I wish to use a two gate latch to control a 555 timer.The goal is to turn off the output by shutting down the IC.Will this happen if I disconnect the positive supply or the ground so there will be no current flowing trough the timer?Or,if this won't work,which way should I do it?The 555 is set in the astable configuration.
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2 Answers
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Pulling Reset to ground will ensure that the output is held low.
answered Jul 24, 2015 at 15:10
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\$\begingroup\$ And during that time,the chip won't oscillate? \$\endgroup\$ Jul 24, 2015 at 15:12
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3\$\begingroup\$ For sure it won't oscillate. /RESET drives the DISCH output low so it discharges the timing capacitor. \$\endgroup\$ Jul 24, 2015 at 16:31
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If you use a 4001 (quad 2 input NOR) you could use 2 gates for your latch circuit and one of the remaining gates to synchronise control with the LOW output state of the 555. This avoids the production of spikes (short pulses) at the output of the 555.
The NOR gate SR circuit would normally have a HIGH on the Q output. This forces the control gate output LOW regardless of the output from the 555. D1 is always reverse biased so the capacitor is charged (through R1 and R2) and discharged (through R2) as normal.
When Q is taken LOW if the input to the control gate from the 555 is HIGH (capacitor still charging) then nothing happens until the capacitor reaches 2/3 Vc and the output of the 555 goes LOW. At this point the output of the control gate goes HIGH and keeps the capacitor voltage HIGH even though R2 is trying to discharge it. The result is the output of the 555 remains LOW.
If Q is taken LOW during the LOW output of the 555 it will (quickly) recharge the capacitor through the 100R and diode and keep the output LOW.
When Q is taken HIGH the output of the control gate immediately goes LOW and the capacitor will start to discharge through R2 until it reaches 1/3 Vc and restart the oscillation.
