| 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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