# Delorean project

#### Rissy

Joined Nov 23, 2015
106
I have no idea. I'm still confused, because if I have to reduce the whole lot of inputs AND the VDD to 5V, then I guess I'll have to do something. But with what, I'm not sure.

What's the point in the datasheet saying it accepts between 3V and 18V, if you're then restricted to trying to match the regulated 5V outputs if you have the intention of linking them for a particular logic base?!

Will potential dividers be good enough, or will I have to add in regulators, or both?

#### Alec_t

Joined Sep 17, 2013
14,312
Potential dividers will be fine for the logic-level gate inputs, but the Vdd supply would benefit from a regulator, which would stabilise the logic level switching threshold and help with spike/noise suppression.

#### MikeML

Joined Oct 2, 2009
5,444
The allowable range of the supply to those CMOS chips is 3 to 18V. Pick a supply voltage, call it X. The allowable range for any input is -0.5V to X+0.5V, and no lower or no higher...

Typically, with CMOS, to be a valid logic 1 (high), the input must be between 0.6XV and 1.0XV; to be a logic zero (low) the input must be between 0.4XV and 0V. If the input is between 0.4XV and 0.6XV, the logic state is indeterminate.

#### Rissy

Joined Nov 23, 2015
106
Pick a supply voltage, call it X. The allowable range for any input is -0.5V to X+0.5V, and no lower or no higher...
Ah! I had read this differently. That's a lot more acceptable! Thanks Mike!

Potential dividers will be fine for the logic-level gate inputs, but the Vdd supply would benefit from a regulator, which would stabilise the logic level switching threshold and help with spike/noise suppression.
Which regulator would you prescribe for an automotive solution, where the inputs are expected to be between 12-14V for 1, and 0V for 0?

Would I benefit from bringing the whole lot down from 12V to 5V anyway, as this would provide more security of performance based on the battery/alternator output, which everyone knows, can be as little as 10-11V, (especially if you're stuck at the side of the road with your hazard lights on) and go up to as high as roughly 14V when the engine is running.

Is there a good regulator out there which will support this voltage range and give me a stable 5V output which I can rely on taking care of the circuit for me? If so, this would be more stable than using a bunch of voltage dividers, and I could use one on each input as well as the VDD source... (I presume)

#### Alec_t

Joined Sep 17, 2013
14,312
Here is a discussion which is relevant.

#### Rissy

Joined Nov 23, 2015
106
Here is a discussion which is relevant.
This is interesting, thanks Alec,

What is "ESR"?

Because i'm not fussed on what voltage I use (as long as the circuitry works!), then if I don't need to drop everything down to 5V (Mike's advice), then I can pick a more middle ground. Say 9V. This will reduce my dropped voltage value, and therefore resultant heat loss from the step down.

Because I have four inputs which are from the 12V car system (which includes the VDD to power the circuit), then I could put a regulator on to each of these. Everything else is generated from within my own circuit, and therefore will run at whatever power I drop the four inputs down to (except the NAND outputs, which seem to come out at 5V regardless!?). The whole circuit is simply designed to switch a PNP transistor, which is being used to switch on and off an automotive LED, which can be put straight on to the 12V system direct (although i'm maybe going to put in an inline resistance to try and limit the current flow through the PNP....if necessary)

#### Alec_t

Joined Sep 17, 2013
14,312
This will reduce my dropped voltage value, and therefore resultant heat loss from the step down.
If all you're doing is driving a few NAND gates and a LED, then heat dissipation will be negligible.
except the NAND outputs, which seem to come out at 5V regardless!?
A CMOS NAND output should be 5V, only if Vdd is 5V.

#### Rissy

Joined Nov 23, 2015
106
A CMOS NAND output should be 5V, only if Vdd is 5V.
The LTSpice simulator is suggesting otherwise!?

Heat loss is still an issue if your dropping the input voltage down by quite a bit for the output though, is it not? According to the discussion on your link.

#### Alec_t

Joined Sep 17, 2013
14,312
The LTSpice simulator is suggesting otherwise!?
See post #32. Were you using the CD4000_v library and a Vdd-labelled supply?
Heat loss is still an issue if your dropping the input voltage down by quite a bit for the output though, is it not? According to the discussion on your link.
Assume your LED draws 20mA (from the '12V' supply). The CMOS gates draw next to nothing, apart from ~2mA to drive the base of a transistor for switching the LED on. The total power dissipated by the regulator will therefore be about 5mA * 14V + 2mA*(14-5)V = 88mW. It will hardly get warm.

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#### Rissy

Joined Nov 23, 2015
106
I tried out hooking up my first nand chip tonight. I put 12v in on VDD and one nand. I tied all other inputs down to 0v. My chip didn't seem to respond like it should. My test led came on with no input. Went brighter with one input active to 12v and went back to the first state again with both inputs at 12v!? It should have gone out at that point!?

I tried the same again on another completely different chip (I have 4 in total here) and the same result again!?

They both also got very hot very quickly!!!??

#### djsfantasi

Joined Apr 11, 2010
9,160
What particular NAND chip are you using? What are you using as a test LED? Do you have a schematic? There is missing info from your verbal schematic.

#### Rissy

Joined Nov 23, 2015
106
I've confirmed two fried chips (for whatever reason). I've proven the other two remaining chips completely. That means 2oo3 I bought from maplin were no good. I've now ordered a bunch more from
eBay.

I'm using CD4011BE chips. I've got a simple 12V red led here (non automotive)

I've now proven my circuit at least in a rudementry way for the logic switching an NPN and the led part of my circuit, so I guess now all I have to do is sort out the stable voltage part and then bolt the two together and test again.

#### Alec_t

Joined Sep 17, 2013
14,312
The CD4000 gates can source/sink only a few mA (check the datasheet) so aren't good for driving LEDs (except high efficiency ones) directly. How much current does your LED need? There is no such thing as a '12V LED' unless it has a built-in current-limiting resistor. Does yours?
Be aware that CMOS ICs (including the CD4011BE) are static-sensitive so are easy to fry.

#### djsfantasi

Joined Apr 11, 2010
9,160
Draw a picture of your schematic and take a picture on a smartphone to upload here if you don't have the software to do it. Alec_t has a very good point about the current draw, but it sounds like you are using a transistor to switch the LEDs with the logic output. But do you have a base resistor? Many of these questions would be answered with a schematic. If you haven't uploaded a picture, read this blog

#### Rissy

Joined Nov 23, 2015
106
See post #32. Were you using the CD4000_v library and a Vdd-labelled supply?
I forgot about this, i'll admit. I was just using the normal CD4000 library objects. I'll have to look up the "V" version and use that I guess.

There is no such thing as a '12V LED' unless it has a built-in current-limiting resistor.
I was as surprised as you are about this. I've had LED's lying around the house for year (last time I was mucking around with electronics during my student years), and these were labelled as being 5V LED's (green ones and red ones), but when I realised they weren't lighting up with 5V, I chose to sacrifice one, by shoving 12V up it. Low and behold, it worked just fine!? So i've scratched out 5V and replaced with 12V on the label of the baggies.

Draw a picture of your schematic and take a picture on a smartphone to upload here if you don't have the software to do it. Alec_t has a very good point about the current draw, but it sounds like you are using a transistor to switch the LEDs with the logic output. But do you have a base resistor? Many of these questions would be answered with a schematic. If you haven't uploaded a picture, read this blog
I'm sorry, I'd like to stick to verbal questions about issues i'm having. I'm playing this circuit design close to my chest. (Never know who's watching lol!)

I'm thinking to go with first off, a potential divider, to first drop the 12V (variable) to 14V, to something closer to 10.9V-12.72V, and then use a diode followed by an LM7809, with 0.33uF and 0.1uF capacitors, as advised in the documentation. This should then regulate to 9V without much in the way of required heat loss with the required drop in voltage. I'll use this setup for each of my required 12V-14V inputs, as well as the VDD.

Does this sound like an acceptable solution?

(I'm not sure if this will work or not for signals which are expected to switch on and off completely?)

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#### MikeML

Joined Oct 2, 2009
5,444
...I'm sorry, I'd like to stick to verbal questions about issues i'm having. I'm playing this circuit design close to my chest. (Never know who's watching lol!)
...
Then you won't get any more help from me. Get yourself a consultant, and PAY them!!!

#### Rissy

Joined Nov 23, 2015
106
Then you won't get any more help from me. Get yourself a consultant, and PAY them!!!
Oh!

Why?

This is a personal project of mine, it's not a business enterprise or anything. I've just had issues with people in the past using my ideas and efforts for their own doing in the past, so I don't want any repeats of that, especially if they DID use my personal projects to make money (which i'm not interested in.

#### Alec_t

Joined Sep 17, 2013
14,312
I'm playing this circuit design close to my chest
The normal reason for doing that is if you wanted to commercialise a design!
The ethos of these forums is that we share information, for the benefit of all. I, and other members, give our help freely and publicly. If someone else chooses to copy/commercialise something we post, so be it. The fact that the info has been published on a provable date would be evidence if that someone claimed it as their own intellectual property.

#### Rissy

Joined Nov 23, 2015
106
Well i'm not here for commercial reasons. I'm just fed up with certain people in my community being snidey and claiming other people's efforts as their own and taking credit for it.

I'm mostly there now I think, so my questions are more specific on certain areas of my overall project. I'm still keen for help (that's why i joined) but if i can't get it here, then i'm sure i can muddle through or get help from colleagues. But i'd much rather be part of this community.

I'm sorry if i'm offending anyone, but I don't want to put my whole circuit up on here.

#### djsfantasi

Joined Apr 11, 2010
9,160
Then at least answer my base question? What do you have for base resistance on the transistor? And what is the current drawn by your 12V LEDs?