analogltspiceopto-isolator
Circuit 1:
Circuit 2:
The only difference between Circuit 1 and Circuit 2 is`that in Circuit 2 a parallel 2.2k resistor R8 is added as a load next to R6.
Why adding this R8 is halving the output voltage here?
I tested this in real with a scope, and the voltage remains at 12V pulse when I add R8. So unlike in simulation R8 didn't change the output voltage. In real implementation the grounds of the input side and output side are isolated.
edit:
Please try using the setup as shown in the figure. The 1 K value on pin one is fine, but increase the value of resistor to 100k between Pin 4 and 3.3 V. When there is no 12 V on pin 1, The output (Pin Number 4) will be high, and when there is 12 V on Pin 1, the transistor will turn on shorting the Pin 4 to ground (VCE will be less than 100-200 mV). please try this and reply if not possible.
I am using the same setup, different kind of coupler and it is working.
A number of considerations, having read the data sheet.
Follow fig 17 a lot closer than you are showing in your simulation diagram. 100nF on input and output supplies, RC lowpass filter with a capacitor across the inputs. Choose the capacitor value such that it does not impact your signal accuracy too much, while providing a low impedance at RF. At least 100pF according to the pages of hints an tips in the data sheet.
That part is designed for a colossal transient common mode rejection. As such, any attempt to put a can round it may well degrade that. With a good layout, so short leads between the IC supply terminals and the 100nF, short leads to the low impedance 20ohms and a capacitor across the inputs, the IC will be very hardened against RF interference. Follow the tips about how to connect the -ve signal input, with respect to other connections to that node.
Try the part out without kludgy metalwork first, but with good layout and appropriate capacitors. You can't screen your way to success if your layout is bad, or if you are lacking the decoupling suggested in the applications info. Once you have the latter, seperate screening is rarely needed.
Caution, 1000v across R1 is 10 watts, 5W average for a sine wave. Make sure it can dissipate that, while the tempco is within your accuracy specification.
Best Answer
The simulation model will likely have a typical current transfer ratio of based on the data sheet and your real test may be using an opto isolator with a very high CTR. With a very high CTR the output opto-transistor remains saturated and you largely get the same output peak-to-peak voltage as you got when you didn't load with the extra 2k2 resistor.
The minimum CTR for the 4N26 is 20% but typically this can be 50% and if you had a really good one it could be a lot higher such as over 100%. The model parameter can be changed and you should be able to resimulate and obtain a result that is closer to what you measured.