Currently working on a Gyraf G7 mic build and need some help with PSU transformers. The schematic is pictured at the bottom of the post.
I have bought 2 transformers.
115/230V - 9V dual output
115/230V - 15V dual output
Wiring diagram is attached at the bottom - it's the exact same for both transformers.
I'm in the UK so am working with 230V mains.
I use brown/blue on the primary to give me 230V input.
From the schematic I see that I have to connect the output of the first transformer to the output (now the input) of the second transformer (reversed) and also to a bridge rectifier. Can I simply use one the 2 seperate outputs of the first transformer for this (i.e. red/black to transformer 2 and yellow orange to bridge rectifier) or do I need to use only one output (red/black), splice off it to the bridge rectifier and leave output 2 (yellow/orange) untouched?
Also, do I just mirror the outputs from transformer 1 to connect to transformer 2 (reversed)? And then use what would have been the "primaries" (now outputs) - (brown/blue) to connect to the second bridge rectifier?
Sorry if anything I've said has been confusing, I just prefer to put forward my own logic and then have someone explain to me where I've gone wrong!
I have bought 2 transformers.
115/230V - 9V dual output
115/230V - 15V dual output
Wiring diagram is attached at the bottom - it's the exact same for both transformers.
I'm in the UK so am working with 230V mains.
I use brown/blue on the primary to give me 230V input.
From the schematic I see that I have to connect the output of the first transformer to the output (now the input) of the second transformer (reversed) and also to a bridge rectifier. Can I simply use one the 2 seperate outputs of the first transformer for this (i.e. red/black to transformer 2 and yellow orange to bridge rectifier) or do I need to use only one output (red/black), splice off it to the bridge rectifier and leave output 2 (yellow/orange) untouched?
Also, do I just mirror the outputs from transformer 1 to connect to transformer 2 (reversed)? And then use what would have been the "primaries" (now outputs) - (brown/blue) to connect to the second bridge rectifier?
Sorry if anything I've said has been confusing, I just prefer to put forward my own logic and then have someone explain to me where I've gone wrong!
You can do all u have written without any problem. Just see that the diodes on the high voltage are rated to 135x2 volts.
Gajanan Phadte
Gajanan Phadte
🙂
That took me a long time to understand what you are saying.
I don't think it matters if the current requirements are low and within the VA rating of the single winding but it seems more elegant to use both and use one winding for the heater supply and the other for feeding the "reversed" transformer.
That took me a long time to understand what you are saying.
I don't think it matters if the current requirements are low and within the VA rating of the single winding but it seems more elegant to use both and use one winding for the heater supply and the other for feeding the "reversed" transformer.
I see a few things I think need comment.
1.) both transformers are quite low VA. These will have high regulation. This leads to high voltage on the output when unloaded or very lightly loaded. If the regulation of these are 10% then the output voltage of the 9Vac isolation transformer will rise to about 9.9Vac, when fed with 230Vac. In the UK the supply voltage is 240Vac, but can vary from 216Vac to 254Vac.
Feeding 9.9Vac into the 15:230Vac will give higher than fully loaded voltage. But:
2.) a 230:15 has turns ratio of 230:16.5, not 230:15 which does not take account of the regulation. Another way of putting that is that the 15Vac transformer will have an output of 16.5Vac when unloaded, due to that 10% regulation. By turning this 15Vac transformer back to front you have a 16.5:230 (by turns ratio) being fed with 9.9Vac. The output when unloaded will be 9.9/16.5 * 230 ~= 138Vac. Now correct that for a 240Vac mains supply by applying a 240/230 correction factor. The unloaded voltage will be ~ 144Vac.
3.) Using one output winding of the 9Vac, will reduce the VA available by about 50%. Using one winding of the 15Vac, will reduce the VA by about 50%. Combining both these effects leaves the VA @ about 50%. Now we have a 16.5Vac transformer being fed with 9.9Vac. This reduces the VA to a lower value. Expect the final VA to be about 50% * 9.9/16.5 = 30%.
If your transformers are 100VA then the effective VA available at the final 144Vac (unloaded) to be ~ 30VA, when using one half of the LV windings.
I would seriously recommend that you do not use only half the windings. Work with series connected 18Vac (9+9) and 30Vac (15+15) rather parallel the LV windings.
4.) the heater supply is quite a high current. This would suit using the 9Vac windings in parallel, rather than the series I suggested in 3.) You have to choose depending on the transformer VA.
5.) will a single 317 cope with the heat and current feeding the heaters?
1.) both transformers are quite low VA. These will have high regulation. This leads to high voltage on the output when unloaded or very lightly loaded. If the regulation of these are 10% then the output voltage of the 9Vac isolation transformer will rise to about 9.9Vac, when fed with 230Vac. In the UK the supply voltage is 240Vac, but can vary from 216Vac to 254Vac.
Feeding 9.9Vac into the 15:230Vac will give higher than fully loaded voltage. But:
2.) a 230:15 has turns ratio of 230:16.5, not 230:15 which does not take account of the regulation. Another way of putting that is that the 15Vac transformer will have an output of 16.5Vac when unloaded, due to that 10% regulation. By turning this 15Vac transformer back to front you have a 16.5:230 (by turns ratio) being fed with 9.9Vac. The output when unloaded will be 9.9/16.5 * 230 ~= 138Vac. Now correct that for a 240Vac mains supply by applying a 240/230 correction factor. The unloaded voltage will be ~ 144Vac.
3.) Using one output winding of the 9Vac, will reduce the VA available by about 50%. Using one winding of the 15Vac, will reduce the VA by about 50%. Combining both these effects leaves the VA @ about 50%. Now we have a 16.5Vac transformer being fed with 9.9Vac. This reduces the VA to a lower value. Expect the final VA to be about 50% * 9.9/16.5 = 30%.
If your transformers are 100VA then the effective VA available at the final 144Vac (unloaded) to be ~ 30VA, when using one half of the LV windings.
I would seriously recommend that you do not use only half the windings. Work with series connected 18Vac (9+9) and 30Vac (15+15) rather parallel the LV windings.
4.) the heater supply is quite a high current. This would suit using the 9Vac windings in parallel, rather than the series I suggested in 3.) You have to choose depending on the transformer VA.
5.) will a single 317 cope with the heat and current feeding the heaters?
Thanks for the help guys! Both my transformers are 5VA by the way.
From your comments, I think this is how it should be wired:
230V AC -> 9V transformer brown/blue (via fuse and switch). Form a single output via parallel connection (red/yellow and black/orange).
Use this to feed same connections of 15V transformer (reversed). Also from here add wires to connect to bridge rectifier.
Use brown/blue to feed second bridge rectifier.
This will mean that both transformers are used fully as you say Andrew.
From your comments, I think this is how it should be wired:
230V AC -> 9V transformer brown/blue (via fuse and switch). Form a single output via parallel connection (red/yellow and black/orange).
Use this to feed same connections of 15V transformer (reversed). Also from here add wires to connect to bridge rectifier.
Use brown/blue to feed second bridge rectifier.
This will mean that both transformers are used fully as you say Andrew.
5VA is very tiny. Is it enough for your current requirements ? I can see the HV feed is via a 10K so no issue on that one. It's the heater supply one that seems limited and as Andrew eluded to, heater supplys are usually fairly high current unless it's for some micro miniature valve 🙂
I'm afraid that I don't personally know enough to fully understand if 5VA is enough, I have seen a couple of builds using 15VA, but the design page shows both are 5VA. Here's some info from the page, maybe this will be of some assistance:
This is a very simple power supply. You only need a couple of mA at 160V for the anode, and about 200mA at 6.3V for the heater. The PSU is based on standard transformers to make it easier for you get hold on them. First there's a mains-to-9V about 5VA transformer, that is rectified and regulated to 6.3V DC by a LM317T regulator. This is sent to pin 6(+) and 7(0) on the XLR, to power the heater in the microphone.
The 9VAC from the first transformer is also taken to the secondary("the wrong way around") of a 220:15V transformer. Now we have about 135V AC on the primary of this second transformer. We rectify this and remove ripple with a 220uF/200V (or maybe larger)electrolytic. We take this DC thru another ripple filter consisting of a 10K resistor and another 220uF/200V (or maybe larger)electrolytic. Now we have our 160VDC for the anode (HT) voltage of our EF86 tube, and run this to pin 2 on the XLR, the ground goes to pin 1.
Now we need the polarizing voltages: 0V, 80V and 160V. The 0 and the 160 are easy, we have them already from the HT supply. We do the 80V simply by dividing the 160V by two 100K resistors, that will also work as "bleeders" ensuring that when you power off the mic, it should be safe to open after 10 to 15 minutes.
BUT ALWAYS CHECK YOUR VOLTAGES BEFORE TOUCHING ANYWAY!!! ( Burnt out bleeders often results in burned out tecnicians. And we dont want that!)
OK - the three different polarization voltages are taken to a 3-pole switch that selects the pattern you want, and goes to XLR pin 5
This is a very simple power supply. You only need a couple of mA at 160V for the anode, and about 200mA at 6.3V for the heater. The PSU is based on standard transformers to make it easier for you get hold on them. First there's a mains-to-9V about 5VA transformer, that is rectified and regulated to 6.3V DC by a LM317T regulator. This is sent to pin 6(+) and 7(0) on the XLR, to power the heater in the microphone.
The 9VAC from the first transformer is also taken to the secondary("the wrong way around") of a 220:15V transformer. Now we have about 135V AC on the primary of this second transformer. We rectify this and remove ripple with a 220uF/200V (or maybe larger)electrolytic. We take this DC thru another ripple filter consisting of a 10K resistor and another 220uF/200V (or maybe larger)electrolytic. Now we have our 160VDC for the anode (HT) voltage of our EF86 tube, and run this to pin 2 on the XLR, the ground goes to pin 1.
Now we need the polarizing voltages: 0V, 80V and 160V. The 0 and the 160 are easy, we have them already from the HT supply. We do the 80V simply by dividing the 160V by two 100K resistors, that will also work as "bleeders" ensuring that when you power off the mic, it should be safe to open after 10 to 15 minutes.
BUT ALWAYS CHECK YOUR VOLTAGES BEFORE TOUCHING ANYWAY!!! ( Burnt out bleeders often results in burned out tecnicians. And we dont want that!)
OK - the three different polarization voltages are taken to a 3-pole switch that selects the pattern you want, and goes to XLR pin 5
Well VA is volt-amps and supposed to be a way of expressing real power. A capacitor for example would dissipate no power as heat in itself but still draws current. So the VA notation was a way of expressing transformer ratings that would take that into account.
So your 6.3 volt heater which is a simple resistive load would be only "1.26VA" if connected across a 6.3 volt AC winding. Your tranny is 5VA and has two 9 volt windings.
So thats approx 0.28 amps per winding. (0.28*9) * 2 giving 5VA
I would go with Andrews suggestion and parallel the two windings to maximise current availability obtaining a "single" 9 volt 0.56 amp transformer winding.
(When we rectify and smooth AC to obtain DC the load on the transformer changes. It only supplies current when the voltage on the winding exceeds the voltage on the reservoir cap. This causes high peak currents in the transformer winding.)
So your 6.3 volt heater which is a simple resistive load would be only "1.26VA" if connected across a 6.3 volt AC winding. Your tranny is 5VA and has two 9 volt windings.
So thats approx 0.28 amps per winding. (0.28*9) * 2 giving 5VA
I would go with Andrews suggestion and parallel the two windings to maximise current availability obtaining a "single" 9 volt 0.56 amp transformer winding.
(When we rectify and smooth AC to obtain DC the load on the transformer changes. It only supplies current when the voltage on the winding exceeds the voltage on the reservoir cap. This causes high peak currents in the transformer winding.)
Great, thanks Mooly. So my 5VA units are sufficient for this project? I had planned to buy 15VA until I saw that they mentioned 5VA which would allow for a smaller PSU box.
I'll definitely use a parallel connection for the output of the 9V tranny and the same for connecting into the 15V tranny as Andrew suggested.
I'll definitely use a parallel connection for the output of the 9V tranny and the same for connecting into the 15V tranny as Andrew suggested.
Yes a 5VA is fine for such a small current.
One tip when wiring it all up for the first time. Use a bulb tester... a mains filament bulb in series with the mains to limit current. Normally we use 60 or 100watt but for this you could go smaller. Whatever you have 🙂
One tip when wiring it all up for the first time. Use a bulb tester... a mains filament bulb in series with the mains to limit current. Normally we use 60 or 100watt but for this you could go smaller. Whatever you have 🙂
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