Exactly. They go to the extra expense of a grounded bridge to use a cheaper power supply. If the savings in the power supply are bigger than the added expense for the amplifier section, they build a bridged amplifier to increase their profit. If the savings are less, they build the non-bridged. If you count, how many bridged amplifiers are in the market and how many non-bridged, you get a hint, which concept is more efficient in that aspect.RocketScientist said:
You wrotemegajocke said:
Power factor (cos phi) is phase angle. Did you want to write efficiency (eta) instead?megajocke said:
If that is so, how can a higher ripple current can be any good?megajocke said:
I don't even have to calculate. The integral of a curve is the surface area between it and the x-axis. If you look at your own simulator results in post #7, you can see at a glance that the surface area of the non-bridged amplifier is half that of the bridged amplifier.megajocke said:
The savings in the power supply themselves aren't really significant enough by themselves to warrant using it. Add in the halved voltage stress (this is the big one) and ease of sensing current and its advantages come to light. The Crown circuit was patented though, so others came up with different circuits having different advantages/disadvantages. With the advent of class G/H the main advantage, lowered voltage stress, could be accomplished in a simpler way and dissipation lowered at the same time. Cascoding is another way of doing it.
However, grounded bridge designs are still used nowadays by QSC in their PL6.0 and PL9.0 amplifiers. They are 4-step class H so the power supply would have needed 8 different voltages and 6 rail switches if they had been non-bridge amplifiers!
But first you have to know what RMS is and how it's calculated to be able to draw conclusions!
http://en.wikipedia.org/wiki/Root_mean_square
The square root of the mean of the squares of the values. The mean of the squares (calculated by the integral) halves as half the surface is removed. And sqrt(0.5) = 0.707...
You need to be more careful when dealing with non-linear relationships!
However, grounded bridge designs are still used nowadays by QSC in their PL6.0 and PL9.0 amplifiers. They are 4-step class H so the power supply would have needed 8 different voltages and 6 rail switches if they had been non-bridge amplifiers!
No, power factor is real power (watts) divided by apparent power (VA). It is a factor between 0 and 1, not a phase angle. However, it is the cosine of the phase angle in the special case of both current and voltage being sinewaves. In this case neither current nor voltage are sine waves, so cos phi has nothing to do with nothing.pacificblue said:
Total ripple current is lower in the bridged amp, so your point doesn't make any sense.pacificblue said:
pacificblue said:
But first you have to know what RMS is and how it's calculated to be able to draw conclusions!
http://en.wikipedia.org/wiki/Root_mean_square
The square root of the mean of the squares of the values. The mean of the squares (calculated by the integral) halves as half the surface is removed. And sqrt(0.5) = 0.707...
You need to be more careful when dealing with non-linear relationships!
Okay, if splitting hairs makes you happy. So where does the additional apparent power in the non-bridged amplifier come from? What are the power factors of typical bridged and non-bridged amplifiers?megajocke said:
I am really at a loss now. In post #16 you wrote yourself that the non-bridged amp draws less current from each rail. How can its ripple current be higher, then?megajocke said:
I would appreciate it, if you were less condescending. If I want advice from you about what I should be or do, I will ask for it.megajocke said:
pacificblue said:
From the higher peakiness. The power factor on the DC side of the PSU differs by a factor of sqrt(2) between the both cases. Look at those numbers in post #16 where power and apparent power on the DC side is calculated. For sine wave output it is 0.91 for the bridge amplifier and 0.65 for the non-bridged.
The power factor on the primary side of the transformer is the same in both cases, if you wonder.
The non-bridged amplifiers has two such power supplies. The total ripple current is higher because of the higher total RMS current (or apparent power) on the DC side of the psu.pacificblue said:
Okay, but could you please try to read my posts then? If I write that something is a common misconception, there might just be a good reason for that. The RMS values of full wave rectified sine and a half wave rectified differ by a factor sqrt(2). It's not 2 as you have repeatedly stated. The sqrt(2) difference is actually true for any waveform having no DC component and spending the same amount of time being positive as being negative.pacificblue said:
Power factor in linear loads results from the presence of inductive or capacitive components, not from peakiness of the wave form.megajocke said:
The capacitors are linear loads, the speaker is a linear load and the transistors don't change anything about that as long as we talk about class A, B or AB amplifiers. Could the confusion between us have arisen, because you are thinking about class D amplifiers that are switching, i. e. non-linear loads?
I am starting to suspect that you add up the ripple currents of the negative and the positive half of the split supply. Could that be?megajocke said:
I cannot for the life of me find a reference to rectifiers in any of my posts. Weren't we talking about amplifiers and capacitance? I assumed of course that both amplifiers use the same rectification scheme and the same speaker load, etc. The only difference should be the amplifier itself and of course the number of rails and the rail voltage to ensure comparability.megajocke said:
pacificblue said:
The power amplifier is definetely not a linear load to the power supply. Voltage supplied to it is practically constant DC but current varies arbitrarily! No linear circuit element has that property.
Why do you assume the transistors don't change anything? Their collector current can be controlled arbitrarily in their linear region, totally disregarding the way Vce changes. Without a linear relationship between Vce and Ic, they aren't linear circuit elements.
Yes, of course. If the same capacitors are used, that is the two are in parallell on the only rail of the bridge amplifier, ripple current in each will be lower. Energy storage will be the same. You gain lower heating in the capacitors and lower ripple at bass frequencies.pacificblue said:
Now you can decrease the capacitance and get back the original ripple voltage by decreasing the capacitance of both caps. Ripple current is still lower. If you decrease capacitance by removing one capacitor instead, ripple current in that capacitor will be higher than in a single capacitor in the non-bridged amplifier. Consult the datasheet of the capacitor to see if this is ok.
Usually, ripple current isn't going to be a problem because audio amplifier capacitors are selected by capacitance and large ones are needed anyways. Ripple current rating is usually more than enough, especially if using the types with screw terminals.
Neither did I mention rectifiers in my post. I was referring to the current waveforms drawn by bridge/non-bridge amplifiers and their average/RMS values.pacificblue said:
This chart illustrates the difference. PSU peak power is *double* in the half bridge example, figure 1, compared to the full bridge with the same number of power supplies, chart 2.
This is why less total energy storage is needed in the power supply for the same ripple at low frequencies.
This is why less total energy storage is needed in the power supply for the same ripple at low frequencies.
Attachments
underwurlde said:@megajocke: you are certainly putting a lot of effort in here. Kuodos to you...
Yes he is and it's impressive. Capacitor charging currents are both complex and non-intuitive to many--especially when presented with a dynamic load like an audio amplifier.
For example, many I've run into don't realize that bigger power supply caps increase the peak charging currents and hence can increase EMI and/or ground related distortion in an amplifier. They also stress the diodes more and can make some transformers physically noisier. The intuitive understanding, by some, is bigger caps "smooth things out more" so they're better in every way except the turn on surge at power up. But that's not the reality.
Transformer current increases up to a point, but after that the resistance and leakage inductance of the transformer dominates. RMS current won't increase much after that with increasing capacitance as the table on Rod's site shows.
If something like 20% ripple can be tolerated, transformer RMS current will be just a little bit lower but that isn't really very useful.
Rectifying the mains directly for an off-line SMPS might get a bit unpleasant though if large capacitance is used in a low power unit...
If something like 20% ripple can be tolerated, transformer RMS current will be just a little bit lower but that isn't really very useful.
Rectifying the mains directly for an off-line SMPS might get a bit unpleasant though if large capacitance is used in a low power unit...
megajocke said:
That's a good point and an interesting link but the section on increasing capacitor size is rather short on math to prove his claims. As the capacitance goes up the conduction angle is reduced changing the ratio of Irms to Iavg for the worse. Obviously, at some large value of C, other things dominate. But he doesn't go very far to explain his numbers or how to calculate that "crossover" point.
The data in the linked table is at 40 watts of load with unknown transformer properties. As the load goes up, and a larger transformer is used, I would expect the crossover point for C, i.e. where other factors dominate, to go up as well.
The output impedance of big transformers is typically around 0.1 - 0.2 ohms. While the caps have an ESR about 1/10 as much so I'm not sure the ESR matters much. I'm not sure about the parasitic inductance but, at the frequencies involved, I wouldn't expect to make a big contribution. But perhaps that's a bad assumption?
As I recall from simulations with 0.1 ohms for the transformer, you get about 25% higher currents going from 10,000 to 50,000 with a 40 volt transformer. That's not huge, but it's a lot more significant than what's shown in the linked table.
Several authors, such as Self, have documented bigger caps generally increase problems from ground currents, diode noise, and EMI in real world amplifiers. It would be an easy enough thing to measure in an actual amplifier at various power levels. Now you have me curious what the numbers really are? This might be another of those situations where simulations are misleading
I think you are right. In his example with a 0.75 ohm transformer output resistance the "large" capacitance you speak of isn't very large.
Playing with the particular transformer values in a simulator seems to be helpful. That 25% increase in current isn't that insignificant, it gives over a 50% increase in transformer copper losses. Which is a 50% total transformer loss increase in the case of toroids with their practically negligible core losses...
Leakage inductance in a transformer can be significant in EI types, especially of the split-bobbin type. I've seen this in some power supplies: the output voltage of the transformer "sticks" to the rail voltage a while after you would have expected the rectifier to stop conducting and then the voltage abruptly falls back to the sine wave level. (Typically ringing at high frequency if there are not any snubbers)
Playing with the particular transformer values in a simulator seems to be helpful. That 25% increase in current isn't that insignificant, it gives over a 50% increase in transformer copper losses. Which is a 50% total transformer loss increase in the case of toroids with their practically negligible core losses...
Leakage inductance in a transformer can be significant in EI types, especially of the split-bobbin type. I've seen this in some power supplies: the output voltage of the transformer "sticks" to the rail voltage a while after you would have expected the rectifier to stop conducting and then the voltage abruptly falls back to the sine wave level. (Typically ringing at high frequency if there are not any snubbers)
pacificblue said:
This whole discussion is about AC ripple current-not DC. Megajocke is correct about the power factor. Because the current is not a perfect sine wave over the entire cycle in phase with the voltage, like it would be with a resistive load, audio amps have a power factor less than 1. The current waveform into the amp is not a sine wave, but the voltage is a sine wave. So, by definition, they are a non-linear load.
Audioholics states, in the link below "The PF for most consumer amplifiers is 0.65 - 0.72" ... "Most consumer amplifiers utilize large transformer followed by a full wave bridge rectifier and bulk supply capacitors. The current drawn is out of phase with the voltage - the Power Factor (PF) will be less than 1. "
Audioholics Power Factor Description
I am sure there are many more references to power factor and amplifiers as well. Any time the current doesn't follow the voltage you have a power factor other than 1.
Just thought I'd pass this along, sorry if it is redundant.
I found this equation in a magazine (Nuts and Volts I believe?):
C * dv/dt = I
Applying this to a power supply:
C = filter capacitance
dv = allowable ripple voltage (5V sag? 20V sag? How much can you tolerate?)
dt = 1 / 120 (60Hz, rectified, is 120Hz, thus the capacitors get recharged 120 times a second)
I = maximum amount of current you expect your amp to consume
I found this equation in a magazine (Nuts and Volts I believe?):
C * dv/dt = I
Applying this to a power supply:
C = filter capacitance
dv = allowable ripple voltage (5V sag? 20V sag? How much can you tolerate?)
dt = 1 / 120 (60Hz, rectified, is 120Hz, thus the capacitors get recharged 120 times a second)
I = maximum amount of current you expect your amp to consume
That number concerns the entire amplifier, i. e. the audio section, which we are talking about here, the power supply capacitors, which we are also talking about here, the rectifiers, which we are explicitly not talking about and the transformer, which we are also not talking about.RocketScientist said:
Please read megajocke's post #16 again. He states that a non-bridged amplifier's audio section has a power factor of 0,707, while a bridged amp's audio section has a power factor of 1.
In post #24 and some following posts he says that a non-bridged amplifier has to be treated like a half-wave rectifier and therefore has a power factor of 0,65. while a bridged amplifier has to be treated like a full-wave rectifier with a power factor of 0,91. Again, we are only talking about the audio power section.
To make a long story short, he tries to convey that a bridged amplifier has a significantly lower power consumption than a non-bridged amplifier that delivers the same power and therefore needs half the capacitance with half the voltage rating. He thinks that a bridged amplifier changes the waveforms in the power supply rails in a way that improves the power factor and lowers the ripple current so much that it is better than in a non-bridged amplifier inspite of double the current draw.
My position is that a bridged amplifier needs double the capacitance to achieve the same performance of the power supply, because the character of the load is the same, the current draw is doubled and the voltage is lower. Moreover the bridged amplifier has a lower efficiency, because it has twice as many transistors and emitter resistors in series with the load, so that any current that flows through the load leads to double the losses than in a non-bridged amplifier.
pacificblue said:
While I have not followed every post and calculation, I think Megajocke is correct (or at least largely correct). Because a bridged amp draws power for the full cycle from each rail (instead of for just half the cycle), the amplifier will have a higher (better) power factor. This can be seen in the specs for Crown's bridged amps. Crown specifies the power factor (real world) as 0.85 for their bridged amps while a conventional amp cannot exceed around 0.72.
So yes, bridging does change the AC current waveforms in the power supply to be more efficient just as Megajocke has said. For the same power output, you can use a transformer with a smaller VA capacity in a bridged amp. I think this is one of the main reasons so many pro amps are bridged. It saves money and makes the amplifier lighter.
It is the peak power from each power supply that is higher in a half-bridge compared to the full-bridge. (see the chart in post #27) The half-bridge needs two power supplies with the peak power ability of the whole PSU of the bridge amplifier.
At low frequencies this peak power draw lasts for quite a long time, and during that time the amount of ripple depends on how high this draw is. This is why there is an advantage at low frequencies for the bridged amplifier.
Power factor (which is not often mentioned for DC circuits as it is 1 in most cases) was just calculated by the amount of watts and VAs consumed from the power supply. Bringing it up seems to just have confused matters though. It doesn't matter much, just look at the absolute numbers if that feels more comfortable.
Of course there are disadvantages to bridge amplifiers, like that you must have transistors in series. At low supply voltage bridge isn't a very good idea because of this. But at 120V+ you need to series transistors anyways, for most common transistor types.
-------------------------------------------------------------------------------
pacificblue:
I have never written anything about power consumption being different. Neither have I written anything about amplifiers being treated like rectifiers. Pacificblue, where do you get this stuff from? The only time I mentioned rectifiers was in the simulation post. In other posts, I wrote 'rectified' as it is part of the name of the waveforms drawn by an amplifier playing a sine wave. See table on this Wikipedia page:
http://en.wikipedia.org/wiki/Crest_factor
At low frequencies this peak power draw lasts for quite a long time, and during that time the amount of ripple depends on how high this draw is. This is why there is an advantage at low frequencies for the bridged amplifier.
Power factor (which is not often mentioned for DC circuits as it is 1 in most cases) was just calculated by the amount of watts and VAs consumed from the power supply. Bringing it up seems to just have confused matters though. It doesn't matter much, just look at the absolute numbers if that feels more comfortable.
Of course there are disadvantages to bridge amplifiers, like that you must have transistors in series. At low supply voltage bridge isn't a very good idea because of this. But at 120V+ you need to series transistors anyways, for most common transistor types.
-------------------------------------------------------------------------------
pacificblue:
I have never written anything about power consumption being different. Neither have I written anything about amplifiers being treated like rectifiers. Pacificblue, where do you get this stuff from?
The only time I mentioned rectifiers was in the simulation post. In other posts, I wrote 'rectified' as it is part of the name of the waveforms drawn by an amplifier playing a sine wave. See table on this Wikipedia page:http://en.wikipedia.org/wiki/Crest_factor
Energy storage is work. W = P x t = P / f. The power consumption is the same. And the frequency, with which the capacitors are recharged from the rectifiers is also the same, as is the wave-form, because we assumed full-wave rectification. So you obviously need the same energy storage in the power supply, whether the load is a bridged amplifier or a non-bridged amplifier.
If the power remains the same and P = U x I, you need double the current, when you use half the voltage. With energy storage being W = ½ * C * V² that means you need four times the capacitance, when you use half the voltage.
The rms value of the ripple voltage gamma = 1 / (4 * sqrt(3) * f * C * R). Since the power is the same, but the voltage is halved, R is quartered for the bridged amplifier. If you want the same ripple voltage, you need four times the capacitance. Less capacitance means higher ripple voltage. That is the practical application of energy storage.
There is no improvement for lower frequencies in bridged amplifiers. The ripple voltage changes with the amount of current, not with the frequency at which current is drawn.
Ripple current Irms = sqrt(Ichargerms² + Idischargerms²). The charge current through the rectifiers is doubled, see above. The discharge current through the amplifier is sqrt(2) times higher, see your post #16. Therefore the ripple current is sqrt(6) = 2,45 times higher for the bridged amplifier, which means the capacitor(s) for a bridged amplifier need a higher ripple current rating.
Crest factor or peak-to-average ratio is a non-issue in unregulated power supplies. All components in unregulated power supplies can easily deliver peak currents that are much higher than their rms rating. Crest factor is an issue with current limited, i. e. regulated power supplies. In that case the bridged amplifier might have an edge, but we were not posting about regulated power supplies, were we?
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.