I tried using a transistor as a switch to raise the voltage but that did not work. More unes inputs are tied to GND.
When using the transitor and attaching a LED to its output it even les detectable then without the transitor which doesn't make sense. C is tied to VDD E goes to the 4513 and B comes from the 2 Hz signal. Is this right?
I don't know why it does not work.

No, that is not going to work. That is a voltage follower, and the output will be no higher than the input. What you want is an inverter, which would have a resistor (say 10K) between the collector and V+, emitter to ground, and base to the 2 Hz signal via another resistor (100K). The output is taken from the collector.

Bob

 

I have it working, on a breadboard anyway. I used a simple voltage divider circuit and two diodes and that worked to drive the 4518. Thanks for all the help.

 

I have it working, on a breadboard anyway. I used a simple voltage divider circuit and two diodes and that worked to drive the 4518. Thanks for all the help.

??????? Can you post a schematic of that.

Ken

 

I now built on a perfboard, still working . In order for it to work I need to give something for the hacked clock to drive such as a LED, in order for it to run. i.e. if there is no LED between the output and GND then the signal won't drive the 4518 (Hope that was not confusing). I might try using a diode and a resistor and see if that works.

 

First, with the 180Ω/180Ω divider you are powering the 1.5V clock circuit with 2.5V. From the 4518 datasheet, you need a minimum of 3.5V input, with a 5V VDD, for a high input. You are not going the get that with your circuit. A resistor and and LED will only make the situation worse. Why are not using my circuit with the NPN transistor level shifter?

Ken

 

It works... What more can I say. I initially built your circuit and worked for a little while but due to a defect in building (I think) it stopped working. Then out of curiosity and to make sure I did not fry my hacked clock, i built the simple schematic I posted. Then out of curiosity I tested to see if it would drive the 4518 and it worked. What can I say.. I will go with the simplest method that works....but I still wonder why it works when the voltage is less then the minimum high voltage.

 

I have come up with a theory, If we go from conventional electron flow to actual electron flow, then when the LED/diode is conected between the output and GND and the output is connected to the 4518, electrons are flowing from GND to the "output" of the hacked clock and to the 4518. 5V minus about 1.5V (LED)= 3.5V-enoufh to be registered by the 4518.


I am no electrical genuis and have very little knowledge, so could some one wirh expertise confirm this theory?

 

Is no one answering because my theory is plain stupid? Or do you guys not understand what I am trying to say? Or is it something else?

 

Your theory doesn't make sense. Post the schematic of what you say works.

Ken

 

Disclaimer: I am far from knowledgeable about this subject. This is just a theory.

My project works when it "shoudln't". Much better then it not working when it "should".

I am interested, in why it works.

My clarified theory:

Usually we anaylze circuits using conventional electron flow ( from + to -/GND). Using this we determine that the output from the clock would be ~2.5V, because of the voltage divider. The electrons would then flow down through the diode and resistor and to the 4518 to GND. The signal at the 4518 would be ~2.5V. The minimum high level voltage for the 4518 is 3.5V. So using conventional flow the 4518 would not count.

However, it does count as it should, and I have theory as of why. If we anaylze this circuit using actual electron flow. Then electrons would flow from the bottom to the top of the schematic (-/GND to +). Once the circuit was closed/enabled by the hacked clock, current would flow through the resistor and diode and to the 4518 and hacked clock. Here is where things get out of my knowledge. I may be totally wrong here. What I think would happen is there would be about a 2.5 voltage differance on the left side of Connection 1 (C1)., but on the right side there would be enough voltage drop to drive the 4518.

 

Electron flow vs conventional current makes no difference in how a circuit works. They are just two different ways of looking at current. It's like saying "the ocean is near the shore" as opposed to "the shore is near the ocean". If I have time (I have 4 days off ) I'll through together your layout and see what happens.
Ken

 

Alright, I am not very knowledge about multiple voltage drops, but it still seems to me that there would be one circuit through the hacked clock(2.5V), and another circuit through the 4518 (~5V).

My multimeter isn't quick enough to measure the voltage drop of a short pulse, and I don't have a osciloscope.

Thanks for all your help so far, it is greatly appreciated.

 

Ok, Got a chance to breadboard the circuit.
First it turns out there is a difference in the output drivers on some battery clocks. The hacked ones I've been using hold the output low (~0V) and pulse high (~1.5). The one I tried today held the outputs high (~1.5v) and pulsed low (~0V). So I have two different approaches for different clocks. Attached.

The second attachment is my experimental circuit. Note that the direction of the diodes is reversed from your circuit and the resistor is tied to V1 rather that ground. This is because of the particular hacked clock that I referred to above.

The clock chip was tied to a variable supply and the 4011 logic chip (I didn't have a 4510) was ties to +5V. I was actually able to run the clock up to 5V...with no smoke! With V1 < 2.7V it would not drive the 4011's input. With V1 between ~2.7V and ~2.9V the gate was acting in a linear inverter mode. With V1 at and above 3.0V the gate was functioning as it should with 0v to +5V output swing. Inserting your LED in series with the resistor just required a higher V1 (~3.9V) to drive the gate properly.

So what it seems it that you were lucky in pushing the edge in powering the clock way beyond it's design voltage, and also on the actual minimum Vin on the 4518. The spec'd 3.5v Vin is the design limit to assure reliable operation throughout its production. Don't know why you added the LED. Did it blink with each pulse?

And....I learned that the clock chips come in different flavors. And, that "it appears" that they may be able to operate at 5V logic level. I'm running mine for a few days to see if the operation, or pulse period, changes.

This is why mucking about with actual circuit hardware is so much fun.

Ken

 

At first I used the LED to see if it worked, then I noticed that it did not work without the LED there, I eventually replaced it with just a diode.

The thing is every clock is alightly different.

I am assuming that my two different ciruits, voltage drop theory is not valid?

 

At first I used the LED to see if it worked, then I noticed that it did not work without the LED there, I eventually replaced it with just a diode.

Have you tried it with just a resistor?

The thing is every clock is alightly different.

The output polarity is definitely a difference. Not sure all would work on 5V.

I am assuming that my two different ciruits, voltage drop theory is not valid?

I couldn't follow your theory, so I just skipped over it, and went with what you were doing.

Ken

 

I don't know how to describe my theory clearly.

I have not tried just a resistor, but it is too late now.


I have built the entire thing, and it works.

Here is a link to the thread: http://forum.allaboutcircuits.com/showthread.php?p=624848#post624848

THANK YOU.

 

kurtruk,

Just an update. I ran my 1.5V hacked clock at 5v for about a week. No smoke!
I did monitor the output period on a counter. The period was very, very stable at any voltage between 1.5V and 5V. However, the higher the supply voltage, the shorter the output period:
1.5V = 0.9999789 Sec
5.0V = 0.9999051 Sec

Ken