Why can't you adjust volts and amps when calculating power for OHMs? - Page 1
Forum section: electronics / beginners

author: DW1961
author_rank: Frequent Contributor
timestamp: September 23, 2020, 06:40:55 pm
content: I may have asked this question before, but I can't find it. I apologize if so.Instead of increasing voltage, why can't you increase amps to get the same power when calculating Ohms?Example (rounded):8 Ohms3.5 Amps28 Volts=100 wattsI can't lower volts and increase amps to get the same power.I'm using this calculator:https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html

author: newbrain
author_rank: Super Contributor
timestamp: September 23, 2020, 06:54:45 pm
content: Because the voltage across a resistor and the current through it are bound together byOhm's Law:V = I × RWith V in volts, I in ampere and R in ohms.So, if you have 28 V across an 8 Ω resistor the current through it will be 28/8 = 3.5 A no matter what, and vice versa.In fact, two common ways to calculate the power dissipated by a resistor are:P= V × I = V × (V/R) =V2/ RandP= V × I = (I × R) × I =I2× R

author: Electro Fan
author_rank: Super Contributor
timestamp: September 23, 2020, 07:10:46 pm
content: Quote from: DW1961 on September 23, 2020, 06:40:55 pmThe answer depends on what you are trying to solve for, the math, or a more conceptual understanding of Ohm's Law, or building a particular circuit with some particular constraints.At the simplest level (using the calculator in your link), you very well could lower the volts and increase the amps. For example you could use 25 volts and 4 amps and you would still have 100 watts. The way the calculator works you can choose any two values and it will determine the other two.Of course when you select volts and amps the formula is going to drive you to use or a particular resistance, and you would want to make sure that the resistance (a resistor) would handle the required power (in watts). All depends on what your constraints are, or what your design objectives might be. (If the calculator hasn't been working for you just be sure to use the reset button before you type in two new values).Many people start out working on volts, amps, and resistance and then one day when they build something that accounts for all three they find they "let the smoke out" because they were not keeping volts x amps = watts in mind. Resistors need to be sized for the circuit's wattage. Once you get going it will become somewhat more second nature to balance all four values according to your objectives.

author: TimFox
author_rank: Super Contributor
timestamp: September 23, 2020, 07:22:06 pm
content: Since the fundamental constraint isV=IxR, you can only specify two and solve for the third.In general, with a non-linear load, the equation isV=f(I) and you should graph the function (such as for a semiconductor device).For AC, you can use a variable transformer to obtain the desired power in a specified resistor, or to maintain the desired power when you change the resistor, from a given voltage source; this allows you to change the voltage across the load to obtain your desired power.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: September 23, 2020, 07:24:56 pm
content: Sometimes a slightly different explanation helps.You can't adjust voltage and current independently. You can reduce volts and increase current to get the same power. But to do that you have to reduce resistance.Your question may come from an audio problem. My home stereo puts out 240 W into an 8 ohm impedance speaker. I want my car stereo with only 12 volts available to also put out 240 W. Obviously that will require 20 Amps of current. And to get that 20 Amps the speaker impedance would have to be down around 0.6 ohms. (You can work back and find that the home stereo must have a roughly 44 volt supply).You can make these results happen in the referenced calculator by resetting between calculations. Its internal logic isn't clever enough to deal with serial changes in values.

author: DW1961
author_rank: Frequent Contributor
timestamp: September 23, 2020, 07:56:32 pm
content: Quote from: newbrain on September 23, 2020, 06:54:45 pm"bound together by Ohm's Law"Thank you so much. Clear and easy to understand.

author: DW1961
author_rank: Frequent Contributor
timestamp: September 23, 2020, 07:59:11 pm
content: Quote from: TimFox on September 23, 2020, 07:22:06 pmAh ok so interesting there. So could you do that with a DC amplifier going into speakers too--that is, better utilize the power supply rating by adjusting amps and volts how you want them going to the speakers, while keeping resistance the same?

author: DW1961
author_rank: Frequent Contributor
timestamp: September 23, 2020, 08:00:24 pm
content: Quote from: CatalinaWOW on September 23, 2020, 07:24:56 pmIt does indeed come from calculating amplifier power and Ohms.

author: TimFox
author_rank: Super Contributor
timestamp: September 23, 2020, 08:33:20 pm
content: As I said in a reply to one of your other posts, a transformer can be used to couple a speaker impedance to an amplifier that delivers its maximum power into a different impedance, but solid-state amplifiers normally avoid transformers and are designed to drive a "normal" impedance (like 4 or 8 ohms) directly. (Designing a good transformer for the 10-octave range required for high-fidelity audio is a difficult problem, and good audio transformers are heavy and expensive.)As part of that design, you can select the DC voltage applied to the amplifier, possibly using a switch-mode power supply to obtain the desired voltage from a 12 V car system, or the 170 V DC available after rectifying the 120 V AC line. In the line-driven case, the transformer required to isolate the amplifier from the mains runs at a much higher frequency than 60 Hz, and can be far lighter.

author: DW1961
author_rank: Frequent Contributor
timestamp: September 24, 2020, 12:17:29 am
content: Quote from: TimFox on September 23, 2020, 08:33:20 pmDo you think they could be doing that with those little class D amps we talked about earlier in order to get more output power to the speaker than tehy would otherwise? Or would the transformer be too large or costly? These amps' cases are about 6" x6" by 13/4".

author: TimFox
author_rank: Super Contributor
timestamp: September 24, 2020, 01:17:55 am
content: There isn’t room in that small box for a power audio transformer, but there should be a switching power supply operating at a very high (ultrasonic) frequency using a relatively small transformer. A 50 W output transformer for a tube amp is maybe 4 x 4 x 4 inches or so.

author: james_s
author_rank: Super Contributor
timestamp: September 24, 2020, 04:19:52 am
content: That's how car audio amplifiers work. They have a switching power supply that converts 12V into + and - rails of sufficient voltage to push the rated power through the impedance of the speakers.

author: DW1961
author_rank: Frequent Contributor
timestamp: September 24, 2020, 05:02:55 am
content: Quote from: james_s on September 24, 2020, 04:19:52 amWould the continuous power still be limited by the power source's voltage and amperage? Our other discussion was about class D amp chips and the continuous ratings in power you see on Amazon with these class D chips in home audio systems powered by DC. I mean the maximum wattage a 24V 5A power supply can deliver is 120 watts and no resistance.One of them states the class D 3116D Texas Instruments can put out 200 watts total "RMS" out of a 24V 5A power supply. I'm calling bullshit.

author: james_s
author_rank: Super Contributor
timestamp: September 24, 2020, 06:00:34 pm
content: Quote from: DW1961 on September 24, 2020, 05:02:55 amYes you can't get something for nothing. The RMS wattage the amp delivers cannot exceed the power that it draws from the power supply.

author: DW1961
author_rank: Frequent Contributor
timestamp: September 24, 2020, 09:14:46 pm
content: Quote from: james_s on September 24, 2020, 06:00:34 pmOK, becasue I was going to say that that is REALLY intuitive, and almost seems like magic. lolAlso, why is this @ 2 ohms? It seems like 2 Ohms should be higher than 4 or 8 ohms power output.24V / 5A power supply:@ 8 ohms: 72 watts total@ 4 ohms: 100 watts total@ 2 ohms: 50 watts total <----What?

author: Brumby
author_rank: Unknown
timestamp: September 25, 2020, 01:21:52 am
content: Power supply voltage sag.If you try to pull more amps than a power supply can deliver, something MUST give. This could be anything from a destructive response to a lowering of the output voltage.Audio (especially good car audio) circuits are designed with abuse in mind (to a degree). 2 Ohm loudspeaker systems are not uncommon, but the current drawn from a full voltage output would be more than the power supply could possibly deliver, so the amplifier is designed to "give" which reduces the output voltage and, hence, output power.@ 8 ohms: 72 watts total8 Ohms across 24V gives 3A.3A@24V gives 72W@ 4 ohms: 100 watts total4 Ohms across 24V gives 6A- but the supply can only give 5APower is: I2.R = 5*5*4 which gives 100W(V=I.R which is 20V - a 4V voltage sag)@ 2 ohms: 50 watts total <----What?2 Ohms across 24V gives 12A - not going to happen.The supply can only give 5A.Power is: I2.R = 5*5*2 which gives 50W(V=I.R which is 10V - a 14V voltage sag)These numbers are simplistic - but you get the idea.

author: james_s
author_rank: Super Contributor
timestamp: September 25, 2020, 02:33:30 am
content: Also heat and stability. Many amplifiers have difficulty driving very low impedance loads, 2 ohms is getting down near what could be called a short circuit. The output devices will heat up a lot more so you could be bound by thermal constraints, or the amp could be unstable trying to push more than 50W into 2 ohms, or as said above, the power supply might sag.

author: gbaddeley
author_rank: Regular Contributor
timestamp: September 25, 2020, 06:29:15 am
content: And I^2 R losses in the wiring (including speaker leads & connectors) comes into play.

author: james_s
author_rank: Super Contributor
timestamp: September 25, 2020, 07:19:06 pm
content: Quote from: gbaddeley on September 25, 2020, 06:29:15 amIt shouldn't. Even if you are losing half the power as heat in the wires the amplifier is still producing the power.