Hi, friends.
I made two optocoupler boards, the first using 6N137, the second using PC817, both 16 channel. Pins KFlop as IN.
I want to use I/O pins JP4 and JP6 (LVTTL, 3.3V), but resistors 150 Ohm reduces the voltage on the input to a logical 0.
Can programmatically disable these pull-down resistors? I need only high-impedance input without a pull-down resistor on 16 channels.
It is possible to programmatically disable the pull-down resistor
Moderators: TomKerekes, dynomotion
- TomKerekes
- Posts: 2540
- Joined: Mon Dec 04, 2017 1:49 am
Re: It is possible to programmatically disable the pull-down resistor
Sorry no.
Regards,
Tom Kerekes
Dynomotion, Inc.
Tom Kerekes
Dynomotion, Inc.
Re: It is possible to programmatically disable the pull-down resistor
Hi, Tom.
I know I'm wrong somewhere, but I don't know where.
Made the board 6N137, it works well with high impedance multimeter input, but does not work with KFlop board. Here is the schematic (111.jpg) OK, I didn't know the board had pull-down resistors installed )
I made the second board with PC817C (222.jpg), but I also get a high level voltage of about 1V on it. What am I doing on the second board wrong?
ps.
sorry for the paint and bad camera on my phone )
I know I'm wrong somewhere, but I don't know where.
Made the board 6N137, it works well with high impedance multimeter input, but does not work with KFlop board. Here is the schematic (111.jpg) OK, I didn't know the board had pull-down resistors installed )
I made the second board with PC817C (222.jpg), but I also get a high level voltage of about 1V on it. What am I doing on the second board wrong?
ps.
sorry for the paint and bad camera on my phone )
- TomKerekes
- Posts: 2540
- Joined: Mon Dec 04, 2017 1:49 am
Re: It is possible to programmatically disable the pull-down resistor
hi uralpt,
In circuit #2 I don't think you are driving the LED hard enough.
The 150ohms needs ~ 16ma to drive it to a solid high 16ma x 150ohms = 2.4V
Because the PC816C has a CTR of 50% the LED needs 32ma
If you assume the LED voltage is 1.4V then the current from the 820ohm resistor is:
(18V - 1.4V) / 820ohms = 20ma
But the 62ohm resistor steals all of it:
1.4V / 62ohms = 22 ma.
Maybe remove the 62 ohm and reduce the 820ohms to:
(18V - 1.4V) / 32ma = 520 ohms
In circuit #2 I don't think you are driving the LED hard enough.
The 150ohms needs ~ 16ma to drive it to a solid high 16ma x 150ohms = 2.4V
Because the PC816C has a CTR of 50% the LED needs 32ma
If you assume the LED voltage is 1.4V then the current from the 820ohm resistor is:
(18V - 1.4V) / 820ohms = 20ma
But the 62ohm resistor steals all of it:
1.4V / 62ohms = 22 ma.
Maybe remove the 62 ohm and reduce the 820ohms to:
(18V - 1.4V) / 32ma = 520 ohms
Regards,
Tom Kerekes
Dynomotion, Inc.
Tom Kerekes
Dynomotion, Inc.
Re: It is possible to programmatically disable the pull-down resistor
Tom, thanks!
I`ll apply your advice soon. It takes a little time, which is never enough )
I'll write soon.
I`ll apply your advice soon. It takes a little time, which is never enough )
I'll write soon.
Re: It is possible to programmatically disable the pull-down resistor
Hi, Tom.
Thanks for the support, all excellent!
My problem was that I did not supply power to the KFlop ) and the output stage of the optocoupler passed all the current to KFlop.
I checked - the optocoupler works fine with a current of 20mA, since the CTR is> 50%.
But your circuit is better, there is no R1 (820) heater )
Thanks!
Thanks for the support, all excellent!
My problem was that I did not supply power to the KFlop ) and the output stage of the optocoupler passed all the current to KFlop.
I checked - the optocoupler works fine with a current of 20mA, since the CTR is> 50%.
But your circuit is better, there is no R1 (820) heater )
Thanks!