Circuit....DId I mess up here?

[Falcon69]

yes, the grounds are tied together.  I remember seeing a problem with that external circuit operating as you have said, and tieing the grounds together solved that problem.

So where would the TVS diode go?

[Falcon69]

oh yes, R5, saw that.  I will fix.  My simulation program shows it correctly, but I drew the schematic wrong.  I will fix.    thanks

[Falcon69]

oh, i need to mention, that some circuit sthat the mosfet wll be connected to, may not have a 12v optocoupler, but a 5v instead.  If putting a 12v TVS diode, will it still work?  And you said the 12volt TVS diode between the source and gate, correct?  In simulation, I don't have TVS diodes to work with, but when placing a Zener Diode that way, I can't get the mosfet to switch the Optocoupler.

[Falcon69]

Been doing some reading.

Is this what Rufus means to do for the transient voltage protection?

I placed two diodes on the gate of the mosfet. D4 and D5 in the schematic below.  Would that be enough to protect the circuit? That along with D1?

[Rufus]

Been doing some reading.

Is this what Rufus means to do for the transient voltage protection?

I placed two diodes on the gate of the mosfet. D4 and D5 in the schematic below.  Would that be enough to protect the circuit? That along with D1?

No they make it worse.

Consider you are installing the switch, it is mounted on the equipment and because you are wearing nylon socks you left the whole circuit with 500v of static charge with respect to earth. When you plug it in if the OUT pin connects first you earth the MOSFET source while the gate still has 500v on it and the gate gets destroyed. Pretty much the same if the IN pin connects first. Possibly the same problem if your 'other circuit' gets plugged in.

A TVS diode (which is just a fast zener) cathode to MOSFET gate anode to MOSFET source will protect the gate. The zap then gets passed to the output of U3 which is more robust than the MOSFET gate. If you bump up the value of R5 most of the zap energy will be dissipated in it giving the IC an easier time.

Under normal conditions the gate will never be negative or more than 5v positive with respect to the source so a 12v TVS between them will never conduct and have no effect on circuit operation. 12v because it is more than 5 and less than the 20 the gate can stand.

Crap like this is why you will almost never see a pair of MOSFET pins on a connector that plugs into the rest of the world.

If this thing isn't too cost sensitive and doesn't need to be especially fast then you would be better off using an OPTOMOS relay for the output which really does look like floating switch contacts.

[Falcon69]

Never knew about the OptoMOS.  Very Cool, however, the cheapest Mouser has is $1.24 each, too expensive for my application.  In your honest opinion Rufus, would the 12V TVS be just fine? Or would it be better to go with the OptoMOS? Perhaps later, I could use those in another application. thank you for showing me those.  Very cool device, but out of the price range for this project.

I remember reading somewhere, that You do not want to put too many Optocoupler devices in series.  Putting an OptoMOS in there, when the device I am trying to switch is an Optocoupler, could be too many opto devices in series.

I could do the TVS Diode, no problem if that is good enough.  Any suggestions as to which type?  Just an operating voltage of 12volts?  Bidirectional? Unidirectional?

I found these two...

http://www.onsemi.com/pub_link/Collateral/SD12CT1-D.PDF
http://www.mouser.com/ds/2/308/ESD5Z2.5T1-D-241596.pdf

Which in your opinion would be better for this application?

Again, thanks for all the advice.

[Falcon69]

Would it also be a good idea to put one on the CNTRL wire as well? Cathode to CNTRL and anode to ground of circuit?  That way, all inputs are protected?

Maybe a 5.1v or 5.6v one?

[Rufus]

I would go with an OPTOMOS  relay.

Like VO1400AEFTR $0.92 10 off and $0.76 100 off from Mouser and you save the MOSFET and TVS diode.

I would go with that because it will actually work switching 100mA and up to 60v high side or low side even AC. 

Your switch circuit and 'other circuit' don't have to be referenced to each other. There will be no issues connecting them in parallel or series for wired logic functions. The only down side is they are a bit slow but the VO1400AEFTR is pretty good at 0.5ms max switching times.

The CNTRL input doesn't need a TVS. The chip can take the zap because all the energy goes into the high value resistor in series with its input.

[Falcon69]

Okay, thank you Rufus.

I will look into it, but a SOP-4 package i think will be too big for the tiny circuit board I have to work with.  The molds for the casting resin over the board have already been made. SO I can't change the circuit board dimensions.

[Falcon69]

Hmm, if I am reading the data sheet correctly on that OptoMOS Rufus linked.  It is a normally open, which means when no power is supplied to the led in it (lighting the LED), then the mosfet is open?  If that is the case, I need to redo the circuit so the XOR gate puts zero volts out when the green LED in the switch is on.  Or, I need to find an OptoMOS that is normally closed?

Also, the data sheet shows a resistor going to a source VL...what is the VL?

http://www.mouser.com/ds/2/427/vo1400ae-244774.pdf

[Falcon69]

oh wait, maybe NO is what I want.  The mosFET in the OptoMOS let's current/voltage flow when the LED is lit, correct? Meaning, when the Logic gates delivers a 1 to the input of the OptoMOS, the OptoMOS switches on, allowing the current to flow.  Correct?

[rs20]

oh wait, maybe NO is what I want.  The mosFET in the OptoMOS let's current/voltage flow when the LED is lit, correct? Meaning, when the Logic gates delivers a 1 to the input of the OptoMOS, the OptoMOS switches on, allowing the current to flow.  Correct?

When current is flowing through the LED, current is allowed to flow through the mosFETs.

You don't need separately manufactured NC or NO solid state relays; you can just change how you drive them:
If you want a high voltage to make current flow through the LED, connect the LED from your control line down to ground.
If you want a low voltage to make current flow through the LED, connect the LED from your Vcc down to your control line.

[Rufus]

There is no 'input' it is an LED. You can drive it against 0v or 5v to choose which logic level causes current flow, just like the other LEDs in your circuit.

The data sheet shows the circuit used to determine switching characteristics. VL and IL are specified.

[Falcon69]

okay

It looks like the forward voltage is 1.1v TYP.,  the forward current is 0.3 - 3.2mA to switch on.  Using ohms law with forward LED voltage of 1.1v  and 5v source (logic Gate) I would need a resistor between the value of 1218.75 (3.2mA) and 13K (.3mA) to successfully switch on the mosFET. Data sheet shows Typical is 1mA to switch it on which is 3k9 resistor to go between the logic gate output and the OptoMos LED input.  So that is what I need to purchase for the resistor value and all is well, correct? (I am still learning how to read data sheets).

[rs20]

okay

It looks like the forward voltage is 1.1v TYP.,  the forward current is 0.3 - 3.2mA to switch on.  Using ohms law with forward LED voltage of 1.1v  and 5v source (logic Gate) I would need a resistor between the value of 1218.75 (3.2mA) and 13K (.3mA) to successfully switch on the mosFET. Data sheet shows Typical is 1mA to switch it on which is 3k9 resistor to go between the logic gate output and the OptoMos LED input.  So that is what I need to purchase for the resistor value and all is well, correct? (I am still learning how to read data sheets).

Not quite -- it's saying that some VO1400's will turn on with 0.3 mA, others will require a whole 3.2mA, but the typical one will turn on* with 1mA. To make sure your circuit works reliably, you should supply 3.2mA at the very least. The "maximum" means "the maximum value that we 'guarantee' the part will have", not "maximum allowable current" (which is 50mA). For example, you can see the turn-on times further down the page are specified for an LED current of 10mA.

Similarly, you should ideally assume the worst case for the LED forward voltage as well: 1.4V, not 1.1V. If you choose your resistor to supply 3.2mA to a 1.4V LED, and then round the resistor value down to the nearest E12 value, that would be a bare maximum value for reliable operation.

* Albeit very slowly: that row is talking about how much current you need to supply before the part switches on after 10ms -- further down the page is gives a turn-on-time of 52us typical (200 times faster) for an LED current of 10mA.

[Falcon69]

ahh!  Okay

Makes sense now. Does that go for all data sheets?  I can apply this knowledge you just gave me to any data sheet I see like that?

Hmm, I am trying to reduce the amount of mA the switches draw.  I am not sure if the switching times is important.  It is for application of machines, and the switches will make the machines stop if triggered.  But, If we are dealing with Micro Seconds here, I don't think it makes too much of a difference as machines will take longer then that most likely to actually stop.

What do you think?  Shoot for the faster time and use 10mA current, or stick with the typical 3.2mA.  I am guessing that the breakout boards probably don't even swtch that fast, but I could be wrong.

[Rufus]

the forward current is 0.3 - 3.2mA to switch on.

3.2mA is the minimum that will turn on all parts and only just turn on. You should probably aim for 10mA which is the current at which data sheet on resistance and switching times are specified. Guess at 1.2v forward voltage and pick the nearest standard resistor value (390R) it isn't at all critical.

If you want to reduce current use a better red LED. High efficiency or super bright - the one you have seems to need a ridiculous amount of current.

[Falcon69]

Great, thanks!

Now, one last thing, the TVS diode. Needed or not needed, or not really needed but not a bad idea to put on anyway?  It is D1 in the last schematic.  If yes, 5.6volt? 5.1 volt? 5 volt? Bidirectional? Unidirectional?

[Rufus]

Now, one last thing, the TVS diode. Needed or not needed, or not really needed but not a bad idea to put on anyway?  It is D1 in the last schematic.  If yes, 5.6volt? 5.1 volt? 5 volt? Bidirectional? Unidirectional?

If a TVS 5v. TVS diodes are specified by working voltage, break down voltage is higher. Unidirectional.

Not really needed. Zaps on a supply line couple to the other through the decoupling capacitors so circuits between the supplies don't really see them. It might save the circuit if someone sticks 12 or 24v on the supply or reverses it, but, if there is any significant current available it will just blow up the diode then the circuit. You need a fuse as well to protect against that.

[Falcon69]

Awesome, thank you Rufus and others. You have been a great help!  Now to redesign the PCB and get them ordered. It will cost me alittle more, because of that OptoMOS, but, should be an overall better circuit.

I will leave the TVS off, since it is not needed.

Thank You very Much.  I'll post a pic of the finished product in a couple weeks.

Jason

[rs20]

I can apply this knowledge you just gave me to any data sheet I see like that?

All components, like all manufactured goods, have some item-to-item variation. So, the "min, typ, max" thing is extremely common, and yes, you should always be designing to the worst case.

[Falcon69]

Well, I managed to get all the components, including that big SOP-4 package for the OptoMOS, onto the little tiny circuit board. It measures 21.36mm x 23.28mm.  That was some work.  the auto router for DipTrace really sucks. I did it all manually.

[Falcon69]

I've added some TVS Diodes.  Will this work correctly?

I've also changed the chips to 74HC's. I was told it was bad having the other chips.

[Falcon69]

Hmm, I'm going to have to eliminate D3 and the cap just above it.  I don't have the room on the PCB

[Falcon69]

Got the boards in and a couple assembled. They Work!

While testing, I accidentally touched the control wire to +5volts.  WHAM!  The red LED stayed on and the Hall switch wouldn't switch.

What chip did I fry?  Also, If I put a diode on the control wire, cathode out, anode toward circuit, would that solve the problem if it happens in the future?  One like this maybe?  http://www.mouser.com/ProductDetail/Vishay-Semiconductors/BAT54WS-E3-18/?qs=sGAEpiMZZMtQ8nqTKtFS%2fD7QH7JJs%2flZ9YSVt88cA%2fQ%3d


Oh, and it is the circuit above except that D3, D4, D5 and the cap above D3 is NOT there.

FYI, on a side note.  I am glad I'm not assembling thousands of these.  Those little 0603 chips are hard to work with and that 28awg wire as well.