I want to supply a DSP with 1.2V. This DSP needs 2.6 Amps of current at full load. The minimum supply based on the electrical specs of this DSP is 1.16V, which means that the maximum voltage drop caused by power planes, traces and connectors should not exceed 40 mV.

In my case, I found it very hard to achieve this since the distance between power source and DSP is about 8000 Mil (~ 20 cm) and this supply passes by two connectors which add 100 mOhms, so the drop is 260 mV (100m x 2.6A) without counting in planes impedance. I drew a simple schematic for my case shown in the next image:

Simple Schematic to show the problem

My questions are:

  • Is the total distance only 20 cm? or should I add the return so that the actual distance is 40 cm ? ( Much worse :( )

  • How can I solve this issue? knowing that the distance between source and DSP can't be less than 20 cm. Should I add another regulator beside the DSP? or is it better to generate a slightly larger voltage to compensate this drop? (there is other components the need 1.2V supply and are at different distances from the DSP).

  • How can I calculate the plane impedance, shown in the above image as R(Plane)?

# Edit 1:

Regarding point 1, ok, the total distance now is unfortunately 40 cm.

I thought of a solution to reduce the connectors resistance, which are the main factor of high resistance. According to connectors data sheet, the resistance of the pin is 25 mOhms, I have extra free pins, so I'll use 8 Pins to transmit the 1.2V so that the is now divided by 8, but the question now is, I don't know if this resistance is for the pin only or is it the total after mating? and after mating should they be treated as series or parallel resistors?

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    \$\begingroup\$ For number one, the DSP's GND is gong to be affected by the voltage on the return line, so it's going to be that much higher. That in practice means that you'll need to have higher voltage on the input. The whole situation simply screams "regulator next to DSP" to me, but if you expect noise problems near DSP, a switcher is going to be hard to get design and a linear one is going to be hard to find. \$\endgroup\$ – AndrejaKo Jan 22 '12 at 13:45
  • \$\begingroup\$ As for the plane impedance, you could try something like this if you have the board: Get a constant current source, set it to say 1 A, connect it instead of the power supply, short the DSP's Vcc and GND pads and measure voltage drop between the source and Vcc pin and then between the GND pin and source's other end. From this it should be easy to calculate the resistance and most multimeters should be better at measuring low voltages than measuring low resistances. You can make a simple constant current source using say LM317. \$\endgroup\$ – AndrejaKo Jan 22 '12 at 13:49
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    \$\begingroup\$ If you can't measure the resistance, you could try to calculate the resistance of the planes by finding one of the copper resistance tables on the Internet and then calculate the volume of the copper (or surface area, depending on the type of table you get) on the plane and then multiply it by copper's specific resistance \$\endgroup\$ – AndrejaKo Jan 22 '12 at 13:52
  • \$\begingroup\$ Andrejako, that sounds like a good answer to me. \$\endgroup\$ – Kellenjb Jan 22 '12 at 14:09
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    \$\begingroup\$ Connection resistance is for mated pin and socket. If you use N of them then resistance goes down by a factor of ABOUT N. \$\endgroup\$ – Russell McMahon Jan 22 '12 at 14:44

In general, trying to push final regulated power any distance is not a good idea. In your case it clearly won't work. Yes, the return path adds to the total resistance since it is in series with the load. It is strange that you have connectors in the positive supply but not in the ground. If this is a fixed installation, then why not solder wires from one end to the other?

A better way to deal with the need for distributed regulated power, especially at low voltage and high currents as you have, is to distribute a higher roughly regulated voltage and make the final tightly regulated voltage locally. This does two useful things:

  1. The drop in distributing the higher voltage won't matter since is will be regulated anyway to the final voltage. You do have to make sure the voltage at the other is at least the minimum required for that regulator to work correctly, but that headroom is usually easy to build in.

  2. In the case of local regulators being switchers, the higher voltage will have less current, which means it will also have less voltage drop accross the distance, with less power wasted and heat that must be dealt with.

So where does your 1.2V supply come from? You probably have some higher voltage with a buck converter somewhere. Send that higher voltage over the distance and put a buck regulator right at the DSP. Note that this relaxes the requirements on the 1.2V supply on the main board. Two smaller buck regulators will still be more expensive than one larger one, but allowing both to be smaller will help somewhat. It also distributes the heat from any losses, which usually makes that easier to deal with.

Added in response to your comment:

If you really really can't put a local regulator by the load, then the next best thing is to have a sense line coming back. This line reports that actual voltage at the far end back to the regulator on the main board. This voltage is used as feedback so that the voltage at the far end is what is regulated. The voltage at the regulator then will automatically be higher as needed to overcome the voltage drop on the way to the load. The sense line doesn't experience these voltage drops since it has very little current flowing thru it. It is just a voltage feedback signal.

If the ground connection also can have significant voltage drop, then it gets more tricky. Sometimes you use two sense lines and treat them differentially at the power supply. Sometimes you assume the forward and backward voltage drops will be about equal and add a little bit of gain in the sense circuit. Sometimes you just set the output of the supply a little higher to compensate for the nominal total voltage drop and not try to actively regulate around it at all.

  • 1
    \$\begingroup\$ Thanks a lot Olin, you showed me a mistake I did in drawing the schematics and consequently a mistake in calculations! I should add the same resistors on the power line to the return path and recalculate. I agree with you that it's much better for the final regulated supplies to be close to the load specially in these low voltages, but the mechanical constraints forced me to split on to two boards, and there is no enough space to place power regulators near the loads :/ \$\endgroup\$ – Abdella Jan 22 '12 at 14:52
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    \$\begingroup\$ Wow, That's it! My regulator has this sense line, I'll use it. Is there any optimum way to connect this pin to the DSP? because the DSP has several 1.2V pins, should I connect it to farthest pin? I checked the datasheet of the regulator but there was no notes about the connection to complex ICs. Thanks a lot! \$\endgroup\$ – Abdella Jan 22 '12 at 20:39
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    \$\begingroup\$ @Abdella You should connect all of the 1.2V pins together. You then connect the power, and sense lines to them \$\endgroup\$ – Brad Gilbert Jan 23 '12 at 20:32
  • \$\begingroup\$ @BradGilbert The switching regulator I am using has +Sense and -Sense and they advise that the +Sense should be connected close to load, and -Sense connected to GND close to source. The problem is that there is an FPGA also that needs 1.2V, so both the DSP and FPGA are connected to a 1.2V plane. And each of them have tens of 1.2V pins. That's why I can't determine what is the best connection of this sense signal?! \$\endgroup\$ – Abdella Jan 24 '12 at 8:51
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    \$\begingroup\$ @Abdella: If all these various pins are connected to a plane, then there will be little voltage different between them. The main point is to compensate for the voltage drop accross the connectors. Connecting the sense line anywhere in the plane on the final board (after all connectors) should be good enough. Make the connection somewhere near the middle of all the pins, but don't worry about it if that's too inconvenient. The last few mV shouldn't matter as long as you've taken care of the 100s of mV due to the connectors. \$\endgroup\$ – Olin Lathrop Jan 24 '12 at 13:12

Connection resistance is for mated pin and socket. If you use N of them then resistance goes down by a factor of ABOUT N.

You really want the regulator near the DSP. If you have two connectors and they are the main resistance (as you say is the case) then they will vary in resistance with circumstance, age, temperature and more and give an uncertain result.

Clearly if the connectors add 100 milliohm and you have 2.6A then you get 260 milliVolt drop. IF 40 mV is max tolerable voltage then you can add infinite return backplane and still will be over spec by 260/40 ~= 6.5:1. You would need at least 6.5 parallel pin-pairs to reduce that connector only voltage to an allowable level and then have the rest of the circuit and the return path to deal with. IF the 50 milliohm value is in fact a typical average value then you have an almost intractable situation. If there are an equal number of connector at 50 miliohm in the return path then the problem simply becomes impossible.

["Impossible is nothing!" if you make certain sports shoes, but is just impossible here. ]

If you cannot bring the regulator to the DSP then a workable solution is to use remote or "Kelvin" sensing. ie run a voltage sense line from regulator to load that carries no current and adjust the feed voltage to suit. While this is simple to do, you obviously want the sense circuit to NEVER go open circuit (as the voltage will rise to attempt to compensate) and you have to deal with noise etc in the sense circuit. Not hard, but ... .

  • \$\begingroup\$ Thanks Russell, how can I know from the datasheet that the mentioned pin resistance is for mated parts? I can't find it stated clearly, or is this a known way of mentioning it in datasheets? \$\endgroup\$ – Abdella Jan 22 '12 at 20:43
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    \$\begingroup\$ @Abdella - connector contact resistance only makes sense when considered as a connector pair. ie into the contact - across the interface to the other contact - out onto the other PCB. t's the actual board to board transition point whose resistance is being specified. The body of the connector's resistance will be low compared to the intercontact resistance. \$\endgroup\$ – Russell McMahon Jan 23 '12 at 13:52
  • \$\begingroup\$ Yes Russell, you were right, I contacted the manufacturer and they replied as you said. Thanks again. \$\endgroup\$ – Abdella Jan 23 '12 at 14:43

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