Transformer fuse size?   - Page 1
Forum section: electronics / beginners

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 08:36:18 am
content: Hello, could some help me out with the calculation for working out what fuse size I should use on a transformers primary side?Its a 9v-9v 6VA transformers with single 230V primary winding.

author: Rerouter
author_rank: Super Contributor
timestamp: January 05, 2014, 08:39:49 am
content: 6VA stands for the load wattage before the voltage considerably drops, based on that (6/230) you end up with a 30mA fuse (make it 40), if you do go that way, pick a slow blow, as i am going to guess you have a few hundred uF after your bridge rectifier, and this can cause a big starting surge.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 08:44:15 am
content: Thanks Rerouter,Thought it was VA divided by voltage, but 28mA seemed to small to me!Quote from: Rerouter on January 05, 2014, 08:39:49 amYep here is what I plan on powering with the adapter I'm making, but wondering if I need a higher VA transformer.....

author: kripton2035
author_rank: Super Contributor
timestamp: January 05, 2014, 08:54:29 am
content: yep seems really 10 times too small !and for audio usage you should always have a big margin,I would pick up a 100VA transformer for your usage.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 09:06:58 am
content: Quote from: kripton2035 on January 05, 2014, 08:54:29 amIs 100VA not too big?Going to be drawing well under 300mAWhats the maths fo working this out?

author: SeanB
author_rank: Super Contributor
timestamp: January 05, 2014, 09:42:39 am
content: 6VA is usable, but I would look for a 10VA unit instead, a slightly better voltage regulation and higher current rating. As you are going to be using the windings in series I would recommend a full wave bridge rectifier with the centre tap of the 9VAC windings connected to common. Better current draw from each half winding and lower ripple. Fuse would be a 63mA slow blow unit, likely around the lowest value slow blow you will find easily.No idea why there is a second voltage regulator built around U2, you would be better off just not using it, and have some 470uF 16V capacitors for power supply instead. CMRR of the output stage will not be worsened at all by not having this in, and you likely would halve the BOM cost.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 09:51:37 am
content: Quote from: SeanB on January 05, 2014, 09:42:39 amThank you, very much appreciated.Quote from: SeanB on January 05, 2014, 09:42:39 amHave a few of these units built as per the spec, so will differently play about with your recommendations once Iv got this power adapter built up.Regards the two regulators, I assumed one was dealing with positive, and the other negative?And. was wondering if someone could brake down the simplest way to calculate the needed Transformers VA size for projects like this?

author: SeanB
author_rank: Super Contributor
timestamp: January 05, 2014, 10:12:52 am
content: The circuit current consumption multiplied by the voltage at the AC input. Thus if in your example the current is 200mA at 20VAC then it is 4VA. However the VA rating is for a resistive load, so you need to multiply by 1.4 to get a rough estimation of 6VA. At 6VA the transformer will be fully loaded, and the losses will be at the highest, so using the next bigger of 10VA will reduce the copper losses ( thicker wire used) and the larger core will run cooler and with less iron losses as well.

author: woodchips
author_rank: Frequent Contributor
timestamp: January 05, 2014, 10:43:23 am
content: You will always need a slow blow fuse for any transformer due to the inrush magnetising current, nothing to do with the amount of smoothing caps on the secondary.

author: Rerouter
author_rank: Super Contributor
timestamp: January 05, 2014, 10:51:02 am
content: hmm, i dont know why this has not yet been pointed out, but with a 9V AC transformer you only end up with about 11.5V (times by 1.414 for DC, then subtract 2 diode drops) after the rectifier, then your regulators need to drop it another 2.5V as there minimum voltage drop, so unless you wanted to convert this thing to 8V rails, you dont have enough voltage to even work.

author: mariush
author_rank: Super Contributor
timestamp: January 05, 2014, 11:16:14 am
content: I was just going to mention it.With a 9v-9v transformer you'll have maybe peaks of 13v DC ignoring the diode drops (which will be about 0.5-1v if using bridge rectifier) , the 7812 needs 2v above output voltage to give you 12v so you'll only get about 11v at the output. Either get higher voltage, or use better linear regulators (really there are cheap regulators from the 1117 with 1.2v drop to tiny 100-200mA regulators that have 0.1-0.3v drop)Use 4 schottky diodes with around 0.3v drop at 200-500mA and use a slightly larger transformer, around 10-15va. 6va is very small, but may be enough. Use better linear regulators with smaller voltage drop.With a 10VA transformer I'd probably use a 100mA slow blow fuse.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 11:27:12 am
content: Sorry, could some brake down why id drop from 18v to 13v using the 9v-9v?The device is meant to be powered with a 12-20VAC 200mA+ adapter, that's all I'm trying to make here, as the amplifier is already built and I just want to be able to power and recharge the batteries.

author: Rerouter
author_rank: Super Contributor
timestamp: January 05, 2014, 11:34:11 am
content: Quote from: peps1 on January 05, 2014, 11:27:12 ambecause this design calls for a center tapped transformer, so your 9V AC becomes +- 12V (24V) with diode drops on the bridge rectifier, and this is insufficient for the +-12V (24V) regulators as they need their input voltage to be 2-2.5V higher than there output to work,there minimum 12VAC transformer gets rectified to about +-15.5V (31V) so it meets this criteria,

author: SeanB
author_rank: Super Contributor
timestamp: January 05, 2014, 11:46:36 am
content: You need a minimum voltage across C2, C3 of 15V so that the regulator can regulate. This means the transformer peak voltage at the AC waveform peak must be over 15V by the diode drop, thus you need 16V peak out of the transformer. Thus a minimum voltage of 12VAC is needed out of the transformer. Using the 2 secondaries in series will give that, giving 18VAC, but at the expense of increased ripple in the capacitors and increased peak current draw through the transformer secondary. Usable with the 6VA unit you have, though it will run warmer than if it was a lower DC resistance 10VA unit. It will work, but will work better with a 10VA unit with 2 12VAC secondaries. Then you will have lower losses all round.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 12:10:25 pm
content: Quote from: Rerouter on January 05, 2014, 11:34:11 amMy 9v AC becomes +-12V (24)? you lost me already!Quote from: SeanB on January 05, 2014, 11:46:36 amOk think I got all that, though its at odds with Rerouter, right?So....I need to just get a 9v+9v or 18v 12VA transistor to make a 18VAC adapter that will have a high enough voltage on the secondary to get regulated in the first place!QuoteCould you expand on this a little?

author: mariush
author_rank: Super Contributor
timestamp: January 05, 2014, 01:15:24 pm
content: You have an 18v AC transformer, so the maximum dc voltage will be 18x1.414 = ~25.5v but since you're using it in split mode 9v - 0 - 9v , you have - 12.7v - 0 - +12.7v .Then you have a diode on each side, which causes a drop of about 0.5-1v , so you're left with about -12v - 0 - +12vThat's peak DC, it's not constant. The capacitors come to smooth out that and make it as smooth as possible. The formula is something like this Capacitance = Current / 2 x AC Frequency x VrippleFor a 6VA transformer you probably have a maximum of about 100mA on each positive and negative side, so you have C = 0.1/ 2 x 60 Hz x Vripple , Vripple = 0.1 / 120 C = 0.1 / 120 x 0.00094 farads  = 0.88vSo at 0.1 amps of current through each side, you'll have a dc voltage after the capacitors between 11 and 12v. The 7812 and 7912 need 2 volts ABOVE the output voltage to output +12v, -12v respectively.. so they won't output 12v but rather around +/- 10v.

author: peps1
author_rank: Regular Contributor
timestamp: January 05, 2014, 01:25:24 pm
content: Thanks for making that understandable mariush,Just one question, so the transformer itself doesn't become less efficient in split mode, its just the fact you have a diode on both +12v AND -12v ?