pcb - Show Pad Nets Seemingly Not Working in Altium

For some reason, the PCB is not showing net names on the pads

I have all connections (rats nests) visible I have checked the "Show Pad Nets" and "Show Via Nets" under the View Configurations > View Options Panel

I have checked layer colors and everything is on

What other setting might there be which is preventing the net names from displaying within the pads/vias on the PCB layout?

I am running AD15 on Windows 8.1 and this is really bugging me!

This is an example screen shot of the PCB environment, you can see that despite the option for "show pad nets" being enabled, they are not showing.

Thanks for the help

This question is not the same as another related question "Show Pads with their Name on PCB altium " which discussed adding silkscreen designators. It is about why the Altium feature "show pad nets" does not seem to be working to show the 'virtual' net names in the PCB tool within the pads

Show Pads with their Name on PCB altium

I have imported a schematic in PCB and placed components. But one thing is missing, its not showing pads names as they are appearing in schematic. It can only show Designator and Comments of component in PCB but not the pads name.

How can I make them appear ?

Thank you

Added screenshot, how to do like this. Like 1 appearing next to pin 1, 2 appearing next to pin 2. But when I add a header, nothing appears, only the entire component designator appears.

Relay "sticking"

We are experiencing a problem where the mechanical switch of a SPST NO relay (COTO 9007-05-01) is not opening back up when the coil is de-energized. Current keeps flowing through the switch until we give the relay a (somewhat gentle) flick.

We measured voltage and current on an o-scope (a Tektronix current probe for the current), and everything appears to be in order: On/Off voltage connected to coil appears good, On/Off current through the coil appears good, on-current through the switch well below rated current, open voltage across the switch well below the rated voltage, no unreasonable spikes, the coil has a fly-back diode installed...

We even replicated our setup with different everything (including relays) and got the same results.

Any suggestions on what to do/check?

Answer

I am in agreement with @BruceAbbott's comment, above. I suspect that although your static load current is well within the relay's capability, the surge current when it closes, probably due to charging caps and/or firing up an SMPS, is welding the contacts. I suspect this because I have seen precisely the behaviour that you're describing, down to the sharp tap on the relay body causing the contacts to release.

My solution was to cause the device that I was switching (that was powered by an SMPS) to soft-start using an LCR network between the output of the relay and the input to the device.

If the cause of your problem is SMPS inrush current, and you have control over the SMPS, you may be able to put it into a "soft start" mode, either because it has the capability and just needs to be configured to use it. Alternatively, try my solution. Apologies, it was 10 years or so ago, I can't recall exactly what I used; a small series resistor (<10R) and large inductor (100uH?) and various smallish caps (1-10nF) to ground sounds vaguely familiar.

motor - 3-Phase Inverter Control with SPWM(Sinusoidal PWM) for 3-phase sinusoidal output

I intend to control a motor with 3 120° phase shifted sinusoidal signals. I already implemented the SPWM to output a sinus shaped voltage, but i am a little bit confused at how should i implement it with my custom design inverter. The inverter has 3 legs/6 Mosfets which can be controlled indivudally.

I will quickly describe what i have in mind and would appriciate any comment/clarification on if it is a good idea or not.

With a 3 phase sinusoidal signal, the outputs are always a permutation of + + - or - - + (momentary sinus values). If you add them up you should get 0. And the inverter legs are HHL or LLH accordingly(Of course LLH or LHL. etc is possible). Lets take the HIGH-HIGH-LOW case :

Here i thought i can identify the different phase (LOW in this case) and control the other 2 with PWM (according to a look up table). What i am confused at is, is it really necessary to seperately control the 3. leg(which should be low ) ? Cant i just turn on the low side switch fully on, while the other 2 are positive and being controlled by a varying PWM duty cycle. This will function as a current return path and since the sum has to be 0, it will automatically take the value it is supposed to take. And in the LLH case, H-leg will be on as long as the other 2 are L and controlled by PWM. This will be alternated with a state machine.

Please let me know if the way i am thinking is correct. I am aware of more advanced methods like SVPWM (Space vector) but would like to stick to SPWM.

power supply - Are normally open contactors virtually safe from failing closed?

Quick question: are normally-open contactors (which I assume are simply high breaking capacity relays) virtually safe from failing closed?

I want to cut off a 400V 3-phase supply using an emergency stop, and I have 2 options:

1. Use a 3-pole emergency stop, and wire the three phases to that button. I don't like the idea of 400V being so close to the user's hands.


2. Wire the contacts of a 3-pole normally-open contactor on the supply line, and wire the emergency stop on the 230V coil.

Three phase is used for heating, not for a giant chainsaw, but the emergency button should still do its job should it be pressed. So is option 2 commonly accepted, and safe enough for an end product? Any norms stating this on which I could rely?

Edit implementing the answer

I am quite keen on the idea of forcing the user to fix a failure in the redundant system, but I'm trying to limit the number of components as well. Is this a good comromise? I found this contactor, which has 1NC and 1NO auxiliary contacts that we can use.

Since the current in the coils is already very small compared to the switches capability, I made them single pole, if that is acceptable.

In particular, I'm not sure whether there is a possibility that the auxiliary contacts do not follow the main contacts (meaning the main contacts may fail closed or welded without the auxiliary contacts staying open when normally closed), could you confirm?

Answer

No, they are not safe from failing closed. No switch is, mechanical or solid-state.

Therefore, an E-STOP or other safety switch requires at least two overrated contacts that operate independently, so that one getting welded does not prevent the other from operating.

For your option #1, you would wire these in series, but it's not recommended because of the reason you stated and because there's no way to tell that one has failed until they both go.

For your option #2, you would wire these as part of two identical circuits, with all relevant safety switches wired in series, and the two channels wired in parallel and kept separate electrically. (no crossover wiring) Then each channel drives the coil of its own contactor, and the contactors are wired in series to control the load.

Also with option #2, you can now create a latching/lockout circuit using the two safety channels and the contactors' auxiliary contacts so that a separate button is required to turn it back on once the safeties are satisfied, and only if both have dropped out. This forces you to fix a stuck contactor before they both become stuck.

Per a comment, here's one possible version of option #2:

^{simulate this circuit – Schematic created using CircuitLab}

Or, if you have sufficiently deep pockets, you could buy a safety rated PLC and do all of this in software with even more fault-checking and detailed diagnostics.

Please note that I am in a different industry now, and so there may have been some legal changes since I left. Check the latest electric codes, OSHA regulations, etc., before trusting this (or anything else really) to an operator.

Questions on Operational Amplifiers

What I've learned so far from Navy school is that operational amplifiers are used as a substitute for transistor amplifiers and they contain three main circuits: the differential amplifier, common collector amplifier and the push-pull amplifier. They are represented by a triangle pointing to the right. My question is, why is the voltage gain \$-\frac{R_f}{R_i}\$

As far as I know, \$R_f\$ provides negative feedback to \$v_{in}\$, but why are we not treating \$R_{in}\$ and \$R_f\$ as part of a voltage divider like a non-inverting amplifier? I can see how they would reason that the incoming current is \$-\frac{v_{in}}{R_{in}}\$ without the voltage divider

Answer

Of course, both resistors act as a voltage divider. However, you most consider the fact that you have TWO voltage sources at the same time (Vin and Vout). Hence, you must apply the superposition rule for calculating the voltage at the midpoint between both resistors:

\$V_{n1}=V_{in}\dfrac{R_f}{R_i+R_f}\$ and \$V_{n2}=V_{out}\dfrac{R_i}{R_i+R_f}\$ with \$V_n=V_{n1}+V_{n2}\$.

Now - because the opamp has a (very) high open-loop gain Aol (approaching infinity) you can set the differential voltage Vd=Vp-Vn between both input terminals to zero. As a consequence, we have \$V_n=V_{n1}+V_{n2}=0\$ which finally yields \$\dfrac{V_{out}}{V_{in}}=-\dfrac{RF}{Ri}\$.

operational amplifier - Gain of an active filter without calculating the transfer function with two OP-AMP's

I need to calculate the gain of the filter without explicitly calculating the transfer function, ie I have to replace the capacitors with an open circuit. The problem is that I've been trying to solve the problem but I do not know how to start. Thanks.