Electron Reversing Device

[poynt99]

Well my meters just wont read a voltage that is going high then low,so i collect the peak to peak voltage and store it in the 150uf cap,so as i have a stable voltage to read.
I then placed my scope across the output of the FG output,and make sure the cap is correct.But we must remember to add the voltage loss from the diode aswell to that cap voltage-and this will vary slightly depending on what diodes you use ofcourse.

Ah ok, now I understand. You are reading the positive peak voltage.

I would also like to say thanks for the help and input you have given so far-its been a great help

You're welcome. I hope my long post explaining things made sense to you?

ETA: If you leave the bottom meter as shown, you can easily see which current is the negative one, then just flip it's polarity in your head for your computation. It would be interesting to see which current is higher in amplitude and which is lower...don't you agree?

[tinman]

Ah ok, now I understand. You are reading the positive peak voltage.
You're welcome. I hope my long post explaining things made sense to you?

ETA: If you leave the bottom meter as shown, you can easily see which current is the negative one, then just flip it's polarity in your head for your computation. It would be interesting to see which current is higher in amplitude and which is lower...don't you agree?

I do agree with you on that,so i will leave the meters as you have them depicted when i do my test.

Now here is the untouched schematic for the FG.
As you will see,i only removed the switch and line for the sawtooth wave side of thing's.
So this would in no way change the square wave operation.
The circuit is just a cheapy from jaycar-probably made in china.
But it dose what i wanted it to do for the time being.

[poynt99]

OK, that FG circuit now makes more sense.

You were right, there was no effect with the parts you removed from the schematic. The sqaure wave output is from the normal pin 3 of the 555.

For the "sawtooth" output, they are simply buffering the timing cap wave form....not very effective, but an interesting and simple idea. It will look like a curved sawtooth, more than a linear one.

[poynt99]

tinman,

When I explained the correct method of computing the average FG current, I was thinking in terms of measuring this with an o-scope, then performing the computation using both peak currents. I made an error here because you will actually be using two separate meters to measure the two separate currents, negative and positive.

So for this method, you were right. You will simply add the two currents together. Why? Because with a 50% duty cycle, the meters will be averaging each polarity of current separately. So for example, if we have +40mA and -50mA (peak), the two meters will actually indicate +20mA and -25mA (average) respectively. Add them together and you have the correct 45mA average current.

Sorry for any confusion.

[tinman]

tinman,

When I explained the correct method of computing the average FG current, I was thinking in terms of measuring this with an o-scope, then performing the computation using both peak currents. I made an error here because you will actually be using two separate meters to measure the two separate currents, negative and positive.

So for this method, you were right. You will simply add the two currents together. Why? Because with a 50% duty cycle, the meters will be averaging each polarity of current separately. So for example, if we have +40mA and -50mA (peak), the two meters will actually indicate +20mA and -25mA (average) respectively. Add them together and you have the correct 45mA average current.

Sorry for any confusion.

Now poynt99-your adding confusion to my confusion lol.
But on a more serious note,my first thought's were as you just stated above.
As we were not long ago talking about the meters averaging current,it made sence to me that we add the two together.
But when you posted your comment about deviding by 2,that also made sence.
As it is a square wave that is either on or off for half a cycle,i thought the meters might read the peak current of that half cycle-so your comment about /2 made sence aswell.
But i was comeing back to put forth my original thought of adding the two together,as there would be a 0 current draw on the meters for half a cycle-so the meter would average between peak and 0 current draw.
But i see you beat me to it-so we are on the same page now.

I think this is one situation where people can get fooled into thinking they have something OU.
If we had of gone for the devide by 2,we could have been making that very error we try to eliminate-and we could of had an OU machine that wasnt OU.

I will put the scope across the meter,as i know the shunt on ma is 1.5 ohm's.This should tell us exactly how accurate our meters are,and what they are reading.

But for my first input-output measurments,i will add the two DMM readings together and use that figure as a starting point.
This will give us a close indication as to how much more we have to get out of the circuit to get to where we want to be.
It may even tell us that there is nothing in the circuit at all aswell.