I'll be building a small class AB power amplifier and I've decided that it's output at full power will be +/-10Vpk (for each of the 2 channels), which means I need to be able to supply at least 2.5A (for 8 ohm loads). I was looking through Digi-Key for a 9V (12.7V after rectification) and was a bit confused by their ratings since they quote them in series and parallel (for example).
Since I want to have +/-12V supplies I'll connect the the two secondaries in a center tap configuration, but in this configuration should I choose a transformer able to deliver >2.5A in parallel or series?
In the example I gave I'm assuming I should use the parallel rating for choosing it, otherwise I'll have to go with this one which quotes 2.78A in series. The reason I'm confused is because I think the series rating would only be used if you're connecting the secondaries to get 18-0, while when you're doing 18-0-18 you're "in some way" (sorry for not being able to explain my thoughts clearly) putting them in parallel.
Since I want to have +/-12V supplies I'll connect the the two secondaries in a center tap configuration, but in this configuration should I choose a transformer able to deliver >2.5A in parallel or series?
In the example I gave I'm assuming I should use the parallel rating for choosing it, otherwise I'll have to go with this one which quotes 2.78A in series. The reason I'm confused is because I think the series rating would only be used if you're connecting the secondaries to get 18-0, while when you're doing 18-0-18 you're "in some way" (sorry for not being able to explain my thoughts clearly) putting them in parallel.
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If you are using 8 ohm loudspeakers they will draw just a bit under 8 watts each. Now as an AB amplifier at full power is only about 45% efficient that would require 16 watts for both or about a 35 watt supply. Now technically transformers are rated in VA which is not quite the same thing but close enough for audio purposes. So a transformer rated for 9 volts would need 4 amps and the same transformer at 18 volts would be 2 amps. Double that if you want to use 4 ohm loudspeakers. Note many would use larger transformers and in cheap gear rated in peak power a smaller transformer with bigger filter capacitors would be used.
Your assumption is mistaken. For the example transformer you'll see that it has dual primaries and dual secondaries.
The dual primaries allow for either 115v or 230v input.
The dual secondaries allow for several output configurations. Your interest is the series current rating.
Here's why: For a center-tapped configuration, one secondary's "low side" will connect to the other secondary's "high side." This becomes the center tap, being at half the voltage of that measured across both secondaries. The same current then can be seen to be flowing through each secondary in series.
A parallel connection would consist of each secondary high side connected and each low side connected. The same voltage appears across each secondary, with an equal amount of current through each secondary, ie twice the current output of the series connection.
The dual primaries allow for either 115v or 230v input.
The dual secondaries allow for several output configurations. Your interest is the series current rating.
Here's why: For a center-tapped configuration, one secondary's "low side" will connect to the other secondary's "high side." This becomes the center tap, being at half the voltage of that measured across both secondaries. The same current then can be seen to be flowing through each secondary in series.
A parallel connection would consist of each secondary high side connected and each low side connected. The same voltage appears across each secondary, with an equal amount of current through each secondary, ie twice the current output of the series connection.
You are right, 2.5 A is the peak current at full load. But fortunately, the average dc current, which is more appropriate for choosing a transformer, is much smaller, it is 2.5 A / Pi = 0.8 A.
For simplicity just assume a factor of 1.25 in AC to DC conversion. For example the transformer in the first link:
AC: 2 x 9 V, 1.39 A, 25 VA
DC: 2 x 11.25 V, 1.112 A, 25 W
Conclusion: The current is fine (1.1 A is clearly more than 0.8 A), but the voltage is short (10 V peak at +/- 11.25 V supply might get difficult).
Also note that the transformer can get overloaded when a 4 Ohm load is connected.
For simplicity just assume a factor of 1.25 in AC to DC conversion. For example the transformer in the first link:
AC: 2 x 9 V, 1.39 A, 25 VA
DC: 2 x 11.25 V, 1.112 A, 25 W
Conclusion: The current is fine (1.1 A is clearly more than 0.8 A), but the voltage is short (10 V peak at +/- 11.25 V supply might get difficult).
Also note that the transformer can get overloaded when a 4 Ohm load is connected.
The classic center tap configuration means the two secondaries are wired in series. Use the current specified for series connection.
My AC to DC conversion in post #5 is little too optimistic. After rethinking I now go with these figures:
VDC = 1.2 * VAC
IDC = 0.69 * IAC
Example:
AC: 2 x 9 V, 1.39 A, 25 VA
DC: 2 x 10.8 V, 0.96 A, 20.7 W
My AC to DC conversion in post #5 is little too optimistic. After rethinking I now go with these figures:
VDC = 1.2 * VAC
IDC = 0.69 * IAC
Example:
AC: 2 x 9 V, 1.39 A, 25 VA
DC: 2 x 10.8 V, 0.96 A, 20.7 W
Dissi,
you can't get 20.7W continuous dc from a 25VA transformer.
The maximum continuous DC current from a capacitor input filter is approximately half the AC current rating of the secondary.
That makes the ideal maximum continuous power output, including the losses through the rectifier approximately 70% of the VA rating.
When one subtracts the diode losses, this reduces the available power even lower.
Expect not more than 60% to 65% of the VA rating of a 9Vac 25VA transformer, i.e. ~15W of continuous DC.
That 15W will make the transformer run hot.
I regularly advise that for continuous duty, the DC output current should be ~50% of the maximum.
That reduces the long term, high reliability, power to around 8W continuous for that 25VA example.
you can't get 20.7W continuous dc from a 25VA transformer.
The maximum continuous DC current from a capacitor input filter is approximately half the AC current rating of the secondary.
That makes the ideal maximum continuous power output, including the losses through the rectifier approximately 70% of the VA rating.
When one subtracts the diode losses, this reduces the available power even lower.
Expect not more than 60% to 65% of the VA rating of a 9Vac 25VA transformer, i.e. ~15W of continuous DC.
That 15W will make the transformer run hot.
I regularly advise that for continuous duty, the DC output current should be ~50% of the maximum.
That reduces the long term, high reliability, power to around 8W continuous for that 25VA example.
For a ClassAB amplifier with a conventional output bias one can use a transformer VA rated from 1times to 2times the maximum total output power of the amplifiers, in domestic type listening. Parties are not ordinary music/domestic listening.
A 25VA transformer can power many/one channel/s that have a total maximum power of ~12.5W to 25W
Two 8W channels fits into that range, provided you do not adopt a high bias version.
A 25VA transformer can power many/one channel/s that have a total maximum power of ~12.5W to 25W
Two 8W channels fits into that range, provided you do not adopt a high bias version.
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There is often very little difference in price between one transformer and the next size up. In this instance the bigger the better. A 100VA transformer here would have a lower output impedance and would probably do a better job.
I have a question for people who know something about transformer current rating system.
I have a power transformer of a commercial amp. It has a 6,3Vac secondary, current rating is unknown to me but i would like to know how much it can be used for.
I made some measurements in order to figure out the current rating:
1) Unloaded the voltager accross the secondary reads 6,36Vac
2) Loaded with 11R resistor (~560ma) the output voltage dropped to -2% from unloaded state, to 6,23Vac
3) Loaded with 5,2R resistor (~1,15A) the output dropped to -4% from unloaded state, to 6,11V
4) The secondary DC resistance read 0,2 ohms with a multimeter (0,1R accuracy). Calculated resistance of the primary is approx. 0,215R based on the measurements.
What do you think would be the amperage rating approximately of such secondary winding?
I have a power transformer of a commercial amp. It has a 6,3Vac secondary, current rating is unknown to me but i would like to know how much it can be used for.
I made some measurements in order to figure out the current rating:
1) Unloaded the voltager accross the secondary reads 6,36Vac
2) Loaded with 11R resistor (~560ma) the output voltage dropped to -2% from unloaded state, to 6,23Vac
3) Loaded with 5,2R resistor (~1,15A) the output dropped to -4% from unloaded state, to 6,11V
4) The secondary DC resistance read 0,2 ohms with a multimeter (0,1R accuracy). Calculated resistance of the primary is approx. 0,215R based on the measurements.
What do you think would be the amperage rating approximately of such secondary winding?
Hi,
If its the only secondary the best thing to do is compare
weight and size to similar VA specified transformers.
VA/V will give you A.
Its impossible to tell from your information, the core
may saturate at any % voltage drop, typically ~ 10%.
The number is often specified as regulation, e.g 8%.
Typically the bigger the transformer, the lower it is.
rgds, sreten.
If its the only secondary the best thing to do is compare
weight and size to similar VA specified transformers.
VA/V will give you A.
Its impossible to tell from your information, the core
may saturate at any % voltage drop, typically ~ 10%.
The number is often specified as regulation, e.g 8%.
Typically the bigger the transformer, the lower it is.
rgds, sreten.
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If your transformer is an EI type, you can estimate VA from the core area.
Measure the core.
VA = core area squared, times 31, times Flux.
core area in square inches, Flux in Tesla, assume 0.7T as a starter.
Measure the core.
VA = core area squared, times 31, times Flux.
core area in square inches, Flux in Tesla, assume 0.7T as a starter.
Where does the winding wire's thickness/secondary winding's resistance come in? It has to affect the outcome of winding's current capacity.
The transformer has many, many secondary windings and is also potted. 🙂
The wire thickness determines the heat that requires dissipating and the voltage drop under load. It does not directly affect the VA.
I would say the trafo (El) is at least 300VA, but this is just a quess. It is big and heavy (~7-10kg), but the VA cannot be questimated from that since it has so many secondaries.
What do you think Andrew, is the 1,2A ok for the transformer based on the ~0,2R secondary resistance and ~4% voltage drop, which results in ~0,25 watts of dissipation in the secondary? I'm just wondering how could that ever harm the transformer.
The reason why I ask, is the secondary's load was originally 0,6A but I want to increase it to 1,2A 🙂. That's why I made the forementioned measurements at those given points. I was quite suprised because the secondary winding seemed quite overkill for the 0,6A current requirement (not very usual in commercial things, overkill) since the output voltage dropped so little when doubling the amperage. But some slight doubt remains, I don't want to smoke the trafo.
measure and get an estimate of the VA rating.
There is a good chance that the TOTAL VA of all the windings adds up to the VA rating of the core. Often the core is less than the secondary.
Once you have EACH winding wire diameter, you can assign an approximate current rating to each. From this, with each winding voltage, you get a rough idea of the total secondary winding ratings.
You cannot just double the current rating of a single winding. It will run excessively hot (4times as much heat) and likely to damage the insulation.
There is a good chance that the TOTAL VA of all the windings adds up to the VA rating of the core. Often the core is less than the secondary.
Once you have EACH winding wire diameter, you can assign an approximate current rating to each. From this, with each winding voltage, you get a rough idea of the total secondary winding ratings.
You cannot just double the current rating of a single winding. It will run excessively hot (4times as much heat) and likely to damage the insulation.
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It's a potted El-transformer, where I can't access the core since the steel cover/shield can't be removed. 🙂
the standard EI has a regular proportion for the lamination sizes relative to the outside width and height of the core.
In an EI with lossless stampings, the I parts fit into the spaces in the E legs.
The outside legs of the E are 1unit wide, the I parts are 1 unit wide, the middle of the E is 2units wide. That adds up to 6units. Measure the width and divide by 3. That gives you the width of the E core. Measure the stack thickness to get the other dimension of the core.
As a check, the I length is 6units and this is cut from TWO E section, so the depth of the E slot is 3units. Add on 1unit for the back of the E and another 1unit for the I and you have 5units for the height of the transformer. Measure it. It should be 5/6ths of the width.
In an EI with lossless stampings, the I parts fit into the spaces in the E legs.
The outside legs of the E are 1unit wide, the I parts are 1 unit wide, the middle of the E is 2units wide. That adds up to 6units. Measure the width and divide by 3. That gives you the width of the E core. Measure the stack thickness to get the other dimension of the core.
As a check, the I length is 6units and this is cut from TWO E section, so the depth of the E slot is 3units. Add on 1unit for the back of the E and another 1unit for the I and you have 5units for the height of the transformer. Measure it. It should be 5/6ths of the width.
Yes you can IF you don't exceed the curren rating of the winding 🙂. Who said commercial amps can't have overspec'ed transformer windings?
The thermal load of the winding wire increases from ~0,077W to ~0,31W (0,6A to 1,2A at 0,215R secondary resistance).
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