What is the role of 1N4148 on my circuits?

iimagine

Joined Dec 20, 2010
482
If you use the output pin 3 to charge the capacitor and get rid of R1, then you dont need those diodes, because the charge and discharge time will be the same
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Dodgydave

Joined Jun 22, 2012
9,468
It's a Pwm generator, D1 is the Capacitor C2 charge time, and D2 is the discharge time, the output will be a Square Wave with variable Mark to Space ratios.
 
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Tonyr1084

Joined Sep 24, 2015
5,090
Those two diodes are steering diodes. Since current flows through only one way, and each is pointed in opposite directions, yes, they control charge and discharge times. The capacitor charges through D1. Depending on where the trim pot is set will affect charge time. Then during discharge, the cap flows through D2, again, through the trim pot. If the pot is set at 50% then you will get a 50% duty cycle, meaning output high and output low will operate at equal time periods. If you swing the pot over to the 75% point to the left, you will have a 25% slower charge and a 75% faster discharge. The resulting output would be 25% high and 75% low.
 

crutschow

Joined Mar 14, 2008
25,446
you dont need those diodes, because the charge and discharge time will be the same
Only for the ideal 555 model that you are using, or if it's the 555 CMOS version whose output goes to the positive rail.
A real bipolar 555's output does not go completely to the rail so the charge time will be less than the discharge time in your circuit (below).

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This can be compensated by adding a resistor from CV to ground.
That value may have to be tweaked to get 50% in the real circuit.

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