# Using asymmetric secondaries of power transformer with bridge rectifier for bipolar supply

I have a power transformer that has one secondary rated at 15v 1A, and another 15v 500mA. I would like to use it to create a bipolar supply.

If I use two separate rectifier bridges, one for each secondary, I will get bipolar with different current load capabilities.

What happens if I join one lead of each winding, and use it as a "center tap"? See schematic. I think this would be ok if the two secondaries had the same current capability. But what if they are different? Obviously, for this to work (vs catching fire), the correct winding leads would have to be connected, that is, the low wire of one connected to the high wire of the other.

I am thinking that this could mean that the result ia a bipolar supply capable of 750mA, with the DC voltage being 15*1.414-Vrectifierdrop for each of positive and negative terminals, with respect to COM.

Do you think that would work?

• No, you will be limited to 500 mA. Commented Mar 24, 2017 at 22:05
• @Mark Colan .Your scheme does have fewer diode drops than the orthodox double bridge .But if you place an even load on the +/_ 15 the transformer currents will share .But your transformers are not equal so you would like a 2:1 share which is not going to happen .If the load is unbalanced then each transformer is half waving which is not good .The double bridge will of course give 500mA .Your proposed scheme will not be much better ,the comment from winny is close . Commented Mar 24, 2017 at 22:17
• With your schematic you get one diode drop on the plus, and one on the minus. Current capabilities are 1A for the plus, and 500mA for the minus. Commented Mar 25, 2017 at 0:24
• I see plenty of designs in which a single bridge is used with a center-tapped transformer. It does not seem unorthodox. Separate bridges for the asymmetric transformer would mean 1A and 500mA for the separate poles. Commented Mar 25, 2017 at 18:34
• If I am limited to 500mA on both poles, OR if one is 1A and the other 500mA, could you explain why? Commented Mar 25, 2017 at 18:37

During 1/2 cycle, the diode of the top(+) raw supply is charged from 1Amp secondary, meaning it has larger wire. During the next 1/2 cycle, the diode of the top(+) supply is charged from 0.5 Amp secondary.

I'd expect the 60Hz ripple to be a bit stronger, because that 1Amp secondary alternatively charges the Top raw supply, and then 8.33milliSecond later is charging the Bottom raw supply.

• That's what I thought would happen, too. AND the reverse for the (-) diodes. So I think you are saying my idea would work? Would you expect 750mA to be available for each pole? As for "ripple a bit stronger"... compared to what? If I used a 30vct transformer, wouldn't the ripple be the same? Does using an asymmetric transformer make a difference to ripple, for example when there is a load drawing 750mA from each pole? Commented Mar 25, 2017 at 18:29
• This does not answer the key aspect of my question - whether 750mA would be available on each pole without overworking the transformer. Commented Mar 25, 2017 at 18:36
• I think the transformer will output slightly MORE than 1amp from the 1amp winding, and slightly LESS than 0.5 amp from the 0.5amp winding; the 1amp winding has to carry more than its share of the current, because of 60Hz ripple. If both windings contributed equally, you'd only see 120Hz ripple. Commented Mar 26, 2017 at 3:01
• Transformers are blobs of metal and insulating (fishpaper, in older such) sheets. Be sure to bolt the mounting feet to a large piece of chassis, so the heat can exit. Commented Mar 26, 2017 at 23:30
• Thanks for your comment on heat-sinking the transformer with a chassis. It is an important point that I had not thought of. Commented Mar 27, 2017 at 10:13