3
\$\begingroup\$

I am wanting to create an amplified speaker system with a subwoofer.

I am fairly new to electronics and this is complicated for me so please be as descriptive as possible and patient with me. The internet has been my guide, so some of my knowledge may be incorrect. Any additional information required I am happy to provide.

I am understanding most of the amplifier circuitry and am following this tutorial on creating it for the correct components.

Edit: Now using D class thanks to user110971 recommendation

I was wanting to use this subwoofer that takes to a max power input of 100W, as stated on the document.

My issue is finding/building a DC voltage regulator that will suit this speaker using a standard Australian 240V power socket. It would be best if someone could direct me towards a pre-made regulator if possible as I don't want to breach any government regulations.

Part of my problem is that I am unsure of how to obtain the correct voltage/ampage from the watts required. (FYI I know about the W=I*V formula, just cant see a way of solving my issue with it as I only have W)

Any links to tutorials providing an overview of building regulators or constructing an entire amplifier including regulator would be much appreciated. Even giving me directions to general electronics books that I can obtain would be helpful.

Thanks,

BFB

Edit:

Many thanks to user110971, Justme and Swedgin for thier help on my issue.

A quick recap of the solution for anyone wondering the same (I wish more people would do this XD)

As outlined by user110971 I needed to use a D class amp for such high wattage as 'the transistor would melt'.

Again, user110971 provided feedback on using the SMPS (Switching Mode Power Supply) as a voltage regulator.

Justme clarified my misunderstanding of the speakers power draw (Its 50W)

Swedgin provided me with useful links I recommend checking as well as alterations to the circuit to improve its performance.

Thanks all for the quick and helpful support!!!

BFB

\$\endgroup\$
8
  • \$\begingroup\$ Your real issue is that if you use a class A amplifier, your transistor will melt. \$\endgroup\$
    – user110971
    Commented Dec 9, 2019 at 10:44
  • 1
    \$\begingroup\$ so what would you recommend? \$\endgroup\$
    – Bigfoot Blondy
    Commented Dec 9, 2019 at 10:46
  • \$\begingroup\$ class D is the usual choice. At least class AB. However 100 W is a lot. Class AB has efficiency of around 50 - 60%. This is still a lot of power you are turning into heat. \$\endgroup\$
    – user110971
    Commented Dec 9, 2019 at 10:58
  • \$\begingroup\$ Shopping or buying questions. Read it. Questions to avoid asking. Read it. \$\endgroup\$
    – Andy aka
    Commented Dec 9, 2019 at 11:05
  • \$\begingroup\$ Thanks for the quick reply @user110971, I will take a look at class D. I will still need a regulator though. I can understand why it would release a lot of heat, but how do voltage regulators such as mains USB chargers and computer chargers not heat to this extent? is it due to the power draw? \$\endgroup\$
    – Bigfoot Blondy
    Commented Dec 9, 2019 at 11:12

1 Answer 1

Reset to default
0
\$\begingroup\$

For starters, I'd buy a premade module based on the TPA3116 chip. This gives you a subwoofer channel and a left and right channel. A good reason to buy one premade is that it's difficult to design the filters with feedback in the class D amp. Also it's cheap.

You can add your own bass/treble circuit in front of the module. Also, you could replace the potmeters (which you should) with smd resistors and add a volume circuit before the module if needed. So that the class D board has the largest possible gain. I noticed that the potmeters on these modules are of below average quality and can introduce noise in the signal.

As for the powersupply. Something like this will work, maybe look for one with a bit more amps if you want to be safe. With this powersupply you can also supply your own circuits with a simple regulator since your input voltage to your circuit will be 24V.

5A should be enough for only driving the subwoofer. If you use the 2 main channels, I'd buy one with at least 240W (24VDC 10A).

As for the calculation of Justme in the comments:

P = V*I = V^2 / R = R * I^2 (by Ohm's law).

P is fixed (P=100W) and R is fixed (R=4/omega), with this you can calculate both V and I. While keeping the ratio V \ I = R = 4

\$\endgroup\$
6
  • \$\begingroup\$ Thankyou very much @Swedgin!!! You just helped me heaps as to what to look for by sending me both links. I may very possibly use both of the devices you gave if i cannot find suitable alternatives. The tips you gave on the removal/move of the potmeters was great too. \$\endgroup\$
    – Bigfoot Blondy
    Commented Dec 9, 2019 at 12:23
  • \$\begingroup\$ @BigfootBlondy Glad I could help. Added one more PSU as 5A will be tight. \$\endgroup\$
    – Swedgin
    Commented Dec 9, 2019 at 12:35
  • \$\begingroup\$ I am adding two other regular speakers, I just need to add their wattage requirements to the total yes? They will probably use less power than the Woofer. I have almost already blown my budget lol. I am pushing the original question a bit, sorry, but the total speaker resistance is used to calculate the amps required using I=V/R yes? I just don't want to build a $200-300 system that blows up ;D \$\endgroup\$
    – Bigfoot Blondy
    Commented Dec 9, 2019 at 12:49
  • \$\begingroup\$ @BigfootBlondy Yes, but you also need to include the formula for the power. Ohm's law only give the ratio between voltage and current. \$\endgroup\$
    – Swedgin
    Commented Dec 9, 2019 at 13:01
  • 1
    \$\begingroup\$ @BigfootBlondy Also note that these are theoretical values and will differ in the real world. This is due to that the woofer isn't really a resistor but also has reactive components as the woofer is actually a coil with inductance. It also depends on the frequency of the signal. Just a FYI, if you decide to measure it out once everything is complete and you see different values. \$\endgroup\$
    – Swedgin
    Commented Dec 9, 2019 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.