flip-flop question

Why not build an astable multivibrator with the 555? You will then get a constant stream of ticks.

Or post your schematic on alt.binaries.schematics.electronic

I need to be able to start/stop the ticks. From what I've seen you can't really do that with astable. Am I correct?

I need to be able to start/stop the ticks. From what I've seen you can't really do that with astable. Am I correct?

No, you can drive the reset input and stop the oscillator with a simple toggle switch.

If I get you, I think you want to enable/disable the 555 using the same pushbutton, is that right?

If so, you can use a couple of inverters (cmos 4049 perhaps) along with a normally open push button and and resistors/capacitors to do this.

.-----------------------------------------. | | | |\\ ___ |\\ ___ | '----| >O-|___|--o----| >O----o---|___|---' |/ 100k | |/ | 100k | | 4049 | 4049 .-. | | | 330k | | | | T '-' | --- | '-----o o----o | --- 1uF --- | | === GND (created by AACircuit v1.28.6 beta 04/19/05

The pushbutton is normally open. The capacitor is charged to the output of the right inverter. When you push the button, it changes the value of the input of the right inverter, causing it to change its output. This is independent of the starting state.

Use the real output (not the output after the 100k resistor) to drive the reset of the 555.

Regards, Bob Monsen

The only constraint I have for this circuit is the momentary switch for user input.

=A0 =A0 =A0 =A0 |

=A0 =A0|

=3D

That looks great, but where do I put the trigger and reset lines from the 555?

The only constraint I have for this circuit is the momentary switch for user input.

Okay, a T flip flop with a debounced push button switch as its clock input. The output goes to the reset input of the 555.

You can form a T flip flop if you take a D flip flop and connect the Q* output back to the D input.

What do I connect the trigger input of the 555 to? The D input of the flip-flop?

That looks great, but where do I put the trigger and reset lines from the 555?

You set up the 555 as an astable multivibrator, and let it free run. You connect this to the reset pin of the 555. When it is "high" output, the 555 will output pulses for you. When it is "low" output, the 555 will wait for a "high" output, and output no pulses.

The typical setup for an astable is to have two resistors and a capacitor. Connect the resistors in series with the capacitor between the voltage supply and ground. Connect the discharge pin between the resistors, and both the 'trigger' and 'threshold' to the cap:

-----------o-----------------. Vcc(12v) | | | .-. | | | R1 | | | | __ '-' '---o| |o-----. | .----o| |o-----)--o output ----)----o| |o---. | | reset ----)----o|__|o- | | .-. | | | | | R2 | | | | | | | | '-' | | | | '-------------o-)--o | | | --- C1 | --- | | GND | | ---------------------------o--'

Discharge through R2 is .693 * R2 * C Charge through R1 and R2 is .693 * (R1 + R2) * C

So, f = 1.44/(R1*C + 2*R2*C) and d = 100 * R2 / (R1 + R2)

(created by AACircuit v1.28.6 beta 04/19/05

Regards, Bob Monsen

What do I connect the trigger input of the 555 to? The D input of the flip-flop?

It's an astable multivibrator...build that first, get it oscillating then work on holding the reset active to stop it.

--- Yes. Assuming you're talking about digital inputs and outputs, the whole thing can be done with a dual "D" type flip-flop and three NAND gates. View in Courier:

C1 Vcc>-+-[100nF]-+--------+-------+ | | |R1 | | | [1MR] | IN1 +---+ | | | | 4011 | | | U1a GND | U1b +-B U2a S1| O | +--+--+ +--+--+ NAND Y--+ | O +--|D S Q|-----+----|> S Q|------A | | | | | | _| +--A +------|> R | | +-|D R Q|-+ NAND Y-->OUT | +--+--+ | | +--+--+ | +--B U2c [10K] 4013| | +----|----+----B | |R2 +--[Rt]--+ | NAND Y--+ | | | | +-A U2b | +-[CR1>]-+ | | | [Ct] | | | | | IN2 GND>-+---------+----------------+ Select Rt and Ct for a time constant longer than the press-release cycle of S1.

On power-up, the S(et) inputs of U1a and B are pulled high momentarily by C1 and R1, forcing U1a-Q high, U1b-Q high and U1b-Q\\ low.

Since the signal on U1a-S will stay high longer than the edge from U1a-Q, U1b-Q will remain high and U1b-Q\\ will remain low until U1a times out and S1 is pressed.

U2a, b, and c comprise a 2-to-1 multiplexer, and with U1a high and U1b low, the logic signal appearing on IN1 will appear on OUT until S1 is pressed.

When S1 is pressed, U1a-Q will go high and will toggle the outputs of U1b, now causing the signal on IN2 to appear on OUT.

When U1a-Q goes high it also starts charging Ct through Rt, and when the voltage on U1a-R gets high enough it will reset U1a and discharge Ct quickly through CR1, which can be any common signal diode, say a

After that, every time S1 is pressed the cycle will begin anew and the signal on OUT will alternate between the signal on IN1 and IN2.

If you decide to build this circuit, be sure to connect the inputs of the unused NAND to either Vcc or GND.

On Tue, 25 Mar 2008 09:09:06 -0500, John Fields wrote:

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