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I am using a motor driver L6235 from ST:application note.On page 6, it gives me a layout suggestion: enter image description here

it says that the SIGNAL GND and POWER GND should be connected in a single point: the negative part of C6. I think the reason is that signal and power part should have the same voltage reference and a single port has enough impedance to seperate noises from power GND. Is it reasonable?

And my question is that what if I have three L6235 in one system. They have the same logic supply and power supply.And if I have only one single point for POWER and SIGNAL GND, it will have following drawbacks:

enter image description here

and if I have multi-point connections it will have equal voltage reference, but the impedance between POWER GND and SIGNAL GND will be small. the noise may go to SIGNAL GND. enter image description here

So, how do i make choices?

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Given that you want three motor controllers I have taken the liberty of re-drawing the circuit from ST: -

enter image description here

Note the wire in red from the logic supply going directly to the star point on the right of the circuit. You'll need to ideally do this for controlling three motors from the same MCU.

The diagram you have for the three controllers in parallel is better than the three controllers in series and using the start point ideology is better still. Therefore, with reference to the modified ST circuit, all three motor controllers should be conected at the star point and the MCU also.

To prevent further issues with the motor drive circuit I'd feed MCU control pins to each of them through 1k resistors too. Each motor controller should utilize a local earth plane for local GND connections with this earth plane making one connection into the previously mentioned star point.

This may be regarded as overkill but if you want it to work faultlessly I'd take this route.

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  • \$\begingroup\$ why do you think i should'all three motor controllers should be conected at the star point and the MCU'?i think they have the same digital signal gournd. How can a 1k resistro in serise prevent further issues? \$\endgroup\$ – oilpig Jun 16 '13 at 1:19
  • \$\begingroup\$ Star point makes sure that there are no motor currents flowing through the digital 0V either to the GND connections on the chip or the MCU. You original circuit shows this - I've just extended the idea for multiple motor controllers and a common MCU. The 1k resistors protect the MCU should the FETs kick voltage through their parasitic capacitance during switching. \$\endgroup\$ – Andy aka Jun 16 '13 at 9:22
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When thinking about how to ground things, I find it helpful to think about where currents flow. Remember, ground isn't magic. It's just a copper conductor like every other copper conductor in your circuit.

Also, remember that circuit means you can start at a point, follow a path, and end up back at the start without retracing your steps. Current only flows in circuits. There are a fixed number of electrons and other charge carriers in your electrical device, and none enter or leave. You can only push them around in circles. It's easy to forget this when we use ground and power rail symbols to simplify our schematics, but if you draw them out as they really are, it's clear that current always flows in a circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

current flows in a circuit, through the battery, even when we don't draw it

If you have a large current coming from the power supply, then you also have a large current going back to the power supply through ground. The goal of a good grounding scheme is to anticipate where these currents will be, and keep them away from places that will be problematic. Wherever a large current is flowing, there will also be a change in voltage, because your ground will have some resistance and some inductance too. Since your signals are usually defined by a difference in voltage from ground, if you change "ground", you also change the signal. That's problematic, so keep large currents away from your signals.

If you connect your signal and power grounds in only one place, then you know all the return current for your power devices must go through that one point. For every unit of charge that enters the power devices, another unit of charge must leave it at this one point where the grounds connect. This makes it very easy to understand where the currents will flow: you know they will not be flowing through your devices that measure the sensitive signals.

If you connect the grounds in two places, it becomes much harder to say for sure what will happen. Where the currents flow now depends on the relative inductance and resistance of the multiple paths and so on. Now maybe you are sharing a current return path with something else, and as that something else switches on and off, your ground moves up and down. Of course, having no other reference, you can't tell the difference between your ground bouncing up and down and all your signals bouncing up and down, and the result is noise in your signals.

So, when deciding how to connect things, here's what you do. Ideally, all the power supply connections, both of them, aren't shared with any other device. Everything gets one wire, not shared with anything else, running directly to the + side of the power supply, and another one to the - side. You can then be sure there are no currents, and thus no changes in voltage, except the ones you cause. When this isn't feasible, ask yourself: if I share this path with something else, what currents will it cause? Will that be a problem?

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  • \$\begingroup\$ I think some of your comments are not right, ‘If you connect your signal and power grounds in only one place, then you know all the return current for your power devices must go through that one point.’POWER supply and SIGNAL suplly are seperate in my schematic so they have own return path. Connecting in one point can not lead current from power devices to go through this point. Some one tell me that we do this for the same voltage reference but i do not know the reason. \$\endgroup\$ – oilpig Jun 16 '13 at 1:04
  • \$\begingroup\$ @oilpig if they are connected at only one point, then through what other point could the current go? The current won't jump through the air, or teleport. \$\endgroup\$ – Phil Frost Jun 16 '13 at 1:39
  • \$\begingroup\$ they are connected at one point in the schematic, but they come from the two different ouput of one DC power supply. so, most of power current will go back through its own GND to the GND of DC power supply and the rest of current will through this connecting point, signal GND and finally to the GND of DC power supply. \$\endgroup\$ – oilpig Jun 16 '13 at 4:28

protected by markrages Jun 14 '13 at 3:36

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