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Old 23rd December 2012, 02:32 PM   #1681
gootee is offline gootee  United States
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That's a rather-beautiful statement, Harrison.
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Old 28th December 2012, 09:01 PM   #1682
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Tom,

I am busy with the power supply (transformer) design of our new amplifier using the predictions of your spreadsheets and will early in the new year post some practical measurements against the predictions. Thank you for all the effort you have spent regarding this thread. I the same breath a very prosperous New Year to you and all the members reading this thread.
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Old 30th December 2012, 11:02 PM   #1683
gootee is offline gootee  United States
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Have fun, Nico. Spice would be more powerful. Maybe it would be a good time to become more familiar with it. I can send you a ready-to-run LT-Spice model, if you want it.

The only spreadsheet of mine that I trust completely is any version of this latest one, which actually solves the differential equations. The older ones are probably useful but remember that they were based on approximations, which break down more and more as they approach some extreme (e.g. when ripple gets large, especially).

Thank you for the new year wishes and know that I hope that the coming year is your best year, so far, in every way.

Tom
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Old 3rd January 2013, 06:33 PM   #1684
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Happy new year guess its time that diagram got fixed
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Old 9th January 2013, 01:53 PM   #1685
Bigun is offline Bigun  Canada
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Default This is an important factor

Quote:
Originally Posted by AKSA View Post
In this I'm reminded of subjective bass response of amps. If you have an amp with a single output pair, bipolar or mosfet, and relatively high Zout, you can actually improve subjective bass response by interposing a small 0.15R resistor into the speaker output line. This promotes overshoot on the voice coil - that is, loss of VC control - but adds a subjectively stronger bass response. Hardly correct in engineering terms, but quite useful when sculpting bass presentation. Perhaps something similar is operating here?
I have seen accounts where people have found subjectively better sound (to their taste) from amplifiers that are slightly under-damped. The power supply dynamic response is under-damped. In other words, transients are allowed some overshoot. This can be achieved through design of the power supply, and the choice of capacitors has a big bearing on the transient response. I believe it can also be achieved with design of the feedback compensation.
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Old 11th January 2013, 02:13 AM   #1686
gootee is offline gootee  United States
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Default Calculating Required Capacitor (Capacitance) Value

[This is adapted from a post at creating 1 watt rms power from 5 watt chip amp? , where I also derived equations involving simpler linear capacitor currents.]

The power supply is The Signal Path:

The audible signal in an electronic music reproduction system is electric current that comes directly from the reservoir capacitors of the power supply (and occasionally from farther upstream, if the rectifiers are conducting), or, whenever possible, from the decoupling capacitors nearer to the power output devices (which includes discrete transistors, chipamps, and vacuum tubes), as allowed by the modulated resistance of the power output devices, from which it flows to the speakers and then back to the power supply.

For accurate reproduction of the input signal that is used to modulate the resistance of the power output devices, the reservoir and decoupling capacitors, and the power and ground conductors, must be able to accurately provide the commanded current flow.

To try to ensure that the capacitors and conductors can always accurately provide the commanded current signal, an amplifier designer must ensure that it is at least theoretically possible for their chosen configuration of capacitors and conductors to satisfy the worst-case current amplitude and timing demands that might occur.

The capacitor equations that I have previously derived were all in terms of linear (straight-line) Δi changes in the current needed from a capacitor, during a time interval Δt, and their relationships to capacitance, ESR, and simultaneous changes in the voltage across the capacitor.

Since music signals are not usually straight lines when plotted vs time, those earlier equations might be useful for planning the handling of worst-case fast transient current demands, such as square-wave edges, but they might not be as useful for ensuring that our capacitors can meet the demands of low-frequency (bass) music signals, for example, or other music signals.

All real-world music signals, and most other types of real-world time-varying signals, can be broken down into a sum of single-frequency sinusoidal components. So if we could calculate, for example, the characteristics of the capacitance and conductors needed to meet the electric current demands for the reproductions of a single sinusoidal signal, then the calculation method should eventually be able to be extended, to enable dealing directly with more-complex "real world" signals. And until then, it will still be useful for calculating a required capacitance, for any frequency of signal.

This time I will start at the beginning and derive everything from scratch, the right way.

The ideal capacitor's differential equation is:

i(t) = C dv/dt

or

dv/dt = i(t)/C

where dv/dt is the time rate of change of the voltage across the capacitor and i(t) is the current flowing into the designated-positive-voltage lead of the capacitor at any time t >= 0.

We can integrate both sides of that equation, in order to have v(t) instead of dv/dt:

(9): v(t) = (1/C) ∫ i(t) dt + v(0)

where v(t) is the voltage across the capacitor at any time t (>= 0), and the integral must be understood to be evaluated from 0 to t (because this forum doesn't enable me to put the limits by the integral sign).

Re-arranging a little, we get

v(t) - v(0) = (1/C) ∫ i(t) dt

which can be expressed using "delta" notation, as

(10): Δv(t) = (1/C) ∫ i(t) dt

where Δv(t) is the change in the voltage across the capacitor due to the capacitor current i(t) existing since t = 0, at any time t >= 0.

Since the capacitor's ESR is a simple resistance, which adds a positive component to the capacitor's voltage for positive current i(t) into the positive-designated end of a capacitor, we can include the ESR terms in the Δv equation and validate them "by inspection":

(11): Δv(t) = (1/C) ∫ i(t) dt + ESR∙i(t) - ESR∙i(0)

That equation for Δv should be valid for any real-world current signal, i(t), at every time t >= 0.

Therefore, we could select i(t) = a∙sin(ωt), noting that the integral with respect to time of a∙sin(ωt) = -(a/ω)∙cos(ωt).

Substituting i(t) = a∙sin(ωt) into equation (11), we have

(12): Δv(t) = (1/C) ∫ a∙sin(ωt) dt + ESR∙a∙sin(ωt) - ESR∙a∙sin(0)

Integrating and evaluating from 0 to t gives:

Δv(t) = (-a/ωC) cos(ωt) - (-a/ωC) cos(0) + ESR∙a∙sin(ωt)

and we get

(13): Δv(t) = ( a / ωC )∙( 1 - cos(ωt) ) + ESR∙a∙sin(ωt)

where a is the peak value of the sine current from capacitance C, w is 2πf, and ESR is the capacitance's equivalent series resistance.

I'll leave out the math but equation (13) also provides a way to calculate the maximum Δv. It usually occurs when the cos term = -1, and is

(13a): Δv_MAX ≤ a / (πfC)

which occurs when when t = 1 / (2f) [and at multiples of 1/f]

or

(13b): C ≥ a / (πfΔv_MAX)

where a is the 0-to-peak amplitude of the capacitor's sinusoidal current waveform, f is frequency in Hertz, and C is capacitance in Farads.

Example: If you have a 100 Watts output into 8 Ohms, of a 30 Hz sine wave, the RMS current must be sqrt(100/8) = 3.5 Amps RMS. So the peak current is sqrt(2) x 3.5 = 5 Amps peak (and 40 Volts peak).

According to equation (13a), the maximum voltage dip caused by drawing a 5-amps-peak 30 Hz sine wave from a capacitance C must be

Δv_MAX(30Hz, 5A pk) = 5 / (3.14 x 30 x C) = 0.05305 / C

or

C = 0.05305 / Δv_MAX(30Hz, 5A pk)

If we want Δv_MAX(30Hz, 5A pk) to be 1 Volt, then C = 53050 uF, for example.

---

For a decoupling capacitor, we might consider only the "leading edge" of the sinusoidal current wavefom, in which case we could think of the 0-to-peak amplitude of the sine wave as Δi, giving (from equation (13))

(14): Δv(t) = (Δi/ωC)∙(1 - cos(ωt)) + ESR∙Δi∙sin(ωt)

But since for decoupling we might only care about the signal's leading edge, or rise time, we can consider only the part of a∙sin(wt) between t=0 and the first peak, which occurs when ωt = π/2.

But cos(π/2) = 0 and sin(π/2) = 1, which gives us

(15): Δv = (Δi/ωC) + ESR∙Δi (for initial rise of sine wave, only)

And ω = 2πf, giving:

(16): Δv = Δi∙(ESR + (1/(2πfC)))

Solving for C and setting the proper inequality gives

(17): C ≥ Δi / ( 2πf∙(Δv - (ESR∙Δi)))

Equation (17) gives the capacitance value, C, that would be required in order to supply the current for the first quarter-cycle of a sine signal of frequency f (in Hz), with 0-to-peak amplitude Δi Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts.

To use equation (17), it will be easier to first set ESR to zero, calculate a C value, find an estimate for ESR for that C value at the frequency being used, and then re-calculate the C value with the ESR value.

If we still want to just account for the whole sine wave, we should be able to simply double the C value given by equation (17), since we're considering only the positive or negative half-cycle, but not both, and the other half of the half-cycle of a sine wave is symmetrical and thus encloses the same area (its integral), i.e. the same amp-seconds value, as the first half.

[Not bad for an old guy who had forgotten almost all of the mathematics he learned, which was over thirty-five years ago.]

Cheers,

Tom Gootee

Last edited by gootee; 11th January 2013 at 02:19 AM.
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Old 11th January 2013, 02:47 AM   #1687
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We can see that if Δv - (ESR∙Δi) = 0, in equation (17), then C would be infinitely large.

To understand "why", we can write it slightly differently:

Δv = ESR∙Δi

or

Δv / Δi = ESR

But notice that Δv/Δi is the impedance we would need to stay below, in terms of the impedance that the active devices would "see", if the voltage should never change by more than Δv, for any Δi.

The ESR is the minimum impedance we can get, with a capacitor. So if the ESR is already as large as Δv/Δi, then we have already failed to meet the most-basic need of the power-distribution circuit, which would be required in order to meet our Δv/Δi specification.

That simply means that we need to try to make the ESR be much less than our desired Δv/Δi. Otherwise, the required capacitance value will be excessive.

Last edited by gootee; 11th January 2013 at 02:57 AM.
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Old 11th January 2013, 10:09 AM   #1688
DF96 is offline DF96  England
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Nice to see some real engineering in the forum!

This calculation does for the last capacitor in a PSU the same sort of job which a ripple calculation does for the first capacitor. For some SS power amps they may be the same capacitor, in which case I think the ripple calculation dominates.

Your delta-V is set by the amp PSRR and the required headroom, just like for the ripple calc. By oversizing the transformer you can reduce the cap size. However, it could be that for some amps the gain depends on supply rail voltage so delta-V may need to be kept small to reduce IM.
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Old 11th January 2013, 03:50 PM   #1689
Bare is offline Bare  Canada
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Originally Posted by AKSA
In this I'm reminded of subjective bass response of amps. If you have an amp with a single output pair, bipolar or mosfet, and relatively high Zout, you can actually improve subjective bass response by interposing a small 0.15R resistor into the speaker output line. This promotes overshoot on the voice coil - that is, loss of VC control - but adds a subjectively stronger bass response. Hardly correct in engineering terms, but quite useful when sculpting bass presentation. Perhaps something similar is operating here?


Which is Why the idea of Speaker Wires affecting sound has a small basis in reality :-)
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Old 11th January 2013, 05:38 PM   #1690
DF96 is offline DF96  England
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Quote:
Originally Posted by Bare
Which is Why the idea of Speaker Wires affecting sound has a small basis in reality :-)
With the emphasis on "small".

Of course, if the NFB loop is doing its job then the PSU cap ESR is not seen by the speakers.
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