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Old 13th December 2003, 08:28 PM   #1
satyr19 is offline satyr19  United States
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Default IC sockets and noise

I'm upgrading the op-amp in my cd player's analog stage, and I'm wondering if I should use an IC socket with it, in case I want to swap later.
Does anyone know how IC sockets perform in terms of adding noise or corrupting signals?
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Old 14th December 2003, 11:26 AM   #2
sreten is offline sreten  United Kingdom
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Using a socket is a very good idea if your going to be experimenting.

There are the best quality sockets (gold-plated turned pins)
and the more common not so good sockets.

IMO the best quality sockets will form very good and reliable
connections the first few times they are used, ideally a proper
IC insertion tool should be used.
I don't know what will happen if the socket is re-used a lot.

sreten.
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Old 14th December 2003, 11:52 AM   #3
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If you're going to use sockets, be sure to use the turned pin ones.
They look better, it's easier to install IC's in them, give better contact, don't cost that much more, ...
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Old 14th December 2003, 01:11 PM   #4
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If you do this, it can make sense to fit 2 sockets on top of each other, and use the topmost, replacing that when it becomes troublesome. That way the one soldered to the board get less use, and does not need to be replaced.

But it all depends on how much swapping you intend to do.
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Old 18th December 2003, 05:41 PM   #5
satyr19 is offline satyr19  United States
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Aren't turned pin sockets for surface mounting only? Or am I thinking of something else..
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Old 28th December 2003, 08:47 AM   #6
pieris is offline pieris  United States
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This response is a little late, but the original question was not really addressed, so:

IC sockets will increase the swept area between signal lines and ground. The effects of nearby electric fields (esp from digital signals or AC) will increase as you increase this area.

Additionally, the IC socket will add inductance on all pins and capacitance between pins. This is on the order of nH and pF (fairly small). It will cause the opamp distortion to increase, but with small gain configurations it's probably not noticeable.

A hand waving explanation: inductance is added to all the pins, inductance means delay, opamps work in a feedback configuration, more delay in the feedback loop is going to give you distortion.

One (high quality) socket is probably fine, if there are no digital signals nearby. That said, stacking two IC sockets together, as Havoc recommended, will probably give you a noticeably deleterious effect as opposed to mounting directly on the board.
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Old 28th December 2003, 11:37 AM   #7
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Maybe you could get a tiny little ZIF socket (if they actually exist) like the kind used on an eprom programmer.
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Old 28th December 2003, 01:42 PM   #8
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Quote:
A hand waving explanation: inductance is added to all the pins, inductance means delay, opamps work in a feedback configuration, more delay in the feedback loop is going to give you distortion.
I think you vastly over estimate the added values of sockets! We use sockets in uprocessor circuits with busses running at 100MHz, at it continues to work. So the loads are very small.

IMNSHO most distortion of opamps today comes from output stages not able to supply current. In lots of distortion measurements, the antique NE5532 comes out above as it can hold its spec while supplying current to loads of 1k and lower. While most modern amps their distortion goes off the graph anco the load drops below 50k.
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Old 28th December 2003, 05:04 PM   #9
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if it isn't socketed already, don't add a socket -- mount the device "dead bug" style (i.e. with its feet sticking up in the air) on the bottom of the printed circuit board. just push the pins onto the traces and tack down. this will minimize trace lengths and it will be pretty easy to remove if you don't like the effect.
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Old 29th December 2003, 11:53 PM   #10
pieris is offline pieris  United States
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Havoc, I think I did too (overstated effects of distortion due to added inductance). And your point is well taken that which op-amp is chosen will far outweigh the socket, or lack thereof. However, that a socketed part "works" in the digital world is of course no testament to the effect it might have on nearby analog circuits.

A socket will increase the loop area susceptible to B-fields, which proportionally increases the induced voltage (Faraday's law). And it's pretty noisy inside most consumer electronics devices. Two sockets, of course, doubles the noise. Is the noise significant?

(some math below concludes it is)

Assume a digital signal with a rise-time of 2ns, into 10pF, with a swing of 5V. The instantaneous current (I=C*dv/dt) is 25mA. Now, ground planes are sometimes hard to come by in cheap products, so let's say this trace runs over 2cm of bare board. By the Biot-Savart law (a special case of Maxwell) the magnetic field at a point (our socket) 10cm from the center of the trace is 10nT.

(for perspective, B-fields in the average home are 100-500nT, but most of that is varying at a leisurely 60Hz).

10nT seems small but remember this field is being created and destroyed in just 2ns (dB/dt=5T/s). Faraday's law states that the voltage induced in a loop is proportional to the rate of change of the magnetic flux. How big is our loop? A standard DIP socket adds 0.4cm of height to the chip. A DIP package is 0.8cm wide. We'll use this rectangle. The EMF induced in this loop is, applying Faraday's law, 160uV.

1 LSB of a 16-bit DAC into a full scale of 2V (line out level) is 30uV. Let's say for safety that I was off by a power of 10, that's 16uV or 1/2 LSB of noise . I'm not even counting E-field noise, which is where 60Hz pickup will happen. Keep in mind this noise is in addition to the noise (due to the same flux) that would exist without the socket.

Now, even if all my math is correct, I admit there's still hand-waving in my favor: I've assumed the loop in the socket is parallel to the interfering trace, I've assumed the loop impedence to ground is high, I've assumed certain geometries and electrical characteristics in favor of making a point (but not, I think, outside of reason). I think the proverbial horse is dead now. My conclusion: no sockets.

By the way, I welcome a second opinion on the math. Just google for Biot-Savart's law, and Faraday's law, it's quite straightforward.
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