for that detailed explanation. I followed your calculations and they make sense to me. Would you say that little heat will be lost in the voltage regulator? \$\endgroup\$
    – M2T156
    Sep 2, 2019 at 12:08
  • \$\begingroup\$ Just over a Watt.. \$\endgroup\$
    – user216912
    Sep 4, 2019 at 15:41
  • \$\begingroup\$ Thank you so it seems like a heatsink would not be necessary. Thank you so much for your help in my design process! \$\endgroup\$
    – M2T156
    Sep 5, 2019 at 7:34

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