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The bypass capacitor is seen on almost every amp out there but is it really necessary? As in many text book, it serve 2 purpose: 1. Act as short to ground at high frequency which shunt noise from other noise source that couple into the power supply through the power supply trace. 2. Act as a power source for fast transient current at high frequency. But in reality, i see many amps connect the power supply directly to the V+ or V- leg of the opamp, if we want to maximize effect (1), it should be power supply -> bypass cap -> ic legs. So i think most of the time, the bypass cap is there to serve for purpose (2) but, here is the impedance plot of a 0.1uf x7r capacitor from kemet: enter image description here You can see that at audio frequency, the impedance of the capacitor is very high, so when a "high" frequency audio signal need to be produce by the amp, the current should flow from the power supply because it has much lower output impedance at audio frequency, not the capacitor. The capacitor is only good at 1->100Mhz which is far above audio frequency. So what is the real benefit of a by pass capacitor?

Edit: the bypass here is some 100nf or 1uf ceramic capacitor from V+ or V- of opamp to ground. I just want to disscus the benefit of bypass capacitor in audio application, especially with opamp as headphone amplifier. Here is my simulation which take account of trace impedance and inductance, the load is about 5cm from the power supply and is bypass by a 10uf 70mR esr tantalum and 1uf ceramic which is the bypass strategy often seen in headphone amplifier. The blue trace is without bypass and green trace is bypassed. As you can see, the bypass capacitor only has effect from 300khz which is far above audio frequency. enter image description here

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    \$\begingroup\$ Could you add a schematic of where you have the "bypass capactior" Edit, from what I understand, you mean decoupling capacitor? \$\endgroup\$
    – Joren Vaes
    Apr 28, 2017 at 15:32
  • \$\begingroup\$ For low frequency work bypass, or better named de-coupling capacitors, do very little unless the power line itself is noisy. \$\endgroup\$
    – Trevor_G
    Apr 28, 2017 at 15:34
  • \$\begingroup\$ You are not considering the impedance of the supply wires, traces, etc. You are also not considering that "just pepper 100nF everywhere" is the poor-man's alternative to actually placing correct values for their purpose. \$\endgroup\$
    – Asmyldof
    Apr 28, 2017 at 16:24
  • \$\begingroup\$ I have updated my question. \$\endgroup\$
    – Thaitao
    Apr 28, 2017 at 17:08

4 Answers 4

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A typical opamp used in audio has a gain bandwidth product somewhere in the one to a few tens of MHz, so the things usually have quite some gain outside the audio band.

Also, it is instructive to remember the current flows in loops, and that typically the opamps output is referenced to some 'ground return' which is NOT directly connected to one of the opamps power pins (They are often +-15V or so), the bypass caps near the opamp serve to close this loop and minimise the potential for magnetic coupling (proportional to loop area).

A further bit of nastyness is that a class AB opamp will impose half wave rectified current pulses onto the tow power rails as the output switches from sourcing to sinking and vice versa, the local caps help minimise these loop areas too.

Now 100nF was sort of standard for ceramics, back in the day but these days 1uF is readily available in 0603 X7R and is worth using, and wise designers tend to use a local electrolytic as well, a few 10s of uF is useful, do not go for overly low ESR here. Some of us like a modest series resistor before this arrangement as well, it both adds a pole and helps to damp any resonances between the power network and the rather large caps back at the board entry, 10 ohms or so is usually ok.

It is surprising how little series inductance it takes when you have 1000uF across the input to give a resonance in the audio or low ultrasonic region.

I only really got this stuff once I stopped thinking about voltage as somehow being the primary thing and started thinking about current (and its loops) as being at least as big a consideration, for all that the schools seem to concentrate of voltages as being the thing.

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  • \$\begingroup\$ But isn't the current flow from the power supply as it has lower impedance in audio frequency compare to the bypass capacitor and then flow back to its source, which is also the power supply, not the capacitor? \$\endgroup\$
    – Thaitao
    Apr 29, 2017 at 2:52
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If you want to cover audio frequencies you need to use capacitors with higher values than this one. But if you're sampling your audio with a fast ADC, you still want to quash interference at ~MHz frequencies to avoid noise aliasing into your sampled signal. So you will often want to bypass with a sequence of different valued capacitors to cover all important frequency bands.

Because the associated wavelengths are longer at low frequencies, it's not as important to keep the higher-valued capacitors as close to the ICs being bypassed. Often just a few "bulk" capacitors are spread around the board and "shared" between different loads.

Below maybe 1 kHz, if you have local regulation on the board, your power regulator will likely have low enough output impedance that attempting to bypass these frequencies would be superfluous.

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Op Amps run pretty clean since they do not produce switched Coss spike current of logic and have excellent PSRR at low frequency.

The theory is simply current spike peak noise, or ripple current * ESR of decoupling cap. Since OA's are not the offending source of noise, one only needs to worry about long inductive tracks or wires since GBW is ~1MHz whereas logic can be GBW ~ 1GHz with Zout as low as 25 Ohms on 74ALVCxx family.

Thus consider where noise comes from and what BW and how much is needed for clean OA operation at some BW.

One 0.1uF per cluster of OA's may be enough or NOT ... depending on switch inductive relay , motor and CMOS noise.

However if you go have large motor switch current noise, then routing of ground currents and ripple suppression at source of noise is key to better immunity. acontrolled impedance paths, ground plane or twisted pair with CM choke to reduce emanations or conducted noise.

If unknown, then isolates Analog and digital power paths for better ripple reduction.

It pays to learn ESR of each cap type and track impedance and DCR of inductive loads and Coss of Mosfets and CMOS logic. But this takes effort.

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Suppose you have an electronic organ, producing 59Hertz tone. The energy for the loudspeaker must handle that 59Hz, even though the PowerLine provides 60Hz (or 50Hz) without a beatnote. The energy storage capacitor(s) are necessary for that: avoiding interaction between the energy consumption and the energy replenishment timings.

What is resonance of 100,000 uF and 1 meter of wiring inside the organ? radian resonance is 1/sqrt(L*C) = 1/sqrt(1e-6 * 1e-1) = 1/sqrt(1e-7) = sqrt(10)/sqrt(1e-6) = 3.16 / 0.001 = 3,160 radian/second ~= 500Hertz.

How to prevent irksome peaking in the VDD supply of this organ? We dampen. Using Rdampen = sqrt(L/C) = sqrt(1e-6 / 0.1) = sqrt(1e-5) = 3.16 milliOhms, and realizing 6 squares of PCB foil are 0.0005 Ohms * 6 = 0.003 ohms, the need to dampen is achieved in hidden but still effective ways, letting us mostly ignore the need to dampen until we get surprised.

With power == I^2 * R, at 4 ohms and 10 amps, we have 400 watts of 59Hz tone. How much VDD sag occurs between the recharge peaks of the 120Hz fullwave rectified power? Q = VC, I = C*dV/dT, and dV = I * T / C = 10A * 0.008sec/0.1, making the dV sag be 0.8 volts. Depending on Power Supply Rejection of your 400 watt bass-speaker drive in the organ, you may or may not hear beatnotes between 59Hz and 60(120) Hz power.

Now the amplifier, this big (400 watts, inside an organ) is not classA; it is classB, thus when producing a sinusoidal output, pulls faston, fastoff surges from the +-60volt VDD to the organ amplifier. These non-sin transients will make the +-60V rails ring.....unless dampened. The big capacitors (0.1 farad) have internal inductance (plus that external 1 meter of wiring) and ring at higher frequencies. To provide power at higher frequencies, smaller capacitors are placed in parallel, and they ring at their own 1/[2pi*sqt(LC)] frequencies, and must be dampened. That dampening requires progressively larger and larger resistors.

A microcontroller, clocking at 100MHz, is even more challenging. The ADC on the silicon needs quiet, << one millivolt for 12-bit ADC to achieve 12 bits; PI filters are used, to exclude external trash.

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