cd4013 flip flop outputs always on

Thread Starter

488 studios

Joined Nov 10, 2023
13
hey, I built this kitchen timer circuit: https://www.electronics-project-design.com/kitchen-timer-circuit.html but C1=1uF film cap (for longer timing), r2=150ohm and r1=1kohm (for lower time when the potentiometer is at 0, but it draws more current) and I replaced the active buzzer with an NE555 oscillator going to a passive buzzer. and I added cd4013 flip flops to make an led progress bar (25%, 50%, 75%, and 100%). q12 of the cd4020 goes to the set pin of one flip flop, with it's output to the 25% led. Same with q13 and the 50% led. for the 75% led, I have both q12 and q13 going to an and gate made with two 2N4401's, the output going to the set pin of another flip flop, leading to the led. Anyways, my problem is the 25, 50, and 75% led's come on immediately after powering on. I notice they all get slightly brighter at 25, 50, and 75% of the set time. Poking the 25 and 50% flip flop outputs with a multimeter makes its led behave as it should. Metering the 75% led makes it act funny. at first, only the 50% led is on. at a quarter of the time, the 25% led comes on. at half the time, the 75% led comes on, and at 3/4 of the time, they all get brighter. The 100% led works fine, though. I've attached a schematic, and I've grounded the unused inputs/set pins and added bypass caps. Can anyone help me get the progress bar working?
 

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crutschow

Joined Mar 14, 2008
34,669
You cannot leave inputs floating with CMOS circuits.
Due to their very high input impedance the input voltage can be be anything (indeterminant), so connect all unused inputs to either ground or V+.
 

sarahMCML

Joined May 11, 2019
391
There's a couple of other significant errors in the circuit.
U2's output is driving ~12V into U1 pin 10, which is supplied with 5V. The internal diodes aren't going to like that!
U3's VCC cannot get the 12V from Q1 because its base is only getting 5V from pin 13 of U1, so it's probably getting 4.3V at most!
Either supply everything from 12V or 5V!
 

Thread Starter

488 studios

Joined Nov 10, 2023
13
There's a couple of other significant errors in the circuit.
U2's output is driving ~12V into U1 pin 10, which is supplied with 5V. The internal diodes aren't going to like that!
U3's VCC cannot get the 12V from Q1 because its base is only getting 5V from pin 13 of U1, so it's probably getting 4.3V at most!
Either supply everything from 12V or 5V!
I did the 5v because the 2N4401's have a max emitter-base voltage of 6v. Will I damage the 2N4401's with 12v?
 

AnalogKid

Joined Aug 1, 2013
11,120
I did the 5v because the 2N4401's have a max emitter-base voltage of 6v. Will I damage the 2N4401's with 12v?
That is the *reverse* base-emitter voltage. The forward voltage for normal conduction (Vf) is around 0.5 V to 0.8 V depending on the amount of base current and collector current. Everything in your circuit is safe for 12 V operation. You will have to re-size the LED current limiting resistor, because CD4000-series outputs are not very strong.

The 2-transistoR AND gate is cute, but you can reduce the circuit complexity and cost by deleting then, plus their base resistors, and using two diodes. In this case, R8 can be a 10K pull up resistor.

U1 and U2 can be combined into a CD4060.

Both 555 oscillator circuits have significant errors.

Your U3 circuit is not good for a piezo element. There is a better circuit, one that guarantees a 50% duty cycle into the piezo element throughout the frequency adjustment range. It is in the LMC555 datasheet.

Note: Add a decoupling capacitor at each IC, spanning from V+ to GND. These usually are a low cost ceramic type in the 0.1 uF to 1.0 uF range. Keep the leads as short as possible.

ak
 
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schmitt trigger

Joined Jul 12, 2010
916
I did the 5v because the 2N4401's have a max emitter-base voltage of 6v. Will I damage the 2N4401's with 12v?
Why have two voltages when all the ICs in your circuit can operate from 5 to 15 volts? Use a single voltage, whichever you choose.
Also U2’s configuration is weird. What is it supposed to do?
 

AnalogKid

Joined Aug 1, 2013
11,120
First pass at an alternate theory of the crime. This gets the design down to three ICs.Not shown is the 4th flipflop section with the S and R inputs grounded.

The U1 oscillator values vary the output freq from around 1 s to 9 s. That's not very long for a kitchen timer, but I don't wee anything in the thread about a timer range.

UPDATE - Used 4th flipflop to latch the beeper ON signal.

ak


Kitchen-Timer-1-c.gif
 
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AnalogKid

Joined Aug 1, 2013
11,120
OK, another flavor. This one uses a shift register to get rid of that cumbersome logic. That also got rid of the Schmitt trigger gate used for the beeper oscillator, so the 555 is back. The schematic shows a CMOS version, but if you have a large beeper, the bipolar version has a beefier output stage to drive it.

For the U1 oscillator, these component values yield a timing range out to 15 minutes.

Because the shirt register does all of the decoding for the LEDs, everything can be driven from just one counter output. You can use any counter output with the appropriate changes to the oscillator component values.

UPDATE - Used shift register output to latch the beeper ON signal.

ak


Kitchen-Timer-2-c.gif
 
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AnalogKid

Joined Aug 1, 2013
11,120
A potential problem with all of the circuits so far is that the beeper time is equal to the delay time. That is, if the delay timer is set to 15 seconds, the beeper will sound for only 15 seconds, then go off for 15, then on again, etc.

In both of my schematics, unused spare sections can be used to latch the result of the Q14 transition after the delay period, so the beeper goes continuously until power is cycled.

Hmmm . . . I might update both schematic with this change.


Done.

ak
 
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