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I am making my own 2GHz DSO and I have question regarding position of my power supply.

Many companies and hobbyists that are designing their own osciloscopes, not include power supply on their motherboards. I didn't find a reason why is that so. I started to ask myself if it's possible because product have some kind of warranty and its cheaper to just replace the power supply. But that doesn't apply to DIY researchers.

Is it possible because of the EMC? Or maybe for some other reason?

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    \$\begingroup\$ Probably regulatory. If you build your own power supply you have to get it approved by all kinds of regulatory agencies, but if you buy one it's already got all those approvals. I'm also wondering how you intend to build your own 2GHz oscilloscope; that's quite a feat if you can actually do that! \$\endgroup\$
    – Hearth
    Dec 5 '20 at 20:07
  • \$\begingroup\$ Best reason is risk management. If you screw up one of those, the other is not affected. Another reason is that one can impose constraints on another- power supply goes betting on certain PCB, analog and high speed on a different one. Bottom line is that there is not really much reasons to have all of them together. \$\endgroup\$ Dec 5 '20 at 20:10
  • \$\begingroup\$ @hearth have you not seen the unpopulated pcb oscilloscope kits out there? \$\endgroup\$
    – Passerby
    Dec 6 '20 at 4:01
  • \$\begingroup\$ @Passerby I have not! I have some doubts about getting 2GHz out of such a thing, though; seems like the AFE would be too sensitive to hand-solder. I can't really think of a reason why, though, so maybe my gut feeling is wrong. \$\endgroup\$
    – Hearth
    Dec 6 '20 at 4:15
  • \$\begingroup\$ @hearth okay I may be a little optimistic about the kits bandwidth but I did see TI have a reference front end for a 2ghz scope. \$\endgroup\$
    – Passerby
    Dec 6 '20 at 4:22
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EMC will be a major factor. There are quite strict laws in many countries now around how much noise a power supply may inject into the incoming mains. Designing a switch mode power supply that meets the regulations in every country is a specialised business.

For that reason, it's a lot simpler for a company to buy in ready-made power supplies than to try to design them in house.

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Let us compute two different NOISE FLOORS, from inductive_coupling by your power supply.

In both cases we'll use the very simple long_straight_wire to rectangular_loop induced voltage:

  • Vinduce = [MUo * MUr * Area/ ( 2 * PI * Distance)] * dI/dT

which for MU0 = 4 * pi * 1e-7 hanry/meter, or MUr = 1 (air, FR-4, copper foil), becomes

  • Vinduce = 2e-7 * Area/Distance * dI/dT

In both cases we'll assume 10dm (4" distance) between the Power Supply di/dT and your front end circuitry.

And let the victim loop area be 2mm by 2cm (thus you have a fine layout, over Ground plane).

For the 60Hz cause, with the rectifier diodes producing 1 amp surges (maybe 10 (TEN) watts of power) with 10microsecond risetimes; for the SwitchReg situation, assume 1 amp peak with 10 nano Second switching times.

Now let use set a noise floor budget of 1 millivolt.

Math:

  • 60Hz case: Vinduce = 2e-7 * 2mm * (2cm /10cm) * (1amp/10microseocnd)

  • 60Hz case: Vinduce = 2e-7 * 2mm * 1meter/1000 mm * 0.2 * 1e+5

  • 60Hz case: Vinduce = 2e-7 * 2 * 1e-3 * 0.2 * 1e+5

  • 60Hz case: Vinduce = 2 * 2 * 0.2 * 1e^{-7 -3 +5}

  • 60Hz case: Vinduce = 0.8 * 1e-5 ~~ 8 microVolts, << 1,000 uV goal

However the SwitchReg case, with 100X the edge speed, will be 800 microVolts at whatever the SWitchReg uses as its FET switching clock.

Thus the SwitchReg (ignoring any fields of your inductors) is a risk.

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