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Audio power supply using DC-DC converter

  1. Jul 17, 2013 #1
    So I have been using some small audio amplifier ics in the tda series because they are good and easy to use but I have never had to power one with more than 24v. I am wanting to look into using a chip that requires about 35-40v supply but I can't find a transformer that is past 24v and am not looking to spend a lot of money. Recently I heard about dc-dc converters and am wondering if what I am thinking they do is what they actually do. From what I am getting they can take a low voltage at a lot of amps and convert it to adjustable high voltage with less amps. Is this correct?

    So I want to try useing a tda chip that I would normally use a small transformer going to rectifier and using the center tab as ground, But instead sending the dc from the regulator through a dc-dc converter so I have enough voltage to power the chip and still use the center tab as ground.
    I have NO experience with dc-dc converters so I don't know what you can and can't do with them and that's why I am asking if this would work.

    I was looking at a module like this: http://cgi.ebay.com/ws/eBayISAPI.dl...akeTrack=true&ssPageName=VIP:watchlink:top:en
  2. jcsd
  3. Jul 18, 2013 #2
    DC/DC converters are typically pretty limited in power, and since you are looking for higher V - I am Assuming that you want more power? You should be able to find a 48-0-48 V Transformer - have you tried the electrical suppliers like Allied or DigiKey. Of course you could also use 2 of the 24V type transformers - they are abundant - for Audio I am assuming you are using as 24-0-24v ??
    Last edited by a moderator: Jul 21, 2013
  4. Jul 18, 2013 #3
    How much current do you need?
    When you developed the 24V, did you use a full wave diode bridge or a center tapped transformer with a diode on each lead?
    What is the RMS output voltage of your transformer?
  5. Jul 18, 2013 #4


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    DC to DC convertors are used in car audio amplifiers and they are hardly limited in power.
  6. Jul 20, 2013 #5


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    A DC to DC boost converter will do the job.
    A typical selection; http://dx.com/s/DC+BOOST+Converter
    For more power see this; http://dx.com/p/03100275-300w-boost-converter-for-car-audio-black-216871 [Broken]
    Last edited by a moderator: May 6, 2017
  7. Jul 21, 2013 #6
    I used a 24-0-24 and put both 24s on a simple rectifier circuit and put 2 caps between the + & - with the center tab in the center then this ran strait to my audio circuit. For this bigger project I am planning to do the exact same thing except before it hits the amp circuit is goes strait into the dc-dc converter turning 24v into about 35 to power my larger circuit. but having done this can I still use the center tab as ground? idk why I wouldn't be able to but I want to make sure.
  8. Jul 21, 2013 #7
    And I have no idea what power these will draw, I want to use 4 tda7293s, the transformer I have chosen is a 10a 24-0-24 transformer and when it goes through my dc-dc converter that I have chose I will have a max of 8a. hopefully that is enough
  9. Jul 21, 2013 #8
    Instead of using a DC-DC converter, why not use a full wave bride rectifier instead of the rectifiers you are using now? That will double your output voltage at very little additional cost. You will need to verify the power supply capacitors can handle the higher voltage.
  10. Jul 21, 2013 #9


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    These days a transformer at mains frequency, (50 or 60Hz), is very heavy and expensive. It will also need big power supply storage capacitors. You would be better using a switch-mode power supply that will be much more efficient, smaller, lighter and less expensive.

    Get a copy of a data sheet for the TDA7293 such as this; http://www.st.com/web/en/resource/technical/document/datasheet/CD00001887.pdf

    There is an internal 6.5A limit which would suggest a 26amp total is possible with 4 modules.

    The power amplifier and the signal amplifier supplies are separate.

    These amplifier modules are voltage amplifiers. You will need to decide;
    1. What speaker system impedance to use. R ohms.
    2. The maximum power the speakers will handle. W watts.

    From that you can select an appropriate power amplifier supply.
    The voltage needed will be V = Squareroot(W*R). The current will be I = V / R

    As an example, for an R=4 ohm speaker at W=100 watt, it will require a supply voltage of
    V = Sqrt(100*4) = 20 volts. The current in the speaker will be I = 20 / 4 = 5 amps.

    Another example, for an R=8 ohm speaker at W=100 watt, it will require a supply voltage of
    V = Sqrt(100*8) = 28.3 volts. The current in the speaker will be I = 20 / 8 = 3.53 amps.

    Are you using these four modules as two bridge amplifiers or as four separate channels? That will decide if you need a single V power supply or a split +/- V supply.
    These voltages and currents are a good guide to the supply requirements but are subject to interpretation because the signals are sine waves not DC currents.

    From the data sheet, the signal negative power supply is shown as connected to the heatsink tab. If you electrically isolate the tab from the heatsink then no problems. If all amplifier modules share their signal supply then they can all share the heatsink without isolation, but then the heatsink will need to be isolated from the chassis and environment.
  11. Jul 21, 2013 #10
    ok this is giving me some stuf to think about! I was going to have 4 separate outputs I only really want maybe 70watts, heck I am actually fine with 40! into 8ohms. What is a full wave bride rectifier and switch-mode power supply?

    "heck I am actually fine with 40"
    I know this kind of changes things, If I am fine with 40 should I look for a different chip? Any suggestions?
    Last edited: Jul 21, 2013
  12. Jul 21, 2013 #11


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    A full wave bridge rectifier is four diodes connected to convert the AC current from a power transformer into DC for the storage capacitor.

    A switch-mode power supply rectifies the mains AC and stores it in a small high voltage capacitor. It then switches that high voltage at say 80kHz through a very small ferrite isolation transformer that also reduces the voltage of the outputs. By storing high voltage it takes advantage of capacitor energy E = 0.5 * C * V^2, the V squared shrinks the capacitor size significantly. By switching at 1000 times the mains frequency it uses the transformer more often, so it need not be so big. The efficiency is typically 80%.
  13. Jul 21, 2013 #12
    How is it different from the normal 4 diode bridge rectifier?
  14. Jul 21, 2013 #13
    What would be the max current on one?
  15. Jul 22, 2013 #14


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    There is no difference.
    Rectifiers come as half wave or full wave depending on whether they conduct on one phase or both.

    One what?
  16. Jul 22, 2013 #15

    You want 4 outputs each at 40! into 8 ohms. Are 40! and 40" the same as 40V? This would give you a peak current draw of 4 x 40 / 8 or 20 amps. Or will you use the modular circuit shown in Figure 11 of this datasheet such that all 4 outputs feed the same 8 ohms requiring only 5 amps?


    With the transformer you have, 20 amps will be impossible, 8 amps will likely be impossible but 5 amps manageable. Let's look at the 8 amp peak current solution with a bridge rectifier.

    First, if you want 8 amps peak current you'll need 8 amps x 8 ohms = 64 volts, not 40. This means the minimum ripple voltage cannot go below that. In order to keep the ripple that low you'll at least 50,000 uF of capacitance. As you reduce ripple, the time the capacitors have to charge is reduced which means the peak charging current increases, dramatically. The current through the rectifier diodes may go up to 200 amps. Realistically the transformer won't be able put out that much current so its voltage will droop resulting in less output voltage. Less peak output voltage means you have to reduce ripple even more and..., you get my point.

    The advantages of switch mode power supplies over regular power supplies are that they are smaller, lighter and more efficient than ordinary power supplies.

    The disadvantages are that they are significantly more expensive and have more noise on the output.
    Last edited: Jul 22, 2013
  17. Jul 22, 2013 #16


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  18. Jul 23, 2013 #17
    Ok thanks guys I am, still learning and I think like to jump into things a bit too fast sometimes, so I'll work my way up because I didn't know there was THIS much to power supplies. I appreciate the good info and will use it wisely.
    just a couple small questions though.

    ok- filter capacitors for audio where you have 2 (say, 6.8mf) caps with the ground in the middle, and even when just between - and +, why are there sometimes a very small cap such as .1uf in parallel as well?what is so special about 6800.1uf?
    and the filter caps are called decoupling caps right? or is that something else?

    ALSO I am making a very small power supply just for my bread board so I can stop having to replace AAs batteries . Most small voltage regulators are rated at like 1-1.5 amps, will adding a heat sink increase this max current?
    In other words is this just the recommended max current for an un-heat shrunk regulator?
  19. Jul 23, 2013 #18


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    A big electrolytic has a high series resistance and inductance. A small ceramic capacitor of 0u1F has very low resistance and inductance. So the two work well together with the big cap doing long term bulk storage and the ceramic providing capacitance to very fast transients. Linear regulators require a fast capacitor on the input and output to be stable. The 0u1F meets that requirement.

    The 1A voltage regulators, (LM78xx and 79xx), have a maximum current of 1A only when they are on a heatsink that keeps their temperature in spec. The heat they generate is proportional to their current and the voltage they drop. So keep their input voltage as low as possible without the regulator dropping out.

    As an example, the LM7805, a +5V linear regulator, needs about 7V input or more to regulate correctly. If you have a 12VAC transformer secondary, you will have a peak voltage of 12V * Root(2) = 17.0V. The silicon bridge rectifier will drop about 2.2V so the peak storage capacitor voltage will be 14.77V. We need 7V minimum so there is 7.77V free for ripple.
    C = Q / V, Q = I * T, therefore C = I * T / V. I = 1A, V = 7.77, at 60Hz T = 8.33msec (full wave)
    The minimum C = 1 * 8.33e-3 / 7.77 = 1073.uF.
    The average input voltage will be (7 + 14.77) / 2 = 10.9V. The output voltage will be 5V.
    The difference is (10.9 – 5) = 5.9V, so at 1A the regulator will dissipate 1A * 5.9V = 5.9watt.
    The load will only dissipate 5watt. Efficiency is below 50%
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