I happen to have (2) very good, very compact transformers. Each transformer has a single secondary 24VAC output. Each transformer is rated to 900mA.
If I were to connect BOTH transformers in parallel to wall power (120VAC), then I was to run each transformer's outputs to a PCB via a rectifying diode (1N4007) on each output....
Could I sum the voltage to get ~48VDC to a circuit?
I would like to run this type of PS setup for a Ludwig clone but I am not sure about the power theory.
Suggestions??
It seems to me that the secondaries would need to be wired in series - not sure if that's what you meant by "summed".
Here's an example of the primaries in parallel and the secondaries in series:
(http://i1160.photobucket.com/albums/q485/jdansti/A25C426A-DA7D-45E8-A5C8-79BD4EB4859C-4454-0000042E652BA650.jpg)
I'd still wait for one of the brainiacs to respond before using my example above. :)
Edit: I just realized that my example has a shared core. I don't know if it makes any difference if the cores are separate. Hopefully an EE chime in.
Basically, here is what I want to do keeping in mind that BOTH transformers are identical:
Tie both transformer input HIs to the 120VAC input HI.
Tie both transformer input LOs to the 120VAC input LO.
Tie both transformer output LOs to the PCB GND.
Run #1 transformer 24VAC output, via a rectifier 4007 diode to the PCB power rail.
Run #2 transformer 24VAC output, via a rectifier 4007 diode to the PCB power rail.
So essentially, the now 24VDC outputs from both transformers are summed at the PCB = ~48VDC
Essentially, EHX used this same format to power their mains powered pedals however, they did it with center-tapped transformers. I am just using the same format with 2 seperate yet identical transformers.
Would this work or would I need to ties the secondaries in series?
Maybe I'm missing something, but if you rectify both secondaries, you have two 24VDC supplies. If you tie both both Output LOs to PCB ground and both output HIs to the PCB power rail, when you measure the voltage between the power rail and ground, you'll see 24VDC because the two outputs are in parallel. You'll have more current available, but the voltage will be 24VDC.
Is this not like looking at two batteries in parallel vs series? Two 9V batteries in parallel will still measure 9V. Two 9V batteries in series will measure 18V.
Quote from: Jdansti on October 13, 2012, 08:47:32 PM
Maybe I'm missing something, but if you rectify both secondaries, you have two 24VDC supplies. If you tie both both Output LOs to PCB ground and both output HIs to the PCB power rail, when you measure the voltage between the power rail and ground, you'll see 24VDC because the two outputs are in parallel. You'll have more current available, but the voltage will be 24VDC.
Is this not like looking at two batteries in parallel vs series? Two 9V batteries in parallel will still measure 9V. Two 9V batteries in series will measure 18V.
By gosh... I do believe you could be CORRECT!!! :icon_redface:
Now I kinda feel a little bit stupid :-\
Quote from: Govmnt_Lacky on October 13, 2012, 08:50:17 PM
Quote from: Jdansti on October 13, 2012, 08:47:32 PM
Maybe I'm missing something, but if you rectify both secondaries, you have two 24VDC supplies. If you tie both both Output LOs to PCB ground and both output HIs to the PCB power rail, when you measure the voltage between the power rail and ground, you'll see 24VDC because the two outputs are in parallel. You'll have more current available, but the voltage will be 24VDC.
Is this not like looking at two batteries in parallel vs series? Two 9V batteries in parallel will still measure 9V. Two 9V batteries in series will measure 18V.
By gosh... I do believe you could be CORRECT!!! :icon_redface:
Now I kinda feel a little bit stupid :-\
>By gosh... I do believe you could be CORRECT!!! I know-kinda scary. I think I hurt my brain cell.
Now the question is, what happens when you take two transformers and tie their secondaries together in series? Happy 48VDC, or mutual annihilation?
AC voltages add in series (with one caveat, a bit further on) like DC voltages do.
It is a Bad Idea in general to try to parallel voltage sources unless they are designed to be paralleled. Current sources in parallel are good.
The caveat is that AC voltages have phase. In the case of AC power line voltages, the voltages from a transformer are either in series adding or opposing. If you hook both primaries in parallel to 120Vac, and then connect one end of each winding of the two secondaries together, you will either get the sum of the two voltages or the difference, depending on which way the windings are phased. If you get the sum, flipping the connection of ONE of the secondaries will give you the difference.
Neither of these connections are necessarily dangerous to the transformers as long as the "output ends" connect to a load of less than the maximum current of the smallest-current-rated one.
Let's pretend your transformers are not identical, just rated for 24Vac/900ma. If you measure them at no load, they will probably measure more than their full-load voltage. Let's say that one is 28Vac and one is 30Vac. If you hook the two secondaries in series, you will get either 58Vac or 2Vac. If you get 58Vac, you can flip the ends of one of the secondaries and get 2Vac on the free ends. If you get 2Vac, you can flip the ends of one secondary and get 58Vac.
If you parallel the two windings, one wants to make 28Vac and the other wants to make 30Vac. They fight it out, and only the resistance of the copper wire secondaries limits the current that the 2Vac difference makes. This often burns out one or the other transformer, because the currents generally go way higher than the maximum design current.
Series = Good. Parallel = Bad (unless the transformers were specifically designed for this).
In the pretend case, you'd get 48Vac at 900ma from the two in series if you connected them up the right way.
@RG- Thanks. Is it best to 1) connect the secondaries in series and then rectify to DC, 2) rectify first and then connect the two rectified DC outputs in series, or 3) either way is is fine.
There are more options than that.
The possibilities include both half-wave rectification and full wave rectification. In general it's cheaper to series the windings first, then do the rectification, as this uses half the diodes as individually rectifying, then connecting. But both ways work.
I personally would put them in series, then run both free ends into a full wave bridge rectifier. This gives you the same smoothing of the output DC with half the filter capacitance needed for half wave rectification. The capacitance is generally both bigger and more expensive than the diodes.
Thanks R.G. ! :)
You have another possibility here: use a fullwave bridge rectifier on each output then connect the rectified outputs in series. This gives you a 48 VDC output and a 24 VDC output or you could use it as a ±24 VDC supply. Just another option/
Yes - you can either half-wave or full wave rectify either both windings separately or both in series. That's one of the possibilities I thought I was listing, but I guess I wasn't clear.
You could also use the middle connection of two windings in series as a "center tap" and use one bridge rectifier and still get +/- DC out of it, saving a bridge rectifier, although this is not going to work all that well with non-identical windings.
However, you don't get 24Vdc out of rectifiying 24Vac. A capacitor-input filter gives you nearly the peak of the AC waveform, not the RMS value. If you had 24VAC RMS (RMS is how AC voltages usually get specified) then the peak is 24*1.414 = 33.9Vpeak. Since transformers are rated for voltage at the maximum current, the no-load voltage is higher.
For small transformers, it's often 15-20% higher. So "24vac @900ma" is probably about 28.8Vac at low loads, and that gives you a peak of 40.7Vdc into a capacitor filter at no load. This sags down from 40.7 to 33.9 - or less~! - under load from winding resistance, and also from the ripple voltage on the filter caps as they run down between charging pulses. You'll also lose one (for half-wave) or two (for full-wave) diode drops for the diodes as they conduct.
So "24Vac @900ma" could give you 32Vdc to 40Vdc at 500ma to 0ma loading, and two of them stacked could do 64-80V depending on load.
One could use two three terminal regulators to get +/- regulated voltages in a couple of different ways.
Again, go read "Power Supplies Basics" at geofex.
OK...
So I read the replies. I read the literature (as much as I could understand) however,
I am still wondering if I will get the required voltage to my Ludwig build ???
If I run the secondaries in series to the FWB1 rectifier on the Control board, will I get a good DC voltage level out of the rectifier? Or, will it be too low to operate the Ludwig correctly?
I appreciate all of your help thus far. ;D
Oh - THAT'S what you're trying to do.
Secondaries in series will provide plenty of voltage and current. It will provide so much voltage that you may have voltage breakdown problems. You'll have to change all the capacitors which get the un-regulated voltage at the filter to 100V rated capacitors instead of ... was it 63? ... the ones in the parts list, because you could easily get more than 63V and pop a cap.
It also puts a bigger heating load on Q6. The heat generated by Q6 is the current through it times the voltage across it. Going to 75-80V raw DC before Q6 will probably triple the heating. You'll possibly need a heat sink, maybe a bigger device. I haven't done the calculations.
If I were you, I'd do the following:
- Measure the no-load AC voltage output on each transformer. Let's say one is 28Vac and one is 29Vac.
- Pick the higher one and try it all by itself. 29Vac times 1.414 is 41Vpeak, and you'll get 38-39V out of that at the filter cap. This ... might... be enough to run the circuit by itself.
- When you hook it up to the PCB, measure the voltage before and after Q6. If the voltage after it is OK, and the voltage before it is a couple of volts higher, you're golden.
-If that doesn't work, try the other one. The quirks of how much overvoltage they are versus loading and wire resistance might make that one OK.
- I'd go find a different transformer rather than ripping all the caps out and changing to higher voltage ones, which will be bigger.
Quote from: R.G. on October 14, 2012, 05:58:08 PM
Oh - THAT'S what you're trying to do.
Secondaries in series will provide plenty of voltage and current. It will provide so much voltage that you may have voltage breakdown problems. You'll have to change all the capacitors which get the un-regulated voltage at the filter to 100V rated capacitors instead of ... was it 63? ... the ones in the parts list, because you could easily get more than 63V and pop a cap.
It also puts a bigger heating load on Q6. The heat generated by Q6 is the current through it times the voltage across it. Going to 75-80V raw DC before Q6 will probably triple the heating. You'll possibly need a heat sink, maybe a bigger device. I haven't done the calculations.
If I were you, I'd do the following:
- Measure the no-load AC voltage output on each transformer. Let's say one is 28Vac and one is 29Vac.
- Pick the higher one and try it all by itself. 29Vac times 1.414 is 41Vpeak, and you'll get 38-39V out of that at the filter cap. This ... might... be enough to run the circuit by itself.
- When you hook it up to the PCB, measure the voltage before and after Q6. If the voltage after it is OK, and the voltage before it is a couple of volts higher, you're golden.
-If that doesn't work, try the other one. The quirks of how much overvoltage they are versus loading and wire resistance might make that one OK.
- I'd go find a different transformer rather than ripping all the caps out and changing to higher voltage ones, which will be bigger.
It must be coincidence...
I just finished putting the PS caps on the control board of my Ludwig and all of the 47uF caps are 100V rated ;D
In fact, all of the caps on the boards are rated 50V or better.
I'll try to get the unloaded AC output from both transformers and repost tomorrow.
Thanks for all of the help! ;)
BTW- I don't know if you've seen the following, but RG has gone through this exercise before for the Ludwig:
Quote from: R.G. on June 10, 2011, 10:42:34 PM
On the power supply front, here's the recipe for powering this thing from a 21Vac to 26Vac **AC output** wall wart:
Ingredients:
1 - 21.0 to 26V AC output wall wart. At least 200ma rated
1 - 1N4004 through 1N4007 diode
1 - 220uF 50V minimum, 63V preferred, 100V is fine, to taste; electro cap
1 - AC input jack to match the wart
Prep and install the input jack in the selected box. Connect the 1N400x diode with anode/arrow end to pin 1 of the console board, this being the contact that used to go the the transformer center tap in the original. Connect the cathode/bar of this diode to either contact 2 or 3, or both if you like. Connect the (+) lead of the 220uF cap to this same position, leaving the (-) end free for the moment. Finally, connect one incoming AC lead from the wall wart jack to the contact 1 (where the diode anode is) and the other transformer lead to the free (-) end of the 220uF capacitor.
The added capacitor/diode and transformer connection form a voltage doubler with the pre-existing diodes and C11, and pump C11 up to about 45-70Vdc, depending on the transformer, load, and phase of the moon. This should be sufficient to get 33-35Vdc out on contact 17 of the console board.
Test carefully for done-ness and serve with a hot single coil strat on a plate of 4x12 Marshalls.
Thanks for that John. I do remember reading that along the way however, I believe that is prior to the Version 2 board release which has room for AC rectification on the board.
Quote from: Govmnt_Lacky on October 14, 2012, 08:06:45 PM
Thanks for that John. I do remember reading that along the way however, I believe that is prior to the Version 2 board release which has room for AC rectification on the board.
Aha! Well in any case, it looks like you might be able to achieve the required voltage with just one of your transformers.
Measured BOTH transformers.
Wall power - Transformer- Meter
BOTH read 27.5VAC output.
So... am I good with just 1 -OR- would I need to run them in series?
Quote from: Govmnt_Lacky on October 15, 2012, 02:21:30 PM
Measured BOTH transformers.
Wall power - Transformer- Meter
BOTH read 27.5VAC output.
So... am I good with just 1 -OR- would I need to run them in series?
It's really, really close. I make it 37.096Vdc before the regulator (Q6) to make 35Vdc. That's certainly not impossible. I'd say go for it with one on a test basis. You may be OK. If not, you can add the other trannie or make a voltage doubler.
Would using a TL783 be a viable option? It's variable from 1.25V to 125V and rated at 700mA. I don't mean to steer the thread in a different direction, but thought I'd just mention it to at least learn more about the pros and cons.
http://www.ti.com/lit/ds/symlink/tl783.pdf
Quote from: R.G. on October 15, 2012, 04:56:30 PM
It's really, really close. I make it 37.096Vdc before the regulator (Q6) to make 35Vdc. That's certainly not impossible. I'd say go for it with one on a test basis. You may be OK. If not, you can add the other trannie or make a voltage doubler.
Thanks RG and John.... I really appreciate all the help.
So, if I decide to try this with just one transformer.... Should I still use FWB1 and run the AC through it? I am assuming so.
Also, RG... you elude to Q6 as a regulator. I thought that Q6 was an NPN transistor in the Ludwig ???
Finally, when I do try the single transformer... where would I check for the defining voltage to ensure I am getting what I need?
Quote from: Govmnt_Lacky on October 15, 2012, 07:19:52 PM
So, if I decide to try this with just one transformer.... Should I still use FWB1 and run the AC through it? I am assuming so.
Yes. Put in the link to use the bridge rectifier as a full wave bridge rectifier, and connect the transformer secondary leads to the two "AC" pads on the PCB.
QuoteAlso, RG... you elude to Q6 as a regulator. I thought that Q6 was an NPN transistor in the Ludwig ???
Q6 is an NPN transistors that is used as a voltage regulator, albeit a crude one.
QuoteFinally, when I do try the single transformer... where would I check for the defining voltage to ensure I am getting what I need?
Check the voltages to ground on all three leads of Q6. The collector should be 36V or more, the base 35.7 or more, and the emitter 35+/- half a volt.
Thanks again RG ;)
I must admit that I am quite a ways away from actually testing my Ludwig as I am still awaiting parts to show up :-\
Once I get the build up finished and try out these transformers, I will post back with results.
BTW..... PM coming your way RG ;)
OK....
I mocked up the PS section of the Ludwig on breadboard using all of the called-for components (TIP112 for Q6) and my transformer.
I ended up getting 34.25VDC at Q6 Emitter :'( A bit low.
Is there anything I can do to bump it up a bit? Perhaps some adjustments with R26 or R23. Will 34.25V with no load suffice in the circuit?
Thanks for the assistance ;)
Quote from: Govmnt_Lacky on October 23, 2012, 01:59:04 PM
OK....
I mocked up the PS section of the Ludwig on breadboard using all of the called-for components (TIP112 for Q6) and my transformer.
I ended up getting 34.25VDC at Q6 Emitter :'( A bit low.
Is there anything I can do to bump it up a bit? Perhaps some adjustments with R26 or R23. Will 34.25V with no load suffice in the circuit?
Thanks for the assistance ;)
As for whether 34V will work or making some changes to get a little more voltage, I'll leave that to R.G.
If you need a DC/DC booster, the UC2577-ADJ might work, but it's not cheap at around $18.
http://www.ti.com/lit/ds/slus216a/slus216a.pdf
From the data sheet: "The UC2577-ADJ device provides all the active functions necessary to implement step-up (boost), flyback, and forward converter switching regulators. Requiring only a few components, these simple regulators efficiently provide up to 60V as a step-up regulator, and even higher voltages as a flyback or forward converter regulator."
"The UC2577-ADJ features a wide input voltage range of 3.0V to 40V and an adjustable output voltage. An on-chip 3.0A NPN switch is included with undervoltage lockout, thermal protection circuitry, and current limiting, as well as soft start mode operation to reduce current during startup. Other features include a 52kHz fixed fre- quency on-chip oscillator with no external components and current mode control for better line and load regulation."
http://www.mouser.com/Search/m_ProductDetail.aspx?Texas-Instruments/UC2577T-ADJ/&qs=/ha2pyFaduggwaU7%252bB8sFQLVDZIYB1n2MLatPrsmyLw=
Fixed some not-so-obvious wiring issues on the breadboard :icon_redface: And re-measured (this time without the transformer getting toasty :icon_eek:)
Q6-
Base: 35.75V
Collector: 37.25V
Emitter: 35.70V
It was explained to me that in an "ideal" situation, you want the Collector at 3-6V higher than Emitter. This is obviously lacking (only 1.5V higher)
So the question now is... Can I run with this setup (with or without possible PS resistor value adjustments) -OR- do I need to look at running 2 transformers in series to bump up the voltages to get a larger seperation between Q6 Collector and Emitter voltages ???
Hmmmmm.... help for the clueless please ::)
Any chance you have a wall wart around that puts out 16vac to 18vac?
That would be about right to try a doubler on.
Quote from: R.G. on October 24, 2012, 01:14:28 PM
Any chance you have a wall wart around that puts out 16vac to 18vac?
That would be about right to try a doubler on.
Dont have any of those :icon_cry:
I would really like to get these transformers into the Ludwig. I have 2 just begging to be used. If it is a matter of voltage being insufficient, I could always try running the 2 in series.
In the end, if I cannot get this to work within reason I will probably just go ahead and purchase the recommended wall wart but I would like to try Plan A first. ;D
Thanks for all your help with this!!
The only thing I have against two 24Vac transformers in series is that they're likely to cause about 5W of heating in Q6. If you go that way, you'll need something similar to this:
http://www.mouser.com/ProductDetail/Aavid-Thermalloy/507302B00000G/?qs=%2fha2pyFaduguOHMd%2fQeWiVI5mWG%252bm0yKQkNwV4yhhPywG40xOa5m3g%3d%3d (http://www.mouser.com/ProductDetail/Aavid-Thermalloy/507302B00000G/?qs=%2fha2pyFaduguOHMd%2fQeWiVI5mWG%252bm0yKQkNwV4yhhPywG40xOa5m3g%3d%3d)