Kevin Weddle

I tried to start this oscillator by using a battery source. When I connected the battery, the oscillator seemed to be in phase with this other oscillator on the same circuit board. I don't know how exactly they were in phase, but it was close. Does this make much sense to anybody? The only way I know of to get them out of phase is to use an RC circuit to make sure of the phase difference.

Here is the example I can't get to work. What could the problem be? rs232.txt

What determines the startup phase of an oscillator? I have the feeling that the startup is in phase with this other oscillator. Could it be that the oscillator startsup in phase with another oscillator just because they are in close proximity and have the same frequency?

The circuit is correct. The problem is that the classic oscillator has a capacitor charge discharge to it. This makes the square wave not so square.

That is one strange device I must say. It would be interesting to see the circuit. One capacitor with four leads that go to wherever. It could be a parallel path. Have you thought of that? It's like the diode with three leads. They make them that way to realize a certain function. It is actually two diodes that are arranged as pullups to a 5volt line. I think they are inverted diodes. You see them on DIO lines.

If you want to incorporate transistors with logic functions then you need a series resistor and a parallel resistors combination. I normally disregard the sink and leave it open. The source current should fall within the recommendation. You can bias the parallel combination to give you approximately zero volts. Otherwise it is your disgression to have whatever peak to peak you like. I hope this is a start.

I think a short is possible, but not on both sides as you have said. Maybe both sides have a connection to ground. Could you ground a can. Inspect the traces for ground connections.

Hotwaterwizard, how do expect to get 220v from a 9v without a stepup transformer.

With shorted leads, aren't you shorting the two traces together. If they are supposed to be shorted, then why aren't they. They normally don't short two traces together using a component.

That looks like two capacitors rolled into one.

Look at the traces. Are the leads going to different places or are they combined on the board, such as ground? A four leaded capacitor does not strike a chord with me.

The switching power supply would take the 9volt supply and switch the input from 0 to 9. The step down would take this and reduce it. So no. Remember also that the 9volts has to be directly on the transformer without a resistor. A resistor would turn it into an LR circuit which would create a negative transient when the pulse goes high caused by the tranisistor going into cutoff.

What is interesting is the RMS calculation. The equation is squared before integrating. The sin x is much different from the (sin x) squared. You then simply take the square root. The mean is based on the sinx. So to say that it is the root of the mean squared is not accurate. It's actually the root of the integral of sinxsquared which is not the root of the intergal of sinx quantity squared.

I found the GND to give me a ground reference with respect to the signal. I knew it would work. The heat is from the power supply I feel. One of the first things I did was to test an opamp backwards. The gain setting resistors on the noninverting input to output. It worked. The only thing is when the signal is applied to the inverting input, I can't keep the bias. This makes sense because this is the only opamp configuration that you can have that will cause an increase on one input and a decrease on the other input. This pulls the difference far apart and you lose bias. I was able to get half the sine wave with this configuration but the other half was clipped.

I have since found that a third inverter added to the classic oscillator makes for a much better square wave. MP said that a two inverter oscillator migh have trouble starting up. But I found the third inverter to make the square wave better and the startup was never a problem. The output of a two inverter is really bad.

A diode would reduce the voltage, but I don't have a low enough current diode. So I would still need a bias resistor to ground to make the diode operate.

I finally received the oscilloscope. It helps very much in determining pulses. I think I got a good deal as it was only $150. The thing get's very hot too. My only concern is the GND. It does not seem to do anything. Isn't it supposed to show a ground reference trace when GND is selected? Other that, I think the design of the thing is tried and true due to the overmanufacturing of the these devices. It is the crudest of oscilloscopes. What do you think?

Yes. You can not cancel the pi.

It's the first part you will have trouble with; a frequency measuring circuit. This is done with a differential amplifier. You compare the two frequencies, of the same phase, giving you a zero output. The phase part is the hardest. You will need to set the phase up using two generated clocks. One clock is the signal, the other clock is the reference oscillator.

I found a good solution to my problem. I used a parallel resistor to pull down the voltage. Then I used a parallel capacitor to distort the pulse a little. This ensured that pulse did not have the duration because a capacitor tended to round the sharp edge. Now I intend to find the right capacitor and realize the purpose of the circuit.

Yes, the pulse is of a very short duration. I found that a series resistor won't work because of how easy it is to produce a high input. All it takes, in logic, is to produce a no voltage situation since this will reverse the PN junction.

I have a pulse applied to a latch when I don't want one. How can I change the pulse so that it does not trigger the latch? Would a series resistor be better than a parallel resistor? The way logic works doesn't leave me with many options. What I need is a weak source and a weak sink. That sounds like a series resistor.

I operate my transistors at mid current and the beta is fine. This transistor has a beta of about 113.

There seems to be a misrepresentation in the voltage level. 12VAC after rectification has an average of 12V. Once you use the capacitor, the DC value becomes 17V.

I think your PNP is biased at too low a current. It's supposed to be at 100mA.