Kevin Weddle

Are you using just a transformer?It's interesting that voltage and current are used so often in electronics, yet it's the voltage which produces the arc. How does a generator produce a low voltage AC which can be converted to a high voltage AC? In other words, where does it get the current? Is it the physical size of the generator, because the rate gives you amplitude?

I am bewildered by all the different types of regulators on the market. Regulators are simple to construct if you are dealing with one small signal application. What are some examples of the different types of loads one might see in electronics?

You need more gain. Try using a push-pull. What an OR gate uses is a phase splitter. Which works the same as a push-pull. A push-pull is what an opamp uses. The formation of logic begins to be a little trickier, doesn't it.

Has anybody seen the sawtooth wave generator that uses a PUT in parallel with a capacitor to get rapid discharge. The input overdrives the PUT past a short, but allows voltage across the capacitor. When the input drops, the capacitor will rapidly discharge.

This will require the formula wavelength = (.67)100x10E6 / frequency What do you do with the wavelength? You also have to incorporate the load resistance. I would make the length some fraction of the wavelength. I would assume the load to be a short because it will be closer to a short. If you make the length the correct fraction of the wavelengh, you will get the line at it's resonant frequency. You don't want it to appear inductive or capacitive, but do you want it to be parallel resonant frequency or series resonant frequency? I think it should be series resonant frequency because a 50 ohm cable is closer to a short. Any suggestions about measuring the end result.

I must elaborate on an earlier statement. The input capacitor, associated with phase shift, does not inherently change the characteristic of the opamp. Instead, it changes the input signal such that it will not produce 40db roll-off. This makes the graph misleading because it shows the 20db being extended. If you accept that the capacitor is separate, then you will be left with the graph. The graph shows a change in critical frequency thus extending the 20db.

You can't predict how the motherboard will be affected. It will affect the performance before anything. What you have, with a bigger wire, is an inductor between the transistor/resistor and the supply voltage. This might make the gains higher.

You have to account for the time when, after the input saturates, the voltage across the inductor drops. So it would seem that cutoff should happen right after saturation, but it can't. This creates a time after saturation that determines shape of the ouput.

You certainly need an oscilloscope to determine how the siganal is changing. It might vary the pulse width(duty cycle), the pulse time, or the pulse amplitude. This is going to require a more complicated matching circuit. I would consult an electronics guy to develop the circuit.

Shift registers are just serial devices. You would have to wait for them to fill up before taking a parallel output. They require you to clock in the data. The ouput of the A/D is going to be serial with a time base dependent on a clock or the signal. A multiplexar is similar to the shift register except the data won't change while it's filling up. It will require a clock too. Use a multiplexar, clock in the data at the rate it's coming from the A/D, and take the output which will only change after all the bits have been loaded.

From my circuit, which is similar to Siddharths, the input signal is drowned out by a capacitor and the zener gets it's voltage from the ouput through a resistor. Interestingly enough, the voltage of the darlington is high, making the current low. The rated current is 1.5A. I understand the current should be at the midpoint. Also, my circuit doesn't have a pass transistor, it's a regulator.

Remember that an oscillation has to do with a change in polarity across the input. This is what causes the voltage to go in either direction. An oscillation is simply anything around an opamp circuit that would cause the input to change polarity. This could include the feedback. Any good scientist will tell you that a configurable device such as an opamp is just as good as the product. But if your talking DC, then there is no additional phase lag and the opamp works fine.

Has anybody seen the circuit with the PUT across a capacitor across an opamp. This generates a sawtooth. A rapid discharge of the capacitor. I think these 5 layer devices pose a serious physics debate. You know the construction lends itself to PN voltages even though we treat them as a short. There is some serious doping going on with these devices.

Is the signal fed through the capacitor and inductor? This makes for a changing voltage at the negative terminal. To make it work for me, I would ground the 9v negative terminal to start.

I would bet there is a debate regarding sinewave generation. I have seen an example in this forum that uses a square wave, but with an added DC level during the middle part of the transistion. They called this a modified sinewave. I agree that a squarewave can be used to make a sinewave. But I know that an oscillator most resembles a triangle wave.

Has anybody seen the calculations for the more complicated LCR circuits. I understand the basic circuits, but any addition is not well illustrated in the book I have. Take a simple series LC and parallel the capacitor with a resistor. This goes beyond what I have seen being calculated. If a circuit is nowhere near the resonant frequency, can we assume the regular impedance of the capacitor and inductor? We could start with voltage subtraction. The inductive reactance tends to subtract from the capacitive reactance. A parallel resistor will tend to make it an LR or RC circuit XL = 50 XC = 100 then impedance is 50 Make it an RC circuit R=200 then impedance = 300 so the impedance is somewhere between these two values.

The circuit I have gets it's zener voltage from the output. It is a commercial power supply.

Thank you for correcting me audioguru. The only difference is that the zener gets it's voltage from the output through a resistor.

Siddharth, you know the circuit I am talking about. The zener, because it gets it's voltage from the ouput, will drown out the signal at the emitter. The base still has it's signal. So the signal at the PN junction is really onesided. But if you use a resistor, then the function will still be the same. It's kind of arbitrary, but isn't it interesting. What is the impedance that the signal sees? I bet it's just re because of the constant voltage. But this would make for poor biasing.

A sine wave is sometimes made from a triangle wave. This would mean the use of an integrator. I would use a transistor and a capacitor on the collector to construct the triangle wave. Then a simple filter would be required. The filter would be to attenuate the low frequency of the triangle. Use an oscilloscope. Simple.

The regulator is a motorola 513 je 800. I think the je800 is the part. The zener is on the emitter of the NPN control transistor. The collector goes to the control of the regulator. The base is connected to the ouput through resistors. The zener gets its voltage this way. I wonder about the zener. Why is it there? Why is it better than a resistor.

I don't think the voltages are respective. You have to connect the 9v with the 5v somehow. What about the signal at the 9v negative terminal. If you have a signal there, you will have a signal on the other side. You have the 5v connected to the 9v through the photodetector and the 5v connected to the 9v through a 1kohm resistor. What is the voltage at the 9v negative terminal? Is it 5v? That would put the 9v positive terminal at 14v.

Opamp compensation serves another purpose. The input capacitor, often seen with compensation, changes the critical frequency. If you notice the bode plot graph, you will see that the critical frequency can be anywhere. When you add a capacitor, the critical frequency moves from the 40db per decade back to the 20db per decade. This is difficult to see because it shows that the bandwidth is extended. Which is true, when you change the critical frequency. So what I am saying is that the lag through the opamp at high frequencies is due to the reverse collector base of the transistors.

The circuit I was refering to is a commercial circuit. A resistor could replace the zener, but why did they use a zener. This is a simple transistor biasing arrangement with the zener at the emitter. By the way, the difference in a regulator and a zener is not much. The reverse bias zener is just another way to accomplish the same task. This is a physics debate.

I don't think opamps would be very popular if the phase shift were a problem. The problem is that the opamp could become unstable.What does this mean? The input signal can not drive the output to it's voltage level. Am I right? This would be because of the negative feedback of the individual transistors.