What to do when your power source is higher than your power requirements - Page 1
Forum section: electronics / beginners

author: Explodey
author_rank: Contributor
timestamp: August 21, 2018, 10:19:15 pm
content: Hi folks -- super basic question here: I'm building a little project that just uses two of these little amplifier boardshttps://ebay.us/kPrc8I. Due to space limitations, I want to use a 9V battery, but the specs say that the maximum power requirement is 5.5V. In the past, I've used these eboot step down modules for this kind of stuffhttps://ebay.us/WBdBOR, but I'm trying to keep this as simple as possible and I'd rather not add another board if possible. So I'm wondering if it could be as simple as just putting in a resistor with the right value between the power and the boards. Would that work? I don't want to burn out my amp boards. Thanks!

author: rx8pilot
author_rank: Super Contributor
timestamp: August 21, 2018, 10:24:19 pm
content: You have a voltage that is higher than the requirement. The power is a different thing - voltage x current(amps).You likely need a buck converter or linear regulator which can take a DC power source at a certain voltage down to a lower voltage. Resistors will NOT work with an amplifier board at all.EDIT: That board definitely needs a stable-regulated power source. A switching DC-DC converter is the best option for battery power efficiency, linear regulators are best for low noise in amplifier applications.

author: Explodey
author_rank: Contributor
timestamp: August 21, 2018, 11:33:28 pm
content: I had a feeling that my question would betray a fundamental misunderstanding of the problem!I think I'll try to make space for AAA batteries.Thanks for your reply.

author: rx8pilot
author_rank: Super Contributor
timestamp: August 21, 2018, 11:47:20 pm
content: Quote from: Explodey on August 21, 2018, 11:33:28 pmI have not studied your target amplifier PCB, but in general, amplifiers need a regulated power source to meet specifications. Batteries are not regulated. It may not be much of a problem in your case, but worth mentioning.

author: Audioguru
author_rank: Super Contributor
timestamp: August 22, 2018, 12:20:00 am
content: Are your speakers 4 ohms or 8 ohms?Four alkaline batteries in series produce about 6.6V when new which is too high. Three batteries produce 5V when new.Three batteries produce 3V when ready for replacement but then the max output power is very low.The max undistorted output power with a 5V supply and 4 ohm speakers is 2.2W so if each channel is blasting 2.2W then the power from the battery is 4.8W. The max current is 1.2A. Three new AAA or AA alkaline batteries cannot produce 1.2A. An ordinary 9V alkaline battery cannot produce even 0.4A. A resistor in series will increase the voltage and blow out the amplifiers when the output power is low and it will blow out your hearing with severe distortion when the output power is high.

author: ArthurDent
author_rank: Super Contributor
timestamp: August 23, 2018, 02:49:42 pm
content: This is case of trying to use a circuit in a way it probably wasn't intended to be used. If you read the listing for the amp it says: "You are biding for one piece of PAM8403 Audio Amplifier Board.This board could be powered by USB Power supply. Operating voltage: 2.5V-5.5V Maximum output power: 3W*2(5V 4Ω)."If powered by a USB port it probably works the way an amp that costs $0.99 would be expected to work. Probably voice quality is passable but not great fidelity. Running it on 3 AAA batteries would probably be o.k. until the batteries discharge as well. Trying to use a 9 volt battery doesn't make any sense. If you want to use this amp battery powered using AAA batteries you best have a supply of spare batteries on hand. I guess the question is what exactly are you trying to do with this amp?

author: dazz
author_rank: Frequent Contributor
timestamp: August 24, 2018, 01:27:59 am
content: How about a good 18350 Li-Ion battery? It's smaller than a 9V battery, has more capacity and can cope with plenty of current

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 24, 2018, 11:59:53 am
content: Quote from: Explodey on August 21, 2018, 10:19:15 pmthis is already your simplest solution. may i suggest thesmaller and cheaper version. amplifier circuit or ic such as yourPAM8403IC is equivalent to dynamic load. when there is no sound it acts as high impedance load, when there is, it will become low impedance load, meaning by using a resistor to limit current, when there is no sound, 9V still exist at the IC pin and may damage digital circuit inside. if you are lucky and its not, during high load (low impedance), the supply pin can go down to 0V easily making your sound distorted very badly. so your best route is still what you have done before, ie using DC-DC buck converter. not to mention using resistor (or similarly linear regulator) will waste energy unnecessarily.

author: 6PTsocket
author_rank: Regular Contributor
timestamp: August 29, 2018, 11:35:46 pm
content: Quote from: Explodey on August 21, 2018, 10:19:15 pmNot usually. Anything where the current varies during operation would mean that the value of the resistor would have to keep changing to maintain the same voltage to your amps. The cheap and dirty is to put a zener in series with the power supply to drop the extra voltage. A string of diodes of appropriate current rating will drop about 0.7 volts each. You could use a zener as intended, with a resistor to build a simple regulator or use a simple three pin regulator like a 7805.The connections are 9v to the input, ground, and 5 volts out. The TO220 version handles more current and still more on a heat sink. If that isn't enough yo can go to to a 3 pin variable regulator. There are also switching regulator modules. The choice is yours.Sent from my SM-G900V using Tapatalk

author: IanB
author_rank: Super Contributor
timestamp: August 29, 2018, 11:43:45 pm
content: Quote from: 6PTsocket on August 29, 2018, 11:35:46 pmWhy do we care about this? Can you clean up and not leave litter behind?

author: rx8pilot
author_rank: Super Contributor
timestamp: August 29, 2018, 11:49:52 pm
content: Quote from: IanB on August 29, 2018, 11:43:45 pmsent from my Palm Pilot Palm III connected via RS232 to US Robotics 28.8k Dial up on Compuserve.Short and misplld from my mobile......

author: mariush
author_rank: Super Contributor
timestamp: August 29, 2018, 11:58:52 pm
content: The chip's maximum operating voltage is 5.5v with an absolute maximum of 6vIf you do decide to use a 9v battery, then think about how much energy is actually stored inside such a battery... typically you're looking at around 400-500 mAh ... and they're relatively weak at pumping out current, they're designed more for alarm clocks, radios etc.If you have two 1w speakers running them loudly and continuously, that means the amp board will consume about 2.2 watts (the datasheet says around 83% efficiency for the chip) so 2.2w / 9v = 0.25 A of current ... 9v batteries aren't really designed to give 0.25A continuously, the voltage will drop quite quickly.So even if you're using a dc-dc converter to convert the 9v to let's say 5.5v with maybe 90% efficiency, you're probably going to get around 600-700 mAh of battery life.Think about how much power you're gonna send to speakers, how loud will those things be... what speakers are you gonna use? If you're only gonna use 0.5w or 1w speakers, then you can live with smaller voltage, check datasheet :THD+N = 1%, f = 1KHz, RL = 4Ω: 0.85w @ 3.2v , 1.3w @ 3.6v 2.5w @ 5vSo if you're no planning on using more than 1w or so, then you could simply use 3 AAA or AA batteries and have more actual life ... 3 rechargeable batteries give you ~ 3 x 1.1v = 3.3v ... 3 x 1.25v = 3.75v and 4 rechargeable batteries give you up to 5vRechargeable AAA batteries give you around 800mAh so 4 AAA batteries would give you the same amount of energy as a 9v battery without having to use a dc-dc converter.AA batteries would give you around 1800-2500mAh so would last much longer.You may want to keep the output power to under 1w anyway, look at the THD graphs in the datasheet, see page 5 :https://www.diodes.com/assets/Datasheets/PAM8403.pdfWith 3.3v input, you can see the distortions jump as soon as you go over 1w and with 5v the distortions increase as you go over 2w

author: aju11
author_rank: Contributor
timestamp: August 30, 2018, 07:19:22 am
content: You may consider using series pass regulator. low component count and cost.http://www.circuitstoday.com/voltage-regulators

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: August 30, 2018, 08:28:21 am
content: Quote from: Audioguru on August 22, 2018, 12:20:00 am5V peak to peak is ≈ 1.77 V RMS into a 4Ω load => 0.78 W. Or 1.56W max for a square wave. How do you arrive at that 2.2W figure?

author: dazz
author_rank: Frequent Contributor
timestamp: August 30, 2018, 12:06:23 pm
content: Quote from: GeorgeOfTheJungle on August 30, 2018, 08:28:21 amI believe it's because it's a BTL amplifier, so the output rail is doubled to 10V peak to peak.

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: August 30, 2018, 01:01:10 pm
content: Thought so too, but double the voltage would be 4x the power => 4*0.78 = 3.12W, still not 2.2W

author: dazz
author_rank: Frequent Contributor
timestamp: August 30, 2018, 01:07:35 pm
content: Quote from: GeorgeOfTheJungle on August 30, 2018, 01:01:10 pmWell, it's not 100% efficient. According to thedatasheetit can do 2.5W at 5V 4 ohms, 1% THD. So about an 80% efficiency which seems about right

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: August 30, 2018, 01:45:45 pm
content: Oh, I see, that must be, it's a switching amplifier and the figure comes from the datasheet, lol. Thanks!

author: tooki
author_rank: Super Contributor
timestamp: August 30, 2018, 01:56:26 pm
content: It is a class D amp. In my testing of that chip, I found that (at around maximum volume without clipping, into 8 ohm speakers), its current had 8 amp transient peaks. (Average current about 800mA.) So capacitors are essential for operation on power supplies that provide less than that peak current.