Thread: Opto isolator question
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27-06-2019 #1
I'm hoping someone can point me in the right direction. I have an axbb which is great btw, I also bought a cheap 5 axis Bob to connect to port 3 to give me more inputs for the control panel. I thought I would use the additional outputs to light the control switches to show their active state and bought a cheap 4 channel opto isolator for the job. The Leds on the optoboard switch on when I activate the output but I can't seem to get the switch lights to work on the output side of the opto? I connected the output side G to -0v(24v) and the output side to the switch bulb with the other side of the bulb to +24v. Is this correct? Also I checked continuity on both grounds on input and output sides and they seem to be connected....is this right?
Cheers for any help you would like to give
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27-06-2019 #2
I've never used this board but going off normal terminology I would guess that you need to connect your outputs to the U1, U2, U3, U4 terminals and the ground to the G terminals (you may need to common these up if your lights aren't). I also suspect that the 4 jumpers link the input side ground to the output G on a channel by channel basis, so pull thes for proper isolation. And check that your lights don't require too much current, opto's do have a finite current rating.
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28-06-2019 #3
which breakout board do you have ?
a photo or link will help to find out what type of input the BOB has
going by your photo of the opto-isolator board it looks like the opto-isolators photo transistor collector is connected to the output terminal (U1, U2 ,U3 ,U4 ) via a resistor
the photos resolution is too low to read the value
can you either post a photo that shows the 3 numbers printed on the resistors or just say what the resistors are in your reply
John
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28-06-2019 #4
Your thinking is broadly correct, but you don't understand Opto-Isolators. Your chance of success lies with the type of illumination provided by the switches.
In operation, the internal LED translates current to light, and the internal transistor translates light to (essentially) current. The device has a characteristic called Current Transfer Ratio - which for an EL817 (cheap) device (and given the markings on the board a likely candidate for the devices you have) is specified between 50% to 600% (the variance is depending on manufacturing deviations, temperature, etc). Let's imagine a CTR of 200% - not unreasonable. Now, examine the resistor on the input to the Opto Isolator on the driver board, and what looks like a series-connected LED for status-indication. The resistor is likely to be rated to provide a forward current through the Opto Isolator between 1mA and 5mA (once upon a time likely higher, but modern LEDs have better efficiency and the trend is to reduce the forward current) - Let's say 2mA. With a CTR of 200% that means you can only expect 2mA x 200%. = 4mA through the collector/emitter junction of the transistor in the opto-isolator. Now, if your 24V panel indicator/switch is an incandescent lamp I'd expect the operating current to be around 50mA or so - so you're not going to get the required current to illuminate the switch. Additionally, there is a series resistor on the collector to the opto-isolator which is going to limit the maximum current available through the device.
You've a chance of this working if your indicators are, themselves, LED indicators, as these are likely to operate with the relatively low current provided by the device.
A better solution is to wire up any number of NPN transistors, rather than use the opto-isolator board, or even use these after the opto-isolator board if you actually require galvanic isolation (unlikely) - happy to explain further if you want to pursue this option.
***FURTHER INFO***
An image of this board from eBay shows the LED current-limiting resistor is 1k, in series with the on-board status LED, so you have the forward voltage drop across both devices (call this 3.6V), so if driving from 5V (I'll be generous and ignore the drop across the driving device) that gives 1.4V through 1k resistor, or 1.4mA forward current. Now, on the collector there's a 3k3 resistor - which means that at 24V across the C-E on the isolator's transistor, you're limited to 24/3300 = 7mA, before considering the CTR of the device. Realistically, given the LED forward current limiting I'd work on switching no more than 1mA reliably.
*** FURTHER FURTHER INFO ***
I'm concerned of your wiring in the image, and I'd agree with VoiceCoil's post - If the link-options (red jumpers) provide a shorting link from the opto isolator's emitters to the common ground from the input-side, then you need to be wiring your indictors to the upper terminal of each channel - the otto-isolators will be switching the ground level - so the other side of the indicators would be wired to +24V. But for the reasons above this is still unlikely to work.
MikeLast edited by Doddy; 28-06-2019 at 08:03 AM.
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28-06-2019 #5
found this circuit of the 4ch opto-isolator board online
looking at this photo it looks like the resistors are 1K (marked 102) and not the 3K on the circuit diagram !!!
John
PS
once the details of the breakout board are known it will be possible to workout if the 1K resistor connecting the photo transistor to the output terminal will stop the BOB seeing the output switch
PPS
example
two possible ways of connecting the opt-isolator board to a C11G breakout board
revised diagram for your BOB can be made once I have details of the BOB
with the 1K resistor on the opto-isolator board the opt-isolator board limiting the current
I think your going to need to add some form of lamp driver to operate the bulbs on your iluminated switches
an octal darlington array - possibly ULN2803 should work
ULN2803 pdf data
what current does the bulbs take ???Last edited by john swift; 28-06-2019 at 05:58 PM. Reason: add PPS ref to lamp driver
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28-06-2019 #6
Wow guys, I wasn't expecting this much info when I got back! :) it seems I have forgotten stuff since I last dabbled in electronics! I once made a solenoid gun recoil for mame powered via a Parallel port using opto isolator but I forgot about using mosfets. You are all right, I was expecting the opto to act like a switch...that is off or on and it was rated upto 36v on the output side so just assumed I could drive a 24v bulb from the 24v power. Bugger!
I also don't really need the isolator for the bulbs!
I also just found the leftover mosfets but I'm thinking they may be a bit overkill for a bulb. I wonder how hard it would be to fit a LED into the switches?
Cheers guys
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28-06-2019 #7
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28-06-2019 #8
looking at the photo of the opto-isolator board
it has 3K resistors fitted as shown on the chinese circuit diagram
so the output current is even more limited
with a 3K or 3.3K resistor between the gate and source of the Fets you have you could wire the opto-isolator to pull the gate upto +12V (half of the 24V supply ) to switch the FET on
the BOB is like this I have looked at before
JohnLast edited by john swift; 28-06-2019 at 08:57 PM. Reason: add FET lamp driver
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28-06-2019 #9
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28-06-2019 #10
it looks like you have all the parts to try it out with the FET
the combination of the high value resistors and the gate source capacitance is going to slow the FETs switching but for this application it should be OK
I usually use a 100 ohm resistor to connect the gate to a low impeadance driver
(very like the 100 ohm "grid stopper" with valve (tube) circuits )
John
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