# Simple RC time delay circuit of 1 sec to an IC

#### vikrambharath

Joined Aug 13, 2018
17
I need a simple circuit without transistors to give a delay of 1 sec when the power is turned ON.

The voltage is 3.3v
The IC is 4070

#### dl324

Joined Mar 30, 2015
17,012
That limits the options somewhat.

What else is in the circuit?

What is the purpose for the delay?

#### ScottWang

Joined Aug 23, 2012
7,411
Is this a homework or just a personal practice?

#### crutschow

Joined Mar 14, 2008
34,700
Is this to delay the power applied to the 4070?

#### vikrambharath

Joined Aug 13, 2018
17
This is a personal project.

The delay is for 4070IC

I have tried using a transistor before RC circuit, but could accomplish the delay.

The delay to IC so that it won't give any outputs when the input is given.

#### dl324

Joined Mar 30, 2015
17,012
The delay to IC so that it won't give any outputs when the input is given.
I'd use the delay on a gate that held the XOR output low. No telling what the gates will do while the power to the chip is being ramped.

#### crutschow

Joined Mar 14, 2008
34,700
The delay to IC so that it won't give any outputs when the input is given.
For that I would use a 2-input Schmitt trigger Nand gate at the output of the CD4070 with an RC delay at it's input.
LTspice simulation below:
Note that the output of the U3 CD4070 is configured to start out high but U2's output does not go high until a 1 second delay after power is applied, as determined by the R1C1 value.

#### Tonyr1084

Joined Sep 24, 2015
7,964
A transistor is a switch. You can do the same thing with a relay. A large capacitor, a resistor, not over rated to prevent the relay from pulling in, then when the cap reaches sufficient voltage the relay will click in. The switch will (can) hold the input low until it transitions from off to on. Then your signal goes in.

Well, heck! You asked for a non-transistor solution. Gates can do the same thing, but you need something to discriminate between levels. A comparator should also do the job - provided you use it to power a small transistor. But now you're back to using a transistor.

Without knowing more about your circuit - we can spend all day guessing.

#### vikrambharath

Joined Aug 13, 2018
17
The inputs to the gate are from an unreliable source which will give erratic signals for the first one second when the power is turned on.
So in order to avoid those signals, I simply preferred to turn off the gates for the first one second after turning on the power so as to avoid any unwanted outputs...these outputs are fed to a relay which will turn on LEDs.
Due to this unreliable source, when the power is turned on, the first second all the LEDs are flickering.(I used the 4070 as a not gate)
I asked for a minimal design as i purchased an enclosure which is a tight fit for the PCB.
If there are more components to be added, the PCB will not fit the enclosure.

Sorry, I know some sort of transistor has to be used but I thought there would be some simple option which i dont know.

#### crutschow

Joined Mar 14, 2008
34,700
Here's a circuit using a small N-MOSFET to clamp the output of the 4070 low for about a second, as determined by R1C1, when the power is applied:

#### vikrambharath

Joined Aug 13, 2018
17
Thank you everyone.
CRUTSCHOW, The circuit you provided is well.
I want to connect it to vcc of 4070 IC.
Will it work for that setup.

#### ebeowulf17

Joined Aug 12, 2014
3,307
EDIT: Based on comments from @ebp below, I no longer think this circuit is a good idea, because passing signals into the gate while leaving it unpowered could have unwanted consequences.

Maybe something like this would work? There's a glitch at the beginning, and the power ramps just a little before coming on all the way. Not sure how problematic either of those things is, but maybe there's some potential here.

V-Out is meant to be the power pin on your IC.

#### Attachments

• 1 KB Views: 2
Last edited:

#### ebp

Joined Feb 8, 2018
2,332
"I want to connect it to vcc of 4070 IC."

CMOS gates with VDD unconnected can be powered through the inputs. The input protection diodes can easily conduct sufficient current to raise Vdd to one diode drop less than the input voltage. If some other load is also connected to Vdd, the input diode may be damaged due to excessive current. A resistor in series with the input can protect against damage, but not necessarily against powering the IC.

#### crutschow

Joined Mar 14, 2008
34,700
I want to connect it to vcc of 4070 IC.
Will it work for that setup.
Yes.
That's how U3's power is connected, although not shown on the schematic.
CRUTSCHOW, The circuit you provided is well.
I want to connect it to vcc of 4070 IC.
Will it work for that setup.
That will work fine.

#### ebeowulf17

Joined Aug 12, 2014
3,307
"I want to connect it to vcc of 4070 IC."

CMOS gates with VDD unconnected can be powered through the inputs. The input protection diodes can easily conduct sufficient current to raise Vdd to one diode drop less than the input voltage. If some other load is also connected to Vdd, the input diode may be damaged due to excessive current. A resistor in series with the input can protect against damage, but not necessarily against powering the IC.
I totally forgot about that. One of my more confusing electronics diagnostic problems was due to a microcontroller being inadvertently powered through protection diodes. I've edited my earlier post with a delayed power circuit.

If the real issue for the thread starter is unwanted blinking lights, maybe it would be easier to shut down power to the lights instead of the input logic. At this point I feel like it would be really helpful to have a schematic for the whole circuit.

Also, how many inputs and outputs need to be dealt with? If it's just one, the circuit in post 10 looks like a winner to me. If it's multiple i/o paths, then a broader solution would be better, but we need to see the whole circuit to provide good advice in that case.

#### dl324

Joined Mar 30, 2015
17,012
I want to connect it to vcc of 4070 IC.
Will it work for that setup.
No. As mentioned in post #13, the IC will be powered by any inputs that are HIGH. Your only options are to control the inputs or outputs of the CD4070.

#### crutschow

Joined Mar 14, 2008
34,700
No. As mentioned in post #13, the IC will be powered by any inputs that are HIGH.
Okay, I apparently misunderstood his question.
If he wants to power the IC from delay circuit, then no, that won't work as you noted.

Last edited:

#### Tonyr1084

Joined Sep 24, 2015
7,964
You need to debounce the erratic inputs the same as you debounce a switch. Only you want a longer delay than a few milliseconds. I KNOW I don't have a solution, so I'm watching to learn something. But like @ebeowulf17 said:
we need to see the whole circuit to provide good advice in that case.
Only then can someone truly come up with a truly definitive solution to your problem.

#### crutschow

Joined Mar 14, 2008
34,700
After some further thought, I think my statement is post #17 is incorrect and my original response in post #14 is correct.
I believe he is asking if my circuit in post #10 can have capacitor C1 connected to the same Vcc as the 4070 is connected to, and it can be.
The output of the 4070 is clamped by the MOSFET to ground until after the end of the delay time.

#### Tonyr1084

Joined Sep 24, 2015
7,964
My understanding is that he wants to delay the time before the 4070 is turned on. Powered. At which time he wants it to read the input. As I understand the problem, the input is erratic at startup and the data coming through confuses the system (flashing and flickering LED's) He wants these things to have no response for (lets assume) one second. I'm therefore envisioning a circuit that is de-energized at rest. When the system is started power comes on and makes everything do what it's supposed to do - except for the erratic inputs. So really, all he needs to do is delay the transmittal of the inputs. Hold them low for one (assumed) second.

We have to note that the whole system is powering on at the same time, meaning inputs come on the instant power comes on. And that's probably where the issue is occurring. Not so much the powering of chips, but the delay of the sensory input data. Two transistors configured as an AND gate should work (I'd think). One to hold back the data during start up while the other does the countdown. Power comes on and Q1 along with R1 & C1 set up the time delay. Regardless of what's happening on Q2, only when both transistors switch on can the data flow. The signal, the data, the input - whatever it is we're trying to hold back until the system is booted. Signal input is at Q2 base and its emitter is connected to the 4070. When Q1 base turns on, whatever signals are at Q2's base are passed through from the emitter to the 4070. Assuming the signals are of sufficient current to turn on Q2. Which we can't know until we see more detail from the thread starter.

OK, I think I straightened out my mess. Let me know if I still missed the mark.

Last edited: