WEBENCH® Tools/UCC28950: UCC28950 Isolation Power transformer

Hi Raj,

Thanks for contacting us.We are looking for the right contact to address the query.We shall get back to you soon on this.

Regards,

Umayal

Hi Raj

I've just seen this post and I'll take a look at it tomorrow. The TIDA-00951 gives a 14V output so at a minimum I think you would need to change the turns ratio. Have you used the Excel calculator for this device (it's available as a download on the product page) - the calculator will tell you what the turns ratio should be and what the magnetizing inductance should be.

Regards

Colin

Hi Colin,

TIDA-00951 uses an isolation transformer with a variable voltage of 36V - 60VDC from 400VDC. 

What could be the 14V you are referring to?.

Also I prefer to use the Part which is already available in the market instead of going for custom design to reduce the Design cost and time.

Appreciated if alternate part is suggested with higher KVA ratings with the same voltage specs.

Please share your comments.

Regards,

Raj.

Hi Colin,

Any update?

Regards,

Raj

Hello Raj

My apologies for the slow response -

I had a look at the transformer spec. The turns ratio is 3:1 which means that you would have to use pins 19 to 12 as the primary to get 6:1. But this ratio if from the full primary (pins 19 to 12) to the full secondary (pins 1 to 8) and this in turn means that you would have to use a full wave rectifier on the secondary instead of the centre tapped secondary which we assume in the Excel calculator.

Most designers use the centre tapped secondary because it reduces the number of rectifiers needed but it is possible to use a single secondary with a bridge rectifier. The primary side gate signals from the UCC28951 (or UCC28950) will be the same in either case. The secondary side SR control signals are also the same except that each signal must drive a low side and high side FET rather than a single low side FET. This makes the gate drive circuitry a little more complex but other than that, it's not a problem.

A bigger problem is the DCR of the transformer. 7mOhm on each of the secondary windings (pins 1 to 6 and 6 to 10) for a total of 14mOhms, given a current of 42A (estimated from Excel calculator) the DC copper losses will be 25W. The primary windings (pins 19 to 16 and 16 to 13) have a DCR of 24mOhms each for a total of 48mOhms, given a current of 9.4A the DC copper losses will be 4.2W. This gives a total of 29.2W. In fact it will be significantly higher because of the temperature co-efficient of copper (about 40% increase from 20C to 100C) and AC losses due to skin effect and proximity losses.

You won't be able to use this transformer and I'd suggest that you talk with Wurth or another supplier to see if they offer a suitable replacement.

Regards

Colin