Probably a dumb question on Ohm’s law and selecting a transformer. - Page 1
Forum section: electronics / beginners

author: Dan123456
author_rank: Regular Contributor
timestamp: December 30, 2023, 10:43:15 am
content: I was working out what parts I need to build a 25W audio amp and got to the transformer.Using Ohm’s law, 25W at 25V = 1 Amp.However, as I actually need to convert it to +25V and -25V, how do I factor this into the equation?From what amazing people here have said to me in the past to help me understand working with negitive voltages better, I think the answer would be to consider the voltage as 50V (the difference between -25 and +25) so 50V at 25W = 0.5A?Alternatively, I could be completely misunderstanding and I actually need a 2A transformer? Or maybe I’m just confusing myself and the answer is still just 1A?Is my understanding correct in thinking the first option is right? Or if I have messed up, can someone point out / point me to an explanation of where I have gone wrong?Thanks so much in advanceP.s. for the actual project I would probably use is a 24V transformer simply as they are more readily available but I chose to stick with 25V for this exercise for the nice whole numbers

author: Andy Chee
author_rank: Super Contributor
timestamp: December 30, 2023, 10:59:41 am
content: Bear in mind that 25W is your load, you actually need more than 25W feeding on the input. How much extra will depend on your amplifier design.Anyway take a look at this for a dual rail supply:https://www.sound-au.com/articles/rectifiers.htm#s6You can customise the values to your own situation, but the circuit topology remains the same.

author: Dan123456
author_rank: Regular Contributor
timestamp: December 30, 2023, 11:58:20 am
content: Quote from: Andy Chee on December 30, 2023, 10:59:41 amThat’s a very good point and really appreciate the linkUp to this point I’ve just been building in “fudge factors” into my designs (I.e. if I calculate I need a 2.7W resistor, I’ll just pop a 5W in to be safe etc.)However, when I got to the transformer, I realised I didn’t know how to calculate it to start with so wasn’t even sure how much fudge factor to addIf it were a cheaper part, I’d probably just buy one that I know is over rated (say like a 5A transformer) but I’m not sure if that is an absolute overkill 10X fudge or a only a 2.5X fudge etc. in this situation so figured I’d check

author: Andy Chee
author_rank: Super Contributor
timestamp: December 30, 2023, 12:22:38 pm
content: Probably worth reading this link as well:https://www.sound-au.com/power-supplies.htm

author: CaptDon
author_rank: Super Contributor
timestamp: December 30, 2023, 03:23:34 pm
content: If you could find a 48VCT 1 amp transformer it could supply 50 watts all day without overheating and this could provide both the 25wrms to your load as well as power wasted as heat in the output devices. It would allow for some headroom also. One downside is that using that transformer with center tap grounded and a full wave bridge would give you something like 36 volts plus and minus at idle so your design and capacitors must be able to survive that voltage. Your filter capacitors should probably be 50vdc rated. Assuming you could get 30 volts peak or 21.5vrms across 8 ohms your power output would be 57wrms. These voltages are absolutely common for 50wrms audio amplifiers!! Be aware that with music having about a 6 to 10 Db peak to average difference in the real world this would be a conservatively rated 10 watt amplifier with 6Db or a bit more of headroom before clipping would occur. With good efficient speakers that is still loud (and clean!!). If you have room, the main filter caps should be around 10,000uf for each 36vdc rail. Possibly 2 X 4700uf (4 total) on each rail would be good enough. This prevents voltage sag during heavy bass lines. There are some hybrid modules who's part numbers start with STK- that will do 50wrms with very few external components. Look up the NTE1360 or ECG1360. I used them in a lot of projects. There are some rated at 80w or more but I don't have the part numbers. Kawai keyboard amplifiers used them.

author: bdunham7
author_rank: Super Contributor
timestamp: December 30, 2023, 03:54:32 pm
content: Quote from: Dan123456 on December 30, 2023, 10:43:15 amIt's all a bit more complicated than it appears at first glance.First off, if you were just planning on having an output of a 1kHz sinewave, the required current from the positive and negative rails would be equal to each other and also each equal to enough current to produce half the output power on average. But if your amplifier is DC coupled or capable of amplifying very low frequencies, the "on average" part may be less important than "right now" or "peak". Soeach halfwould be required to be able to supply atleast TWICEthe rated power on a peak basis, and that is for a sine wave. Actual audio would be much more. Your filter caps should be large enough to do most of this in the normal audio band.Second, when you use rectifiers and filter capacitors, the current through the transformer is not a nice sine but rather a series of short pulses that start just before the crest of the mains power sine wave and last an amount of time that varies with load. This means that the RMS value of the AC current through the transformer windings is higher than the average value of the DC current sent to the amplifier circuit. How much more depends on load and design values, but fortunately the RMS/DC ratio will go down with increased load so you can just do the calculations (or tests) for the full-load scenario and you'll be OK.Determining the RMS/DC ratio and thus sizing your transformer is difficult to calculate, at least for me. I recently made a linear DC supply for some LED lights and used a full-wave voltage doubler, added primary resistance and some other things to reduce noise and extend the conduction period of the input pulses. I ended up with an effective RMS/DC ratio of about 2:1, so I'm presuming your application will be higher than that. So depending on the exact requirements of your audio amplifier, I think you'd need a 48V center-tapped transformer of 1.5 to 2A capacity, or 75-100VA.

author: CaptDon
author_rank: Super Contributor
timestamp: December 30, 2023, 04:19:54 pm
content: It is true the current flow into the bridge occurs only when the a.c. wave is of higher voltage than what is currently the d.c. voltage on the filter capacitors. The peak current flow tends to be large but is only present for perhaps 50% of each half of the a.c. cycle. A 50 v.a. transformer would work sufficiently well and with musical loads would not tend to overheat at all. As always, conservative ratings are best where possible and practical so the 1.5 to 2 amp rated secondary as suggested is worthwhile if space / money permit. Some allowance should be added in for the preamp / predriver stages which may require lower voltages and will waste a lot of heat in series pass regulation or require a much smaller separate transformer of perhaps 16 to 20 v.c.t. Best wishes in your project!! I have used dozens of the NTE / ECG 1360 hybrid building blocks and they are simple to use. Heat sink them well. Amplifiers with high rail voltages and conservative headroom tend to run hot, even more so when you shift toward AB class bias.

author: Zero999
author_rank: Super Contributor
timestamp: December 30, 2023, 11:06:28 pm
content: Presumably you're talking about a class AB amplifier, not class D.A few points.The amplifier is not 100% efficient. A lot of power, in the form of voltage drop is lost in the driver transistors. It's more helpful to think of current output, rather than power.Music is not a pure sine wave. The amplifier will never output the full power all the time. It depends on the music, but generally it's something like 20%, if I remember rightly.The peak output voltage of a 25V transformer is √2 of the RMS minus a small loss in the rectifier giving 25*√2=35V.To get a bipolar +/- power supply, a voltage doubler is required, which needs huge capacitors. It would be much better to use a centre tapped winding.It might be possible to convert your circuit to single supply. Please post a schematic.

author: MathWizard
author_rank: Super Contributor
timestamp: December 31, 2023, 12:12:21 am
content: A good thing to do with a transformer and DMM is measure the output with different load resistors. I have a board with ~5k, 1k, 100, 22, 10 ohm power resistors, all at least 5W. And a "200W" 10 ohm variable resistor. So with those, I can easily get a plot of a low voltage transformers load line.I have a few, nice transformers from stereo's. I look forward to using them in more projects.

author: EPAIII
author_rank: Super Contributor
timestamp: December 31, 2023, 02:04:55 pm
content: OK, so you don't actually say it, but we are talking about a DC power supply that outputs + and - 25 VDC at 1 Amp. on each side. That is, as you state, 50 VDC with a common or ground in the middle. Here is my procedure for designing DC power supplies. And by the way, I am going to destroy your "nice whole numbers" real fast. But I will stick with your +/- 25 VDC anyway. You must work backwards from the desired output.First there is a question. Is this going to be a regulated supply or just filtered? You didn't say. If it is going to be regulated, then what are you going to use for regulator circuits? Linear or switching? And, most important of all, HOW MUCH HEAD ROOM do those regulators need to operate? If you are going to use just filtered supplies, then there is 0 Volts head room. But a typical linear regulator will use between 2 and 3 Volts to operate. So I will add 2.5 Volts to both rails for the regulators. You then need + and - 27.5 VDC or a total of 55 VDC from the filtered supply that the regulators use. And the regulator circuits will consume some current so lets say your current is now up to 1.05 Amps. You can use the correct amount when you design the regulator circuits, if any.Next there is the rectifier. What are you going to use. Standard diodes will drop around 0.7V. Shockley diodes around 0.35V and a modern, FET based rectifier about 0.1V. So double the value for the type you choose and add it to the 55 VDC. Lets assume a Shockley bridge and add 0.7V for 55.7 VDC.Next your filter capacitors will charge and discharge. There will be some level of ripple on top of the DC Voltage that they store. The equation that is generally used and that I have used many times is:C = I / (2 x f x Vpp)where I = load currentf = input frequency of ACVpp = the minimum ripple allowedAnd remember that for a full wave rectifier f will be twice the line frequency. The filter capacitors will charge up almost to the maximum value that the rectifier outputs and then discharge down to the level given by that maximum minus the ripple. That low value, before the capacitors start to charge up again must be greater than the 55.7 VDC we need for the regulators input after considering their operating head room.You can turn the equation around to calculate the ripple Voltage from the size of your filter capacitors or just pick a ripple Voltage and calculate the needed capacitors. In my experience, a 2 Vp-p ripple is a practical value for a supply like this but that is on both + and - rails so we must add add 4 VDC. And now we have 59.7 VDC.OK, now we get to cut it down some. Up to now we have been talking about peak to peak DC levels. But our transformer is an AC device. And it is specified in RMS values, not peak to peak. So we now multiply that 59.7 VDC by 1/1.414 = 0.707 and we get 42.2 Vrms.So you need a transformer that has an output of 42.2 VAC (RMS) at about 1.05 Amps. One amp is probably close enough. You aren't going to find a commercially available transformer with that exact specification. So the 48V, 1A transformer that was mentioned above will serve quite nicely. And the secondary should have a center tap for that middle, ground point.You can play around with the actual numbers as you like, but for purchasing purposes you are going to wind up with that 48V, 1A, center tapped secondary transformer. It's a bit more than what you really need, but as CaptDon said, it will run all day outputting your Voltage and current with no problems.

author: bdunham7
author_rank: Super Contributor
timestamp: December 31, 2023, 04:57:26 pm
content: Quote from: EPAIII on December 31, 2023, 02:04:55 pmYou've totally ignored the conduction angle, RMS/DC ratio and effective output impedance of the transformer. Your model appears to assume that a 1A rated transformer can put out 1Aaveragecurrent even when the actual current flow only happens in peaks during a small percentage of the time and that it can put out that peak current without any appreciable voltage drop.You might almost get away with this in some audio amplifier situations because the power supply will rarely if ever be loaded anywhere near to 1A on both rails and if it is, it will likely be for a short period of time. But if you just want a power supply that can supply +/- 25V @ 1A, what you're proposing isn't going to work very well.