Latching Output, Output high on powerup & simple power supply for TTL cct from 12v

Thread Starter

blewis9991

Joined Jun 1, 2019
5
I have looked through the posts but may have missed one if similar to my problem.

I have 12v - 13.4v from the automotive supply. I would like to have a latching TTL or some device to always start in the same state on the outputs.

The latch needs to change state based on a momentary DPDT center off contact closure from a mechanical switch. I am assuming there will be some bounce on the inputs given it is mechanical. I want to keep the circuit as simple as possible.

The questions are:
  1. should I try to use a simple zener and resister as a voltage source to bring the 12v down to 5v inorder to use 5v TTL?
  2. Or would even a resistor divider be enough given the low amount of power of the IC?
  3. should I use 5v TTL or would another 12v part be better and not deal with the 12v supply to 5v at all? Can someone recommend a part if so?
  4. would a capacitor across the outputs of the momentary switch and inputs of the latch be sufficient to stop the bounce detection if so what would be a good value? Do I also need a resistor?
  5. What I am trying to do is register the state of a double pole momentary switch. If pushed down one LED will light, if pushed UP a different LED will light. I am assuming that there will be two inputs to the latch to drive the output opposite to each other and that the latch will stay until the alternate push is detected.
  6. The latch has to start in the same state from power up of the supply (in case the battery is disconnected).
Thanks for any advice in advance.

Robert
 

Daniel Sala

Joined May 28, 2015
60
Hi,

I wouldn't count on 12V to 13.4V being that wherever the circuit you'll build is, i.e. it may be a little lower than 12V.

IMHO:

1) I would as automotive can have unpleasant voltage spikes. You might need more stuff to protect the TTL.
2) I personally wouldn't do that for more than one reason (fluctuating Vsupply, spikes = no rudimentary protection for the IC).
3) What about an equivalent CMOS part like anything suitable from the CD4xxx family? CMOS would use less housekeeping/quiescent power as well.
4) Search for pushbutton debounce circuits and focus on the RC ones only (you won't need ones that have 10 transistors and an RC combination).
5) and 6) Use a flip flop IC with Set and Reset inputs and tie one to ground with a pull-down resistor.

Hope it helps a bit.
 

Thread Starter

blewis9991

Joined Jun 1, 2019
5
Hi,

I wouldn't count on 12V to 13.4V being that wherever the circuit you'll build is, i.e. it may be a little lower than 12V.

IMHO:

1) I would as automotive can have unpleasant voltage spikes. You might need more stuff to protect the TTL.
2) I personally wouldn't do that for more than one reason (fluctuating Vsupply, spikes = no rudimentary protection for the IC).
3) What about an equivalent CMOS part like anything suitable from the CD4xxx family? CMOS would use less housekeeping/quiescent power as well.
4) Search for pushbutton debounce circuits and focus on the RC ones only (you won't need ones that have 10 transistors and an RC combination).
5) and 6) Use a flip flop IC with Set and Reset inputs and tie one to ground with a pull-down resistor.

Hope it helps a bit.
Would the CMOS part be more immune to a less regulated supply and work on 9v as a supply.? Then I could use zenor diode to get 9 volts and that would be well within the limit that the battery could drop to. All the CMOS must do is sink enough current too drive a small LED. So since the LED is based on current it doesn't matter about s much about the voltage level. Can you recommend which CMOS part you would use please? Should I use automotive grade parts?Thank you for the help.
 

Thread Starter

blewis9991

Joined Jun 1, 2019
5
Would the CMOS part be more immune to a less regulated supply and work on 9v as a supply.? Then I could use zenor diode to get 9 volts and that would be well within the limit that the battery could drop to. All the CMOS must do is sink enough current too drive a small LED. So since the LED is based on current it doesn't matter about s much about the voltage level. Can you recommend which CMOS part you would use please? Should I use automotive grade parts?Thank you for the help.
Sorry, I overlooked your recommendation on the power. I could use a small IC regulator with a cap.
 

mlsirkis

Joined Aug 11, 2010
22
Sorry, I overlooked your recommendation on the power. I could use a small IC regulator with a cap.
Use two gates of 4011.
https://www.google.com/search?biw=375&bih=635&ei=eZTyXPmGHKSyggeytYnYDA&q=sr+flip+flop+using+nand+gate&oq=sr+flip+flop+ic+pin+diagram&gs_l=mobile-gws-wiz-serp.1.0.0i71l5.0.0..69778...0.0..0.0.0.......0.ZCHzEg9ejns#imgrc=Y22cVuUUp_e82M:

Pull up both inputs. Add a small cap from one input to ground. This will set the power up state. Wire switch common to ground. The switch when toggled will pull one or the other inputs to ground
 

crutschow

Joined Mar 14, 2008
23,512
You don't need to worry about switch-bounce with a latch circuit, since any bounce signals will be ignored once the latch responds to the first input signal change.
 

CharlesWMcDonald

Joined May 16, 2019
233
I have attached a good app note on automotive transient protection.

Another option would be a transistor bi-stable circuit. You would need to add a capacitor-resistor trigger to turn one side on at start up but that would not be difficult. This circuit can be made robust and to operate over a wide voltage range, no regulator required. Since you are using a momentary switch I would add resistors across the base-emitter to insure the transistor stays in the off state. Also, this would be more fun!

 

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Thread Starter

blewis9991

Joined Jun 1, 2019
5
I have attached a good app note on automotive transient protection.

Another option would be a transistor bi-stable circuit. You would need to add a capacitor-resistor trigger to turn one side on at start up but that would not be difficult. This circuit can be made robust and to operate over a wide voltage range, no regulator required. Since you are using a momentary switch I would add resistors across the base-emitter to insure the transistor stays in the off state. Also, this would be more fun!

Thank you for the replies and appnote.

If I use the 4011, I am assuming that I pull up both inputs, is this correct?

The switch is normally off which would have both inputs high to the 4011. Pushing up pulls down one input contact to ground, pushing down makes the other contact to ground.

I will breadboard this once I understand the basics.

Thanks again.
 

mlsirkis

Joined Aug 11, 2010
22
Thank you for the replies and appnote.

If I use the 4011, I am assuming that I pull up both inputs, is this correct?

The switch is normally off which would have both inputs high to the 4011. Pushing up pulls down one input contact to ground, pushing down makes the other contact to ground.

I will breadboard this once I understand the basics.

Thanks again.
Yes pull both inputs up . Add a cap from one input to ground. The cap sets the power up state.
 
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