i want to know if my amplifier sends a 'push' signal down the positive wire, through the voice coil, and down the negative wire
OR
if the amplifier sends a 'push' signal down the positive wire, through the voice coil, and then simply lets it stay there to push the cone outward until the energy is completely discharged (or dissipated as heat).
here is the reason i am asking:
i got some speakers from the junkyard and i chose those specific speakers from an infinity sound system because they come with a seperate tweeter.
i suspected they would be 2 ohms, and when i got the speakers home i found out they are 2 ohms.
i wanted to add 2 ohms to the speaker to keep my cd player safe because it is rated at 4-8 ohms.
i went to radioshack and found some 2 ohm 10 watt resistors.
my first thought was 'my radio is 22 watts RMS @ 4 ohms.. so i need two of these with one on each side'
but i started to think about it again and i thought 'what if the cd player amplifier uses one speaker wire to send a 'push' signal down the wire with 22 watts RMS'
am i 12 watts over the resistor limit?
i connected one resistor to the positive terminal and one resistor to the negative terminal.
the more i think about it, the more i realize the flow of electricity comes into contact with the 10 watt resistor first.. and if there is 22 watts RMS there, then yes i am 12 watts over the limit no matter if there is another resistor on the other side or not.
with that said..
i want to focus on the question of the amplifier sending a 'push' signal down the positive wire.
does that signal go through the voice coil down the negative speaker wire back to the amplifier, or does it stay there in the voice coil?
because i see the following as a possibility:
maybe only 11 watts RMS of 'push' goes down the positive speaker wire, while another 11 watts RMS of 'pull' goes down the negative speaker wire.
if this is true, then i am only 1 watt over the limit and i dont run the amplifier at full blast to see all 22 watts RMS .. therefore i should be safe and sound.
but
this brings up the question.. did i add an extra 2 ohms to the amplifier load for a total of 6 ohms ?
OR
if the amplifier sends a 'push' signal down the positive wire, through the voice coil, and then simply lets it stay there to push the cone outward until the energy is completely discharged (or dissipated as heat).
here is the reason i am asking:
i got some speakers from the junkyard and i chose those specific speakers from an infinity sound system because they come with a seperate tweeter.
i suspected they would be 2 ohms, and when i got the speakers home i found out they are 2 ohms.
i wanted to add 2 ohms to the speaker to keep my cd player safe because it is rated at 4-8 ohms.
i went to radioshack and found some 2 ohm 10 watt resistors.
my first thought was 'my radio is 22 watts RMS @ 4 ohms.. so i need two of these with one on each side'
but i started to think about it again and i thought 'what if the cd player amplifier uses one speaker wire to send a 'push' signal down the wire with 22 watts RMS'
am i 12 watts over the resistor limit?
i connected one resistor to the positive terminal and one resistor to the negative terminal.
the more i think about it, the more i realize the flow of electricity comes into contact with the 10 watt resistor first.. and if there is 22 watts RMS there, then yes i am 12 watts over the limit no matter if there is another resistor on the other side or not.
with that said..
i want to focus on the question of the amplifier sending a 'push' signal down the positive wire.
does that signal go through the voice coil down the negative speaker wire back to the amplifier, or does it stay there in the voice coil?
because i see the following as a possibility:
maybe only 11 watts RMS of 'push' goes down the positive speaker wire, while another 11 watts RMS of 'pull' goes down the negative speaker wire.
if this is true, then i am only 1 watt over the limit and i dont run the amplifier at full blast to see all 22 watts RMS .. therefore i should be safe and sound.
but
this brings up the question.. did i add an extra 2 ohms to the amplifier load for a total of 6 ohms ?
Hi,
Whether or not an amplifier uses transistors or valves (tubes), the term push-pull refers to the topology of the amplifier's output stage. Two 'groups' of transistors are arranged such that each group generates one half-cycle of the audio signal feeding the loudspeaker. There may be one or more transistors per group, depending on the output power capability of the amplifier.
The amplifier's output terminals are usually marked red and black to ensure that a 'positive' loudspeaker cone excursion (forward movement) will be caused by a positive voltage at the loudspeaker’s red terminal with respect to the black terminal.
Your first statement is correct. Current flows from the red (positive) wire, through the voice coil and back to the black (negative) terminal.
Your second statement is incorrect. If the output causes the cone to move outward and let it stay there, something is seriously wrong! Audio amplifiers normally have a designed frequency response, for example from 10Hz to 25KHz. This means that the amplifier will not allow DC to reach the voice coil. The voice coil current will change in sympathy with the audio signal (hopefully very accurately) and only pass the frequencies of interest in the audible range.
Placing 2 ohm resistors in series with your 2 ohm speakers will be safe from the amplifier's perspective, but you will be losing 50% of your output power as heat. A better way, if possible, would be to use four x 2 ohm speakers and wire each pair in series.
In a series circuit, it makes no difference whatsoever which component is seen “first”. When wired in series, the same current will flow through all connected components. With your 22 watts amplifier output, the power will be divided equally between the series resistor and the loudspeaker. This means that you will have 11 watts maximum available at the loudspeaker.
The term 'push' is probably incorrect here and is confused with the push-pull terminology for the amplifier. Current flowing from the positive terminal resulting in energy received by the voice coil does not cause that energy to stay there. It is dissipated as cone movement and, to a lesser degree, as heating in the voice coil.
You appear to be describing the power in the positive and negative half-cycles. If the power being delivered to the loudspeaker is 22 watts RMS, then 22 watts is the power figure for each half cycle. For a sinusoidal signal, the power in a positive excursion is the same as that in the following negative excursion and is simply a continuation.
Your resistors are rated at 10 watts, but may have to dissipate 11 watts. This should not be a problem as long as you're not amplifying test tones at full power as normal music content has a lower average power. If the resistors become to hot, reduce the volume slightly.
I don't understand what you are saying. I understood that you have added exactly one 2-ohm resistor in series with each loudspeaker.
it wasnt my intention to say the cone moves outward and stays there.. because i know that would be DC power.
i wanted to say..
does the electricity simply stay there as the cone moves outwards until the electricity is gone because it was absorbed by the coil (and perhaps splash with the electricity coming from the other speaker terminal), or does it make its way to the amplifier as a constantly clamped ring that simply twists clockwise and counterclockwise with zero loose current between the positive and negative terminals.
nothing loose = a clamped ring
each one loose = both terminals slapping eachother inside the voice coil
first of all.. i dont think you interpreted what i said fully, and you prove this thought later on down the response.
i cant use four speakers wired in series because for one, each speaker has a power requirement that adds up with each additional voice coil added.
i dont know what the stock infinity amplifer's power was.. but i know it was a trunk mounted premium option.
i guess that each speaker is 30-50 watts RMS
adding up four of those speakers is asking for 120 watts RMS (much more than the 22 watts from the cd player)
but since we are talking about how sloppy (or tightly controlled) the current is.. it comes down to the original question again:
will 22 watts RMS slap the cone outwards from the positive terminal and simply pulse again with zero care about what the negative terminal is doing.
or
does the power stay obnoxiously clamped from the positive terminal all the way to the negative terminal, thus allowing more possibility for 11 watts RMS on each side.. because if they are basically shorted with eachother, then it adds up to 22 watts RMS .. there is jst no saying half of it comes from one rail? and the other half comes from the other rail.
look at the clear picture..
not all audio is as simple as a single soundwave.
picture two 60hz waves with 90 degrees of phase difference.
that means each wave has its own rise and fall, and that means one of two things will happen inside the voice coil:
1. the positive shorts the negative side when the two waves come into contact with eachother creating a spark in the coil (or jst a hot spot)
2. the two rails are tightly clamped to eachother and there is no boost because the rails are married and when the two soundwaves touch eachother a stalemate happens.
i know a number 3 is possible where instead of when a stalemate happens, the rings are tightly clamped together, but the same boost happens causing the sparking hot spot in the coil.
i'm such a noob enthusiast, do they call this triple darlington for topology?
(doesnt really matter i suppose because that doesnt tell me if the amp is slapping 22 watts per side or clamping two 11 watts into a ring)
when you say 'half cycles'
are you talking about a wave being cut in half horizontally (left|right across the middle) or vertically (up|down in the middle of the width of the wave) ?
i soldered one end of the resistor to the positive speaker terminal, then i soldered the other end of the resistor to the speaker wire.
i soldered one end of the resistor to the negative speaker terminal, then i soldered the other end of the resistor to the speaker wire.
the ends of the resistors are not touching, they are seperate except for the voice coil coming between them.
how do i know which type of amp i've got because i know of three things:
1. an amplifier that sends 22 watts RMS into the coil from one terminal and lets the other 22 watts RMS from the other terminal rush down the cord and splash into the voice coil like going down a waterslide (or busting through the door without any idea of what is going on inside the house).
2. an amplifier that basically holds a digital silence clamp inside the voice coil where the two terminals are electrically bonded generating some heat in the coil because of whatever millivolts (or milliamps) the amp puts out at idle, as the voltage output ramps up to 11 watts RMS per side on an as needed basis to move the cone the required amount.
the only amount of DC sent down to the voice coil is very small and is nothing compared to the full RMS value.. but if the two rails are bonded with the clamp, then it does happen.
i dont know how to measure only one speaker terminal without the other.
because if one side jst push push pushes, then that is a blinking pulse.
i dont think my multimeter will measure a blinking pulse.
i would expect the needle to jump and fall between each tip of the wave.
and i dont assume it is healthy to do that on an AC setting.. because it needs the other half of the pull pull pull to hold the needle steady.
the wave isnt clipped either, and that means it probably wont register on the DC setting, if the needle moves it would probably only jump little bits that arent anything close to the real voltage amount.
but really..
wow, i am feeling god awful dumb sitting here thinking each terminal can send out a push or pull signal as long as they both do it at the same time.
i mean..
one switched rail of 22 watts RMS for both speaker terminals
or
two switched rails of 11 watts RMS for each speaker terminal
is that the only physical way to know?
what is there to say now that you know i didnt directly solder two resistors together?
i know this can be simple math going by the number of physical rails, but i dont know what the amplifier design looks like.
i jst cant see how two rails of 22 watts RMS can ever EVER touch eachother and not see a rating of 44 watts because those soundwaves overlap eachother quite often.
Any current loop must be closed; current can not simply 'go somewhere' and just 'hang there'. All the electrons going one way MUST come back the other way.
If you would measure the current into the red speaker wire, and then measure into the black wire, they are identical except for opposite directions.
So, as noted above, push pull pertains to the internal amp topology but as far as the speaker is concerned there's just a current flowing through it, from the amp into one terminal and out of the other terminal back to the amp.
Of course this is AC so the current direction flips all the time but the general rule that 'what goes in must come out' holds.
jan didden
If you would measure the current into the red speaker wire, and then measure into the black wire, they are identical except for opposite directions.
So, as noted above, push pull pertains to the internal amp topology but as far as the speaker is concerned there's just a current flowing through it, from the amp into one terminal and out of the other terminal back to the amp.
Of course this is AC so the current direction flips all the time but the general rule that 'what goes in must come out' holds.
jan didden
I think this is confusing and close to incorrect. The + and - (or red and black) are arbitrary designations. It's important to keep them the same between channels, but just as much current flows out (or in) of red as it does with black. + or red isn't "positive," - or black isn't "negative" in the sense of current flow. It's like gender in languages, a handy designator but having nothing to do with objects having particular sex organs.
Sy,
when current flow IN on one speaker lead, then current flows in the opposite direction in the other speaker lead.
These two leads form part of a circuit.
If the current flows out of one lead than at the same moment the current flows in through the other lead. Again they are part of a circuit.
How you can use the phrase "close to incorrect" is beyond my understanding.
when current flow IN on one speaker lead, then current flows in the opposite direction in the other speaker lead.
These two leads form part of a circuit.
If the current flows out of one lead than at the same moment the current flows in through the other lead. Again they are part of a circuit.
How you can use the phrase "close to incorrect" is beyond my understanding.
sounds like a joke about needing to open up the cd player to see a view of the amplifier design.
but
the transistors themselves should be what says whether or not the their portions of the soundwave are allowed to touch or if there must be a gap.
if the transistors do all of the opening and closing of the current, then that means they also control the amount of current because more current = louder volume.
since the transistors control the amount of current released, that is exactly how the two portions of soundwaves can be touching without a gap.
am i jst stuck in a knowledge rut where class ab amplifiers can get the gap between the two transistors lowered to the point of elimination?
you cant run two class b amp topologys through a capacitor right?
that can only be done with a tube i'm thinking.
so does the transistors come before or after the capacitors?
each rail of capacitors touch or each rail of transistors touch?
jst what is it that allows the two halves of soundwaves to actually touch eachother?
i keep looking at this:
File:Electronic Amplifier Push-pull.svg - Wikipedia, the free encyclopedia
i know the timing of each chip is the same and that keeps the soundwave lined up.
but
what fills in the microscopic gap between the two soundwave halfs?
how can one half of the wave make it down the positive terminal and touch the negative terminal without creating a dead short between the two transistor rails or the two capacitor rails?
oh wait..
they've got non-polarized capacitors where it doesnt matter if you solder the positive pin into the positive and the negative pin into the negative.
obviously you cant expect the single capacitor to do all that music, and that means you need to let those two capacitor rails touch eachother, but you need something more than resistors to muffle the incoming electricity.
once the electricity is muffled, it is safe for the other class b circuit to come into contact with the voltage rail.
yall are talking as if two class b amplifiers are put together.
wikipedia says 'In class-AB operation, each device operates the same way as in class B over half the waveform, but also conducts a small amount on the other half.'
seems to be exactly what i was talking about above.
i guess i jst want to know if one half of the amp pushes outwards with bias voltage while the other half pulls inwards.
because if it does, then the wattage cannot all be on one speaker terminal.
if it doesnt, then there is only one wire slapping at the cone and flopping the cone around for whatever is leftover after being slapped.
doesnt it make sense to anybody else..
if one speaker terminal is pulling 75% and the other speaker terminal is pushing 25% .. that should be the theoretical maximum the math allows.
50% and 50% then the cone goes nowhere and its DC.
i dont know what allowed me to think the amp can push on one side while pulling on the other side at the same time.
i guess i got caught up in the 'teamwork' thinking two is better than one.
it means i was right to put 2 ohms on each speaker terminal.. but wrong to think 10 watts was going to double to 20 watts.
funny how something simple can move very far away and then flip to the exact opposite and be all up in your face.
one half of the amp pushes the cone from neutral to outwards
the other half pulls the cone from neutral to inwards
these two dont combine, they simply create double the chunk of time.
i guess i am off to find some resistors with 20 watts or more each.
i think its astonishing that there is information right there for people and maybe they dont appreciate it.
when one half of the amp pushes the cone from neutral to the outwards position, that same half of the amp allows the cone to slowly come back down from the outwards position to the neutral position.
and then..
the other half of the amp takes the cone from neutral position and pulls in on the cone, then slowly lets the cone back to its neutral position.
technically..
both halfs of the amp are doing push pull.
but
each half is divided into its own 180 degrees.
got one +180 and one -180 = the full 360
i'm an enthusiast. have been since like 5 years old.
i've always thought electronics are dope, but they've spent more time making a dope out of me than i've made out of them.
i'm always stuck in some fancy equation i can never get out of.
i think i'm gonna be happy going to school for electronics.
i mean the pieces are only as dumb as legos until you start putting them together to create combinations.
you only gotta learn the equations once, then wait for better components to be released
well maybe it was said as to why i would want to keep a resistor on each side of the voice coil.
electricity always takes the path of less resistance, and if the electrons find the voice coil and magnetic gap to be less resistant than the other 2 ohm resistor.. it can result in a perceived difference of audible output from the speaker.
i dunno if the speaker guru's are saying 'are you still doing that?!' from some oem infinity 6x9's with a tweeter.. but
when it is cheaper to go to the junkyard for a $10 pair of speakers because you cant afford the $30-$40 pair off ebay .. adding a couple dollars worth of resistors to make it work jst seems right.
Within the frequency handling capabilities of the loudspeaker and the amplifier, the cone's position should 'track' that of the voltage being applied. In other words, at 0V, the cone is at it's rest position. As the signal voltage rises positively, the cone moves outwards in proportion to that applied voltage and similarly as the signal voltage rises negatively, the cone moves inwards in the same way. The amount of cone deflection is proportional to the amount of current passing through the voice coil at any given point within the signal waveform. There is no question of applying a current to move the cone and then later remove that current to rely on the cone to fall back to a rest position. That would be a step function.
Your concept of speaker power requirement isn't right. There is no need to drive loudspeakers with an amplifier that matches their power handling. Because a car may have a top speed of 120mph, it doesn't imply you have to drive it at that speed. Driving speakers with an amplifier of a lower power rating is absolutely fine, within the constraint of not allowing clipping to occur. So, yes you can use four speakers with two series-wired on each channel.
Let me explain a basic principle. The current passing through the positive terminal will be identical to the current passing through the negative terminal. Current needs a circuit around which to flow and its magnitude is equal in all parts of a series circuit. The voltage produced at the amplifier output terminals is AC, in much the same way as the mains supply, except that it is varying in frequency and amplitude. If you could freeze time at any given point on a sinusoidal waveform and take a voltmeter to measure the instantaneous voltage, you would see that the voltage polarity changes depending on which half of the cycle you were measuring.
When you combine signals of different frequencies, amplitudes and phases, the resultant waveform will be complex. Some waves will add positively and some will subtract, causing cancellation. As long as the frequency components of this complex waveform are within the handling abilities of the amplifier and loudspeaker combination and do not cause clipping, then this shape will be (faithfully) reproduced by the speaker cone. There is no question of sparking or shorting.
Without seeing your amplifier's circuit, there is no way of knowing its topology. A darlington is just a type of transistor with high gain.
Hopefully my explanations above have answered your "slapping" and "clamping" question.
Consider a graph showing a sine-wave. Along the horizontal axis is time and amplitude along the vertical axis. The wave passes above and below the horizontal line. When it is above the line, those are the positive half cycles. Negative below.
So you have installed two 2-ohm resistors in series with your 2-ohm speakers giving a 6-ohm total impedance per channel. This means your speakers will only produce one-third of the available power. This is pointless if your amplifier will tolerate a 4-ohm load.
[SNIP]
As explained above, the speaker terminals on the amplifier will present an AC waveform to your loudspeaker. Look at it this way: Consider a small 1.5V battery inside the amplifier with its positive terminal initially connected to the red amplifier terminal and the negative to the black. Placing a voltmeter with red lead to red terminal and black to black will show +1.5V on the meter. Now reverse the battery connection. If it's a digital meter, the meter will display -1.5V. Now consider the battery rapidly changing polarity. This would produce a square wave of 1.5V peak. Although we're not dealing with square waves here, the same principle applies.
The term "rail" normally refers to a power supply connection, such as +35V, 0V or -35V. Rails cannot be expressed in terms of watts. To understand the relationship between internal power supplies, output transistors and signal output polarity, it would be advisable to read up on basic amplifier operation theory.
Sy & Janneman,
Taken in context, this seems reasonable to me.
anwaypassible's question was:
"i want to know if my amplifier sends a 'push' signal down the positive wire, through the voice coil, and down the negative wire..."
I took the "push" as meaning a positive swing. In this instantaneous condition, current flows from the red (positive) wire, through the voice coil and back to the black (negative) terminal.
call me dumb.. sure.
but
these are the resistors i got, finally found them on the radioshack website:
NTE 10W2D0 - 10W 2.0 OHM 5% Resistor : Resistors | RadioShack.com
the specifications are 2 ohms 10 watts
but then it says 'voltage: 550v'
doesnt that somehow translate to voltage and amperage = watts ?
i find it a bit hard to believe a resistor that can handle 550 volts is going to die from 6 volts.
but
these are the resistors i got, finally found them on the radioshack website:
NTE 10W2D0 - 10W 2.0 OHM 5% Resistor : Resistors | RadioShack.com
the specifications are 2 ohms 10 watts
but then it says 'voltage: 550v'
doesnt that somehow translate to voltage and amperage = watts ?
i find it a bit hard to believe a resistor that can handle 550 volts is going to die from 6 volts.
amperage = watts / volts
okay
10 watts / 550 volts = 0.0181818181818182 amps
if the resistor can handle 550 volts at that many amps..
anybody know how to adjust that to the amperage limit for 6 volts?
**edit**
22 watts / 6 volts = 3.666666666666667 amps
550 / 6 = 91.66666666666667
so then i should be taking 0.0181818181818182 x 91.66666666666667 = 1.666666666666668 amps for 6 volts
i am basically 2 amps over the limit right?
no worries,
i wont go 'pointless' by wiring up each resistor to eachother trying to get the wattage to double.
no sense confusing the radio, not knowing if it should flow through the resistors or through the coil.
oh gawd
ohms law says voltage / resistance = current (amps)
6 volts / 2 ohms (splashing into the resistor only here) = 3 amps
550 volts / 2 ohms = 275 amps
makes me think i am 272 amps shy of blowing the resistor.
if that is the case, then the thing shouldnt be warm to the touch at all.
who thinks one way is right and who thinks the other way is right?
okay
10 watts / 550 volts = 0.0181818181818182 amps
if the resistor can handle 550 volts at that many amps..
anybody know how to adjust that to the amperage limit for 6 volts?
**edit**
22 watts / 6 volts = 3.666666666666667 amps
550 / 6 = 91.66666666666667
so then i should be taking 0.0181818181818182 x 91.66666666666667 = 1.666666666666668 amps for 6 volts
i am basically 2 amps over the limit right?
no worries,
i wont go 'pointless' by wiring up each resistor to eachother trying to get the wattage to double.
no sense confusing the radio, not knowing if it should flow through the resistors or through the coil.
oh gawd
ohms law says voltage / resistance = current (amps)
6 volts / 2 ohms (splashing into the resistor only here) = 3 amps
550 volts / 2 ohms = 275 amps
makes me think i am 272 amps shy of blowing the resistor.
if that is the case, then the thing shouldnt be warm to the touch at all.
who thinks one way is right and who thinks the other way is right?
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been awake for 19 hours.. been eating twice, sometimes only once per day.
i think i'm jst gonna shove my hand through all the carpet behind the rear speakers and see how hot the resistor feels.
its either gonna burn me or be nothing.
thanks for taking the time to talk about amplifier topology.
i'm thinking higher and higher bias for class ab because i dont know any better yet.
i think i'm jst gonna shove my hand through all the carpet behind the rear speakers and see how hot the resistor feels.
its either gonna burn me or be nothing.
thanks for taking the time to talk about amplifier topology.
i'm thinking higher and higher bias for class ab because i dont know any better yet.
You didn't think through the statement that 'what goes in must come out'. There is only one path for the current: red post to one end of the voice coil to the other end of the voice coil to one end of the resistor to other end of the resistor to black post.
There's no 'path of least resistance' - there's only one path.
jan
I've no idea what you're doing, but maybe this helps:
For a 10 ohm 2w resistor, E^2/R=2 so E^2=20. Max E therefor about 4.47V. At that point, the current is 4.47/10=0.447amps.
jan
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Because the output of an amp into a speaker is an AC circuit - the current is constantly switching back and forth, from above potential to below potential. Essentially push/pull as the OP is trying to understand it, and effectively the truth.
To the OP - resistors are not speakers, and should not be used in any way to modify their impedance - it won't work, and could e dangerous.
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Let me.. ask if i managed to understand correctly:
You would like to know if it is safe to connect a 2 ohm speaker to a cd player that requires 4 ohm load, via a 10 watt 2 ohm resistor.
Yes, it is safe.
But it is inefficient.
Sugested solution was to use 2 speakers in series.
otherwise you loose half of amplifier power wasted as heat.
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