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6

You want as solid ground plane as possible, so you should have the ground on bottom and preferably have no other traces there. For better copper balance you should also have the copper pours on the top layer to fill the empty spaces. If there isn't any better use for it, connecting the pours to the bottom ground plane with vias is a good choice. Though this ...


6

Some comments, My first concern is about the RC filter on the opamp inputs (LMC662), I use classic resistors 4.7Mohm, useful or useless as the probe is now differential, should I use other kind of resistors? 2.2nF capacitors are ceramic. Being your circuit mainly DC for your lab, there is not any specific reason to care about the material of the ...


3

Yes, increase the value of R2. You can't use a simple voltage divider if you don't have a common ground. If you are asking how to "achieve" a common ground, you will need to tell us much more about your system.


3

A copper pour amongst a bunch of component traces is NOT the same as a ground plane. This is because the whole point of a ground plane is to provide currents the shortest, lowest inductance (smallest loop) possible path. This does not happen in a copper pour riddled with component pads and traces since the ground/return currents must take the long route ...


2

I don't see any need for the ground plane in the matrix area. It could serve some purpose for reducing noise, but likely there is no measurable difference. A strip under the USB is probably the best you can do in your case. Make it at least one trace width wider than the diff pair, the wider the better. Going for 4 layers would be the best choice, but if ...


2

There are two things to consider here, ESD and wireless signal strength. For ESD, it can be good to have a metal chassis surrounding the device going to ground to prevent discharges from humans reaching the device. The problem is the antenna's signal strength could be reduced from having a ground plane around it (if the metal chassis is around the whole ...


2

Most of this only becomes critical in 2 categories: High precision, e.g. measuring micro volts High speed, e.g. 100+MHz signals You need to start thinking of your signal paths in loops, there is your signal, and also its return path, either through a power supply, ground, or some other signal line for differential signals. You want this return path to ...


1

Assuming what you want is a Polygon Pour (ground plane around the signals/ components)... The procedure is the following: Go into the "Polygon Manager" When you are in this dialog box you can then make "New Polygon from" And I usually like to create it from the board outline, which will create a polygon pour over all remaining space on the selected layer....


1

The input resistors don't matter much (perhaps the tolerance does- let's assume they're 1% low tempco types) but the capacitors are a matter of some concern. Keep in mind that a mismatch in the RC time constant of the inputs will correspond to a degradation of CMRR (Common Mode Rejection Ratio- the measure of how accurately differential your circuit actually ...


1

Based on the transducer manual, you should just connect it to case (earth) ground. It's probably just for shielding. As for the voltage to current signal converter. I don't see any reason it would need an earth connection. Your power supply negative should be tied to earth somewhere anyway. One last thing, with a component this expensive it's reasonable ...


1

The optocoupler will have a spec over forward current, the recommended current through the photo diode when it is active. You need to adapt the resistor accordingly. It's just Ohm's Law from there, so if the spec says IF=20mA and you have 12V, you need a 600R resistor. The whole point of the optocoupler is to isolate the systems. If you don't need to do ...


1

That ground is actually an error in the schematic. It should be a connection to VCC (logical 1). The intent is that, when the SaH input is very close to analog ground, all of the comparators will output a logical 0. To activate the appropriate condition in the priority encoder (that is, the first row in the truth table), the D0 input needs to be 1. ...


1

Are these equivalent circuits? Not really. The resistance between G and N on CCT#1 is likely to be very low. It may be less than an ohm. The resistance between G and N on CCT#2 is likely to be much higher. With a simple rod hammered into the ground, it could be anything from a few tens of ohms to a few kilohms. Dirt isn't a good conductor, especially ...


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