Electronics > Beginners
LED control circuit, would appreciate feedback and suggestions.
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Zero999:

--- Quote from: doublec4 on June 16, 2019, 03:13:28 am ---Thanks Zero999

I was going to use your circuit to find the "sweet spot" with the pot and then just measure the resistance once I had a suitable duty cycle.

I think maybe the module is somewhat sensitive to the frequency. I'm not sure how to calculate the frequency in your schematic , but perhaps it was too high for the module.

--- End quote ---
I did say how to calculate the frequency, perhaps I wan't clear about what the variables in the formula relate to.

F = 1.44/(RC)

Where:
R is the resistance of the potentiometer, in Ohms
C is the capacitance of the timing capacitor between pins  2 and 6 and 0V, in Farads.
F is the frequency.
This is assuming the potentiometer has a much higher resistance, than 1k.

So in the circuit I posted.
C = 10nF = 10×10-9
R = 100k = 100×103

F = 1.44/(10×10-9*100×103) = 1.44/(1000×10-6) = 1440Hz

So if you want 200Hz, increase the capacitor value to 68nF, which will give about 210Hz, which is as close as is feasible given component tolerances.
doublec4:

--- Quote from: Zero999 on June 16, 2019, 03:39:16 pm ---
--- Quote from: doublec4 on June 16, 2019, 03:13:28 am ---Thanks Zero999

I was going to use your circuit to find the "sweet spot" with the pot and then just measure the resistance once I had a suitable duty cycle.

I think maybe the module is somewhat sensitive to the frequency. I'm not sure how to calculate the frequency in your schematic , but perhaps it was too high for the module.

--- End quote ---
I did say how to calculate the frequency, perhaps I wan't clear about what the variables in the formula relate to.

F = 1.44/(RC)

Where:
R is the resistance of the potentiometer, in Ohms
C is the capacitance of the timing capacitor between pins  2 and 6 and 0V, in Farads.
F is the frequency.
This is assuming the potentiometer has a much higher resistance, than 1k.

So in the circuit I posted.
C = 10nF = 10×10-9
R = 100k = 100×103

F = 1.44/(10×10-9*100×103) = 1.44/(1000×10-6) = 1440Hz

So if you want 200Hz, increase the capacitor value to 68nF, which will give about 210Hz, which is as close as is feasible given component tolerances.

--- End quote ---

Thank you! Perhaps I will try this circuit again with the modified cap to lower the frequency and see what happens!
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