# Change voltage circuit

#### Kev0511

Joined Jul 20, 2004
39
When the input signal 2 is come from a switch then R2 is the pull low resistor when the switch is off, the R3 and C1 are the switch debounce, if the input signal 2 is not come from the switch but a logical signal then you don't need the R2 and C1, if the input signal 1 also is not come from a switch then you don't need R4, if you can accept the original status of voltage changing then you don't need the C2 or change to a smaller as 10uF/10V.

The output voltage that you used the voltage divider, I didn't change the values of resistors and I just thought that you can calculate by yourself, it is not so hard.
Well odds are no i won't need the R2, R4 or C1, since those inputs are pulled high, be it with just a 330ohm resistor with a possible open collector (? need to check) Mosfet / power transistor based trigger to ground and it will not be a quick pulse either.

Also i like your thoughts on using C2, may up it a bit for a more gradual change thou?
And i'm not blaming you on the Output voltage change side, i know you used my numbers, but those were done using a parallel resistor and voltage divider calculators and and trying to get as close as i can via "standard" resistors. So possibly i'll need @330 ohm resistor and a 100ohm pot to fine tune the resistance to dial in a number that would be correct.

Since this one seems to work, and decent well to at this time. It's time to venture onto another option and see what that does. Hey i want to try out everyone's thoughts, it's only fair, if one puts effort to try and help me, it's only fair for me to try what you put together.

Thank you to all again

#### ScottWang

Joined Aug 23, 2012
7,342
Well odds are no i won't need the R2, R4 or C1, since those inputs are pulled high, be it with just a 330ohm resistor with a possible open collector (? need to check) Mosfet / power transistor based trigger to ground and it will not be a quick pulse either.
As the output of CD4013 was just like what I said that the logical signal, it won't generate the contacts bounce, if the input is using the switch or relay, that is not the logical signal from ic and it could generate the contacts bounce, it means that it will send out over ten or twenty pulses, not just one.

Also i like your thoughts on using C2, may up it a bit for a more gradual change thou?
If you have the O'scope then you can do the test to see the voltage changing.

Since this one seems to work, and decent well to at this time. It's time to venture onto another option and see what that does. Hey i want to try out everyone's thoughts, it's only fair, if one puts effort to try and help me, it's only fair for me to try what you put together.
When you do more experiments then you will get more experience and learn more theories, that is a good thing and you could choose which one is better for you.

#### Kev0511

Joined Jul 20, 2004
39
12 days later i''m back on this project!

So i'm still tweaking it a bit, and other then trying to make a prototype board with the circuit on it, and see what happens in a car.

I'm thinking of using the ECU's 5V output as part of the voltage divider. Which in turn will make the circuit smaller (IE no need for a 7805 or LM317 and so forth) Also for another circuit (Fuel Temp) i ca just use a resistor to ground (in this case a 1K8 1/4Watt) to get that signal that it's looking for.

The only thing i can see using a 5V regulator for is possible to have a constant 5 volts for the 4013 instead of being in the 13-14.5V range, but with the pulled to 12V high (through a 330 ohm resistor) on the inputs, would that even work?

#### ScottWang

Joined Aug 23, 2012
7,342
1. What is the 12V power source come from, is it just connected to 14.5V or a 12V through voltage regulator?

2. You should know that the Vcc level of CD4013 should be the same with 12V (or 14.5V), if you provide 5V for CD4013 and the inputs (pin 10, 11 of CD4013) is 12V level then CD4013 got the chance to be destroyed. (What's reason that you can't connecting the Vcc of CD4013 to 12V?)

3. If you have any reason to provide 5V Vcc to CD4013 then you should be careful the inputs voltage level of pin 10, 11, the inputs voltage level should be the same with 5V.

4. For a light current load, if you wish to get a 5V from 12V(or 14.5V), there are two methods that you can do, the first method is using 78L05, it is a very small type TO-92 as 2N3904, the second method is using a resistor (330Ω for 12V, 470Ω for 14.5V) and a 5.1V zener diode then it can be providing around 20 mA current for 5.1V zener diode and Vcc of CD4013.

#### Kev0511

Joined Jul 20, 2004
39
Scott I will get back to you, but all and all I may just leave it powered by the car (12.8-14.5volts) since the inputs should be around the same voltage.

Well I find it mildly depressing when I build the circuit on a breadboard and it works, I go build it one a strip board to try in a car, and the circuit doesn't work?!? I've gone over it a couple of times and I just can't see what is wrong with it.

Hopefully I can figure it out?

#### ScottWang

Joined Aug 23, 2012
7,342
Scott I will get back to you, but all and all I may just leave it powered by the car (12.8-14.5volts) since the inputs should be around the same voltage.

Well I find it mildly depressing when I build the circuit on a breadboard and it works, I go build it one a strip board to try in a car, and the circuit doesn't work?!? I've gone over it a couple of times and I just can't see what is wrong with it.

Hopefully I can figure it out?
Recheck all the parts, sometimes you should using the multi-meter to do the real test, and you can using the 1/2(middle position) method to find out where is the problem, and what is the 1/2 method, that is when the circuit can't works as your expected and then check the circuit from middle (1/2) of the circuit to do the test, to find out the problem is at the left side of middle or the right side of the middle, and also the same, separate the left or right side as 1/2 again, the position of the circuit as :
... 1/7 ... 2/7 ... 3/7 ... 4/7(1/2 middle) ... 5/7 ... 6/7 ... 7/7 ...
So check from 4/7 and then 2/7 or 6/7
Separate the circuit as 4 sections, and check it from 1/2 (middle), if the circuit is a big one then you can separate it to more sections, so it will be easier to find out where the problem is.

#### Kev0511

Joined Jul 20, 2004
39
I know i delay with responses and trys, But sometimes life and laziness gets in the way

I Have no idea, but at this time it seem like BOTH 2N3904's were either bad to begin with, or failed in circuit. Since neither work any more.

1 doesn't even 'switch", and in my breadboard setup, increases the voltage at the Junction of the voltage divider, The other is always "on", i plug it into my breadboard setup, and the voltage instant drops to 1.7V (instead of 1.6V) even with the 10K resistor not going to the 4013, and using a jumper wire to try and trigger either!.

Time to go over the board again (for least the 20th time!) and see if something isn't right, Least at this time the 4013 still seems to work, being it works in my breadboard setup.

Update on the 3904's, Seems like the latch batch i got of them had some bad ones in it, i'm guess least 4? Well see act the same (increasing the voltage at the divider, with one barely working (as above)

NOW i need to try and get some more for stock.

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#### Kev0511

Joined Jul 20, 2004
39
Well Without the 4013 install (have an socket) (also am using the 1-7 side by the way)

If i put 12V to pin 1, the Output goes to 1.6 volts
If i measure pin 3 i get ground, but when i ground input Sig 2 i get 12V
If i measure pin 4 i get 12V (via a 330ohm resistor to 12V) and ground input sig 1 I get ground

From what i gather, it's a working correctly

NOW with the IC installed
pin 1 has 0.575 volts
pin 2 & 5 has 12 volts (which makes since)
pin 3 has 49mv
pin 4 has 12 volts
pin 6 has 0 volts (IE ground)

Now grounding the inputs.

Input 1 grounded, pin 1 goes to 0.624 volts
Input 2 grounded, pin 1 stays the same.
Inputs 1 and 2 grounded pin 1 goes to 0.620?! Ain't it suppose to go to 12V?

Another device i'm going to need to do some testing on!. I have tried it on my breadboard but that's using side 2 (as with the diagram Scott made) and it seemed to work. Can one side fail and not the other?

thanks again

#### ScottWang

Joined Aug 23, 2012
7,342
The below is the CD4013 D F-F Functions simple tester, I haven't test, if you want then you can try it, Led limits to about 4 mA.

I didn't design the auto reset, so it could be lighting the led with any of Q or /Q, if you want a auto reset then you can connecting a 0.1uf /25V to reset pin 10(R2) and +12V, and R4 change to 47K, the circuit was designed to test the toggle function of D F-F, every time when the Sw1 was pressed then the Q and /Q will be change as a toggle status, you can also using Sw2 and Sw3 push button switches to test the set and reset functions, and then all the functions will be tested

* When you testing, some pins of the another un-test D F-F should be connected to Gnd as D, Ck,Set, Reset.

#### Kev0511

Joined Jul 20, 2004
39
Always delayed in my response

Scott; the 4013 I think was just dead, tried it again in my breadboard circuit and it didn't work on either side, also thank you for the testing schematic!

Also learnt after reading more on the test I'm trying to simulate, is I'll need to get the triggered side down to around 0.25-0.5 volts instead of the 1.6ish I thought it was! Easy enough with a change of resistor(s) so that may not be a problem.

Also posting on a iphone isn't fun at all!

#### Kev0511

Joined Jul 20, 2004
39
Wow as a voltage divider I'll need around a >24 ohm resistors ground if I use a 470 on the 5v input! Heh I'm thinking moving up to say a 4.7k and 240 ohm resistors may be in the needed!

Now a mild question

What if I need to pull it fully to ground to get the right value? IE having a voltage divider giving me 2.5 volts, what if I need that value of 0 volts?

#### ScottWang

Joined Aug 23, 2012
7,342
What if I need to pull it fully to ground to get the right value? IE having a voltage divider giving me 2.5 volts, what if I need that value of 0 volts?
If you didn't use the R6 then the output still can't be a 0V when you using an npn bjt or n type mosfet, if you using npn bjt then the Vce has about 0.1~02V, if you using n type mosfet then you can get a pretty low Vds voltage as several mV or over 10 mV as 10mV~100mV, it is depends on which mosfet that you use.

If you really want to get a 0V output then you can using the N.O contacts of a small reed relay to replace R6, and another contact of reed relay connected to Gnd, you also need to cut the Vc of Q2 from R6, and the Vc connecting to the coil of reed relay, another pin of coil connecting to +5V, and the coil of relay in parallel with a 1N4007 diode (reversed), you can try to draw the circuit.

#### Kev0511

Joined Jul 20, 2004
39
If i remember correctly, i will try to do a slightly more better post tomorrow.

Still rereading the specs on this test (one part is confusing thou)

EDITED!: Um i bad information on this page. I needed to convert PSI to inHg to get a correct figures i needed, it's late, and i need to get up early so i'm calling it a night for now

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