This circuit what you provide are not isolated not even opticaly isolated.
If you whant 5V at 1sec on optocoupler trasnsitor side you should make 5V power source on that side, and on LE diode side of optocoupler you should make some oscilator that give power to LED every 1sec.
Pay attention to limit current for optocoupler on both side, acording manufacturer datasheets for that optocoupler.
Optocoupler on its LE diode side will dwar some current needed for operation (about few miliamp, example 5mA to 50mA). Can you draw that current from Garmin?
You can always use phototransistor to sense light form Garmin LE diode and turn some transistor which will turn some relay. This will be completely separated from Garmin and no need for any opening of device or such things. No power consuption for LE diode from Garmin. This will be completly separate device from garmin.
There is some examples :
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My goal is to be able to drive the 50ohm load as this is an input into another system. I don't care too much about keeping them isolated.
One question you whant to pull power from Garmin, for 50R ?
You must know that LED in Garmin have limited current, I supose to 2mA, if you attach 50R in serie with that, you will decrease current and get nothing.
50R load tells nothing if you dont know voltage.
Example :
If you put 50R on 5V current is 5V/50R = 0,1A (100mA)
If you put 50R on 2V current is 2V/50R = 0,04A (4mA)
We gues Garmin use 2mA LED 3,7V.
3,7V / 1500R = 0,0024A (2,4mA)
If you put additional 50R in serie on that 2,4mA (serie with 1500R) you will get 2,3mA.
But there is another problem, you say ok there is enough current, but you should know that LED have voltage level needs to operate. You should increase voltage level if you whant LED to light in serie.
Your first example is correct. The goal is to input the Garmin 1pps @5V into the 50ohm input. Since the garmin cannot provide the current, I wanted to buffer the 1pps signal. I need the 1pps to be in phase as the 1pps will be a sync for the system.
If you look at the circuit I provided, the Vout is the 5V 1pps that will be input into my system which has a 50ohm input.
Garmin should have 3,7V lithium battery, did you check 5V voltage on garmin LED ? :shock:
LED should not have over 2V if green,red,orange (signal showing, or such things), for white led voltage is higher about 3V or over.
If you need synch with garmin led you can do what I proposed in post #2. To use optocoupler on right way, or to use completely separate circuit with phototransistor. In both way you will get synch of LED light and duration of signal.
If you plan to pull power from Garmin I dont recommend that. You should have 3,7V not 5V in device, and my suggestion is to use external power for your external circuit.
If zou plan to consupt power from 3,7V from Garmin then 3,7V / 50R = 74mA. You should know that will shorted battery cycle and working hour of Garmin, or can produce malfunction of Garmin.
As far as I understand, you don't intend isolation, so why using an opto isolator? There's also no need of an OP to buffer a digital signal. A complementary transistor switch (NPN+PNP) will do.
Yes, in fact I won't expect a good effect from connecting the 50 ohm instrument input to the 5 V power supply.
Obviously, the collector terminal of Q2 is the circuit output and R3 a basic load. The 50 load woudl be connected in parallel to R3.
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