Optocoupler fails after 3-4 month.

Thread Starter

shetsachin

Joined Jul 5, 2017
23
I have designed optocoupler circuit which the input signal coming from PLC(+24V) & output goes to DIO(+5V)PC.
I faced a problem after 3 months optocoupler not giving output signal.I don't no why it is going fail.Is it a design problem or something else?
If any improvements require please suggest.
Please help anybody.

opto circuit.jpg
Thanks,
Sachin N
 

ericgibbs

Joined Jan 29, 2010
13,633
Hi,
The light output of the emitter will decrease with age.
A 10k from 24V is only ~2.4mA, that seems low, what is the opto type.?
E
OK I see the Type TLP.
A001.gif
 
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BobaMosfet

Joined Jul 1, 2009
1,850
I also think that the TP needs to make sure that the resistor can handle the wattage being dissipated by the 24V source-- If a resistor fails, the optocoupler will be next.
 

cmartinez

Joined Jan 17, 2007
7,394
I also think that the TP needs to make sure that the resistor can handle the wattage being dissipated by the 24V source-- If a resistor fails, the optocoupler will be next.
As Dodgywave just said, a 1.5k resistor working at 24VDC oughta be good enough. That makes for 24/1500 = 16 mA, which in turn would make 0.016*24 = 0.384 Watts ... so a 1/2 W, 1.5K resistor should be good enough. But I'd probably use 1W, to be on the very safe side of a long-lasting assembly.
 

ScottWang

Joined Aug 23, 2012
7,065
There are two conditions for the current of If, Eric was shown the current for pulse, if the input voltage is fixed on +24V and the current can be smaller then the Vce will get into the saturation region, the current condition shown below.

TLP291_Imin-Vce.jpg
TLP291_datasheet_en_20140922.pdf -- page 4.
 

Alec_t

Joined Sep 17, 2013
12,201
after 3 months optocoupler not giving output signal
.... but is there still the correct current through the opto input diode? If so, it's presumably the output transistor that fails. That is rated for 50mA, but even if driving a digital I/O pin which inadvertently got set as a pull-down output it's unlikely that much current would flow, given that the opto transfer ratio is no more than 100%. Could there be a voltage spike on the supply rail?
 

JohnInTX

Joined Jun 26, 2012
4,554
It might not be component failure. A couple of jabs at the 'Lil Professor using some worst case numbers shows that the voltage swing at the 1K resistor is marginal for many logic inputs. Using 50% CTR and Vf=1.4V I get 1.13mA on the output for 1.13V across the 1K. That's a marginal logic 1 for anything I can think of. I wonder if the TS got lucky early and normal aging, temperature swings etc. pushed things out of that marginal range. Even if the CTR was better, it would still be marginal for many logic inputs. This doesn't take into account any loading/current sinking requirements imposed by the logic input itself.

(24V - 1.4V)/10K = 2.26ma Iled.
2.26ma * 50% CTR = 1.13ma Ic
1K*1.13ma=1.13V

First thing would be to verify that the 1K pulled the input down solidly then set the LED resistor to get a solid logic swing across the 1K. A 5V swing across the 1K would need 10ma ILED at 50% CTR (not counting any current loads from the logic input).

Just thinking out loud.
 
Last edited:

BobaMosfet

Joined Jul 1, 2009
1,850
As Dodgywave just said, a 1.5k resistor working at 24VDC oughta be good enough. That makes for 24/1500 = 16 mA, which in turn would make 0.016*24 = 0.384 Watts ... so a 1/2 W, 1.5K resistor should be good enough. But I'd probably use 1W, to be on the very safe side of a long-lasting assembly.
Assuming this is a DC only application, the resistor must deal with the entire 24V dropped across it. I think you only used have the power-dissipation calculation.

P = E * I
E = I * R

Which becomes

P = I * R * I

Which becomes

P = I^2 * R
P = 0.016^2 * 10000
P = 0.000256 * 10000
P = 2.56W

I'm just using the 16mA figure that was mentioned above. The TP with 2.4mA would be just 57.6mW, which would work with a 1/8th-W resistor, so is unlikely that's the problem. But still good to be aware of for larger energies.
 
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strantor

Joined Oct 3, 2010
5,637
Assuming this is a DC only application, the resistor must deal with the entire 24V dropped across it. I think you only used have the power-dissipation calculation.

P = E * I
E = I * R

Which becomes

P = I * R * I

Which becomes

P = I^2 * R
P = 0.016^2 * 10000
P = 0.000256 * 10000
P = 2.56W

I'm just using the 16mA figure that was mentioned above. The TP with 2.4mA would be just 57.6mW, which would work with a 1/8th-W resistor, so is unlikely that's the problem. But still good to be aware of for larger energies.
I think you missed the part about 1.5K resistor. 16mA is what he would get with a 1.5k resistor. You used the 16mA value and applied it to the OP's 10k resistor.
 

Thread Starter

shetsachin

Joined Jul 5, 2017
23
It might not be component failure. A couple of jabs at the 'Lil Professor using some worst case numbers shows that the voltage swing at the 1K resistor is marginal for many logic inputs. Using 50% CTR and Vf=1.4V I get 1.13mA on the output for 1.13V across the 1K. That's a marginal logic 1 for anything I can think of. I wonder if the TS got lucky early and normal aging, temperature swings etc. pushed things out of that marginal range. Even if the CTR was better, it would still be marginal for many logic inputs. This doesn't take into account any loading/current sinking requirements imposed by the logic input itself.

(24V - 1.4V)/10K = 2.26ma Iled.
2.26ma * 50% CTR = 1.13ma Ic
1K*1.13ma=1.13V

First thing would be to verify that the 1K pulled the input down solidly then set the LED resistor to get a solid logic swing across the 1K. A 5V swing across the 1K would need 10ma ILED at 50% CTR (not counting any current loads from the logic input).

Just thinking out loud.
Thanks for reply,
IF is increased to 16mA say, then if IC is 10mA, CTR will be 62%.
Is it ok,this optocoupler range is 200% to 400%,need to design within range of CTR?.
 

ScottWang

Joined Aug 23, 2012
7,065
Its a signal pulse coming from PLC(10ms)
Vf=1.25V(typ),
If = (24V-1.25V)/10K = 2.275mA
10 mS = 0.01
If_pulse = 2.275mA*0.01 = 0.02275mA
So compares with the current shown in #10 that the If=1mA then
rate% = 1mA/0.02275mA = 43.96
R1 = 10K/43.96 =227Ω, you can choose 220Ω.
So now you have a big range to choose as 1mA~20mA.
 
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