TYPO! They are 8.2kAre R1 and R4 really only 8.2 Ohms, or is that a typo?
HiSo much for a "morning" reply.
Built the circuit below. Upon power-up, Pin 3(4017) lights up. Pressing the PB turns Pin 2 high, and with an additional press Pin 4 goes high. As one LED turns on, the other two LED's are off. However, this is consistent only when I press deliberately. If I press lightly or quickly the LED sequencing order can fail.
Note: R1 & R4 are 8.2k, not 8.2 Ohms
View attachment 168458
I have read that all inverters are not created equal for TTL circuits. These are the Inverters I have on hand:
- HCF4069UBEY(Hex Inverter)
- MC14584BCP(Hex Schmitt Trigger)
- MC74HC04(Hex Inverter High-Performance Silicon-Gate CMOS)
- CD4050BCN(Hex Non-Inverting Buffer)
- CD4010(Hex Buffers (Non-Inverting)
Further observations:
While waiting for replies, I did some more reading. I noticed that the 1uF capacitor I was using was polarized. I think it was a Tantalum cap. I switched it out for a .33uF ceramic cap and I could not replicate the bouncing effects. But!... I also tried the switch with no cap in the circuit and also had no fails. Then I reinserted the polarized cap and could not reproduce the fails either! Perplexed. Not sure if my values for resistors & cap are in the right ballpark, so insight on this would be helpful. From readings, I learned that the diode in the debouncing circuit helps speed up the switching. Also, the inverter should be taking care or the RC ramping characteristics.
Q: Any advantage to debouncing from low side rather than high side? I can't see any, other than what may be simpler for the system/circuit.
ALSO:
- MM74HC14N(Hex Inverting Schmitt Trigger) Now in circuit
Thanks peeps(people)...send me some more knowledge! Must have knowledge!![]()
Isn't that odd! Four .1uF caps from Vcc to GND?Also, use a 0.1uf cap across the power supply pin of each chip just to help stiffen the power supply.
Not odd, just overkill for this circuit.Isn't that odd! Four .1uF caps from Vcc to GND?
Not overkill... just good practice...especially since they’re cheap and the TS might use a wall wart.Not odd, just overkill for this circuit.
But one decoupling cap per IC is often used in high speed logic circuits.

Count-up counters start at all zeros, so yes, any output will be low for it's first half cycle.The output of PIN 1, Q12(divide by 4096) goes low for the first 5min before going high for 5min. I was expecting the opposite.
Your logic is logicalThe 4060 is operational and within 2s of 10min...~6s of 40min. It's not critical that it be so precise, but who's complaining.
The output of PIN 1, Q12(divide by 4096) goes low for the first 5min before going high for 5min. I was expecting the opposite. Thought that the high from U5 4017 and the high from the 4060 into the AND gate would produce a high out, which would feed directly into U4 4017 needing a positive pulse to increment. But it is being inverted by U2C and producing a negative pulse.
Or is my Logic ill-LOGICAL?
circuit is reposted here for convenience.
View attachment 168617
On power up, the output of U2C is initially high because U3D is low and causes C5 to charge up, but POR causes U4 to ignore this. Once C5 is charged, U2C then goes low and the circuit is in its "ready" state.Guess I'm getting confused with the AND gates.
Am I correct in saying that this is the ready-state...power on, no push button pressed?
View attachment 168630
No, it's random.Does powering up the 4017 Decade Counter place the outputs(in our case, S0, S1, & S2) in a particular state?
Yes...its difficult to use fixed value components because of manufacturing variances.The values I have for the 4060 timing are different(4.5uF/13.42k), the 4.5uF being a combination of three 1.5uF ceramic caps. Bottom line, your circuit, containing a variable resistor can dial in a multitude of times.
I agree with Crutschow comment post#53.Q: Does powering up the 4017 Decade Counter place the outputs(in our case, S0, S1, & S2) in a particular state? I am using a jumper wire on the bread board to power the IC. This, of course, seems to emulate a push button action and S1 generally lights up as apposed to S0. Perhaps a power-on delay would work?
HiIt will be important that I am able to select the # of batteries connected to the device(2 - 6) as in the schematic below. I believe I have a spare 6 position rotary switch for use as opposed to using jumper or dip switches. This should not hinder our tests without them. Just hoping to verify that I have wired it correctly. Diode is missing in my schematic from switch to reset of 4017.
They will be hooked up 2,3,4,5 or 6 so that the sequence will cycle depending on how many batteries are selected by a rotary switch. Hope this answers your question.Hi
How will the batteries be connected for charging? Will they always be charged in sequential order?
eT
U3 output would only light an LED for half the time cycle. U5 output would show the time selected continuously.They will be hooked up 2,3,4,5 or 6 so that the sequence will cycle depending on how many batteries are selected by a rotary switch. Hope this answers your question.
Q: Would it be better to tap off U5(S0, S1, S2) or the outputs of U3A, B, C for indication of Time selection.
The switches are wired incorrectly. Reset should occur on the "Q" output equal to the number of batteries. So if two batteries are being charged, reset should be connected to Q2 (count starts at 0), if three batteries, reset connects to Q3, four batteries Q4, and so on.It will be important that I am able to select the # of batteries connected to the device(2 - 6) as in the schematic below. I believe I have a spare 6 position rotary switch for use as opposed to using jumper or dip switches. This should not hinder our tests without them. Just hoping to verify that I have wired it correctly. Diode is missing in my schematic from switch to reset of 4017.