Kevin Weddle

Is it common practice to use higher voltages with transistors if the current is low? I have read that the current should be half the rated current. I suppose if you want to use a higher voltage, then low current is alright, but not optimum.

I want to use a 50ohm cable. This gives you capacitance as well a inductance. Somehow the load impedance will affect the line impedance. If you say the load impedance is closer to a short, then the length of the cable should be chosen to be a series inductor capacitor at the resonant frequency. This is the 50ohm seen by the source. I want to make sure of this with the oscilloscope. A standing wave will appear as an offset of the signal. Am I right?

I believe the output capacitor is tagged on to some power supplies mistakenly. If you were to take the most simple circuit and want a supply that will maintain a voltage, then I doubt it would include a large capacitor. A constant voltage with a change in current is merely a simple task. Most of the finished products in use today use simple voltage regulation. That is the voltage that occurs on the output is fed back, inverted, and applied to the output. This is a simple voltage loop, for lack of a better description.

I have seen a power supply with the zener controlling the emitter voltage of a darlngton. The collector goes to the control. The input is also fed to the control through a resistor network. Why can't a resistor be used in place of the zener in this situation? I am thinking the signal works better on the PN junction against a constant voltage. How does this work? Has anybody seen a zener used for this purpose? Maybe the zener has a very high signal impedance.

I can't find the data sheet on the regulator. I know the circuit was operating at 25vdc with the input, the ouput, and the control all at this voltage. It might be a variable voltage regulator. I must have the data sheet to do anything with it.

The problem with a simple gate is the slew rate. The gain is not high enough through the transition. A low to high transition will not produce a high quick enough. If you make the gain very low with a pulse input, you get a pulse output the with the same reduction that you would have with a sine wave. A reduced amplitude means a lower rate of change.

I am perplexed. Why is the photodetector forward biased? The gate must me CMOS. Am I right? I have another question. The original circuit is what I am concerned with. It has a battery that works in conjunction with the power supply, but without a common ground. The use of devices without a common ground is circumspect, but it still doesn't appear to have the correct voltages.

I want an easy way to determine the matching characteristics of a line. I want the output device to be any resistance, the input device to be any resistance, and the cable a 50 ohm. What is the length? How can we determine this without going to any trouble?

I found the zener to be a 8.5v approx 1.5 A . This would make the control voltage approx. 12v because of the transistor. The input voltage is approx. 70v though. This is very high, but there are large divider resistors which means there is not much current from input to the control. The current comes through the collector of the transistor through the control line. Can a regulator accept an input voltage of 70v down to 12v on the output?

What you are saying is that a large capacitor on the output will allow more current when there is a sudden current burst.

I already took apart the circuit. The large capacitor on the output is open for debate. I have seen smaller capacitors on the output. I think this is correct because the voltage regulator regulates not only the dc, but the fluctuation as well. If you design the voltage regulator to have a small impedance output, then it would be alright. I tend to think it was designed for an impedance equal to the voltage divided by resistance.

I have this power supply that I was given that takes the 120vac and converts it to 50vac. The current of the bridge is 1.5A max. The regulator has a control line, input, and output. The output is fed to the base of a darlington and the collector goes to the control. The emitter has a zener diode that gets it's voltage from the ouput. The input to the regulator is 50vac filtered to 70v dc. I suppose the output is approx. 70v dc. What is the voltage of the zener? Is it 70v? I have already taken the circuit apart so that I can reboard it. The circuit has a large capacitor on the output, which in my opinion is negligent design.

I don't think the outputs are differential like they are supposed to be. The outputs are complementary. The purpose of the differential is to treat the inputs as two signals. Sometimes this is better than using ground because the ground can contain many signals.

I am perplexed at the use of large capacitors on the output of the regulator. I have seen small ones used in their place and I think this is correct. The regulator would have trouble trying to change the voltage with a large capacitor. And remember that the DC voltage is always determined by the load, not the capacitor.

I would guess that the timing is the most important aspect. What is the C function for a delay? You will probably need quite an interface for the computer. The rate of the A/D would not coincide with the computer. Can anybody suggest an interface device. Would this be the job of a PIC? Doesn't a PIC need to be preprogrammed to handle a specific slower device. These are just guesses.

It looks as though it is the front end. Am I right?

There exists more than one design for the opamp. I can't remember which part of the opamp is the high gain. Was it a pushpull at the front end or the middle part?

There are two types of comparators. A voltage difference exists and the output goes either high or low depending on the polarity. The second comparator is a digital device. It compares two digital sequences that are weighted and outputs either a greater than, a less than, or an equal to. I think this is how it goes.

I think a sweep input implies that two signals act on each other. I don't think there has to be modulation. I think the sweep part has to do with the fact that the other signal can have any time base. You can get interesting results when you combine two signals. The TTL output is your normal output and the 500 ohm could be for transmission purposes if the other signal has a short time base. This gets into impedance matching.

I really like how someone posted the opamp schematic next to the state variable filter. I have been able to draw some conclusions about the opamp circuit in the past. I have guessed that the input transistors are low current. Am I right? So the variable filter really stumps me. I think this one of those circuits where the results are just good enough. I still don't see output bias. But I know that once again they will throw anything around an opamp. Sometimes you can get extra current by using low value resistors, but the impedance will be too low. Any suggestions?

I have a homemade inductor that I will use in a power supply circuit. It has a very high current rating, but it does not obey the equation very well. It is made with many strands of magnet wire, combined together, then wrapped around a transformer core. When I increase the frequency 10 fold, it should increase the resistance 10 fold, but it doesn't. I do, however, trust the high current possibilities. I have to wonder what the resonant frequency will be if I add the capacitor to it. The load would then be in parallel to the capacitor.

Did you know that positive feedback is different than phase shift. The phase shift margin needs to be less than 360 degrees, but positive feedback can occur at any time.

This power supply looks a little simple. I would design my own and forget about the opamp, most likely. With so many possibilities, you have to wonder what the ideal operation is. This is the circuit I use presently, but I need a better one.

I have noticed that the BW is very narrow on some opamps. Why is it so narrow? Do you think it is the transistors they use? Maybe the sheer number of transistors and their associated capacitive properties contributes to the overall reduction. So I think this is the frequency at which the capacitance becomes the overall problem. What is the net result of negative feedback in a transistor anyway?

The high frequency aspects of the inductances and capacitances are something I don't want to deal with. You must also know the characteristic of your measuring device. This will be a bridge or simple circuit. The final measurement device will be a DC meter or AC meter. Can I safely say that 10Ghertz is too high and 1 hertz is too low? Maybe the time difference isn't so great given the relativity of things. You will notice that high frequency networks deal of capacitors and inductors that are of reasonable size.