Kevin Weddle

It sounds like you want to use a simple logic device but with an analog signal. Maybe an analog switch. Check to see if you can't use a digital device to enable the analog ouput. Otherwise, you could probably use a couple of emitter followers that are enabled or disabled with logic. You might have to use a couple of transistors to get the voltage right.

Audioguru, we are not talking about rail to rail operation. We are in fact able to reduce the gain with the feedback resistor because the opamp is a symmetrical device. The positive feeback resistor can lower the gain because it is infact still a collector resistor. You will notice that it works a little different though. Just design it backwards using about a 1k load to -18volts.

This is where I have trouble with the datasheets. Sometimes the data can be irrelevant. In this case, they should show the output portion of the circuit with a signal applied. That way you could add your resistance and determine how it will affect the output. What they are actually saying is that your load should be "independent" of the operation.

What I find amazing is how the transition will allow for the charging of the capacitor. The transition is of a certain duration. If I use a long time constant, I think that the transition is of a longer duration. This is how I get an oscillation at the low frequency.

You need to gain first, then filter. The filter should be simple construction, not to say that you couldn't use a capacitor around the opamp like a state variable filter. You can use many stages so that each stage will properly handle the amplitude of the signal which is continuing to be amplified. I might stay away from opamps because you don't need the differential inputs. All you need is a high gain pushpull with arbitrary impedance.

One might get a better feel for opamps if you connect the negative feedback from ouput to noninverting input instead of the inverting input. This is called positive feedback, but the idea of gain is relatively the same. In other words, connect the opamp backwards and realize the similarities.

Referring to the original circuit found on the home page. I can see that you are using the 5volts respective to 9volts through the infrared diode. I don't think this is a reliable situation. I could go into detail as to why, but I will leave it at that unless you are interested. The signal, by the way, may not care that the DC voltages have run amuck because of referencing a voltage through an infrared diode.

Audioguru, thank you for describing my circuit. I used a pullup because the example in my book uses a pullup. The pullup is on the output of the first inverter. I did not happen to notice that I constructed a classic CMOS oscillator, though it appears to be. I have this clock hooked to a BCD counter. The BCD counter should actually be called a counter decoder because it has 10 ouputs for a clocked input. I would have preferred a regular counter fed to a decoder, but the BCD counter seems to fulfill this purpose.

The pullup is on the output of the first inverter. By the way, I stumbled upon the fix. I used a 10M resistor from the input of the second inverter to it's output. I would have thought that a resistor to ground on the input of the second inverter would work. But it seems that the complimetary voltage works better. I guess you can call this arrangement a fed back inverter.

The idea of a D/A is surrounded by the weighted ladder. Each sequence is able to produce a current that results in a voltage. But I am wary of this opamp configuration because of how it is used. The opamp is meant to operate high gain, even though it will operate as a constant current source. I would be tempted to change the configuration a little. You could probably use a transistor circuit and achieve what you want.

I was looking for an ordinary counter, but I only found a BCD counter. It turns out that it works the way I wanted it to. I had planned on using a regular counter fed to a decoder. Each sequence would decode one output. This is what this device does. Maybe we should call this a counter/decoder instead of a BCD counter.

That sounds like a simple switch, but I won't ask why you need to distinguish between the two supplies. You can activate and deactive a circuit using a semiconductor if the voltages allow. I would correct the voltage in order to use a simple semiconductor rather than use a relay.

I built an inverter oscillator with two inverters connected by a capacitor and a pullup on the first inverter. The output of the second inverter is fed to the input of the first inverter. The problem I'm having is that it will only oscillate when I touch the side of the capacitor, which shortens the charge/discharge because of the path to ground from my finger. I tried to add a high resistance to ground, but I can't seem to find the correct value. Can anybody tell me why I need my finger resistance to get the thing oscillating.

I will have to disagree with you about the thresholds. I don't think they are necessary in deciphering the operation of the circuit. But I will continue to read your posting to determine otherwise.

I think I can corroborate. Let's start with what we can easily agree upon. The output of the second inverter goes through the capacitor. This is the first observation. The capacitor will charge to 5 volts, bringing the high that went through the capacitor to 0 volts. Are these two observations correct?

When the output of the second inverter is high, it will charge the capacitor high, which sends the output low. You have the discharge of the capacitor which allows the ouput of the second inverter to go high.

It seems simple enough to me. The input can be grounded or grounded through a resistor. There is the case where you want the input bias currents to be the same, for a zero voltage output. This is the resistor which will allow you to affect the input bias current. I think the resistor is for opamps that don't have a null offset pot.

Sometimes I think that the values chosen for RF applications are just in the neighborhood of being correct. You could narrow your design work quite a bit by choosing component values that are within reason. All you have to do to get it close is to know the impedance and of course properly DC bias it. As far as filtering, why bother. You know the frequency and two bandpass filters is as good as a bandstop.

A high gate capacitance can be a misleading topic. I don't think that it actually charges like a capacitor, but exhibits the frequency response. I may be wrong. What is true is that the gate capacitance will keep the VGS from developing at a higher frequency. By the way, isn't the transfer characteristic of a MOSFET just nice. I like the way it keeps some predictability to the signal.

I think I understand the operation of the circuit with one inverter. But why use two inverters. The last circuit only uses one inverter. The operation with two inverters doesn't make sense because the output of the second inverter would tend to stay high. Could somebody explain why one or two inverters can be used. I have seen two inverters used before, but the operation is a little different.

This is simply the resistors and transistors at work. The transistors are most likely driven by the signal as opposed to being left open. In other words you won't be left with a signal trying to drive the collector of a transistor. The change in voltage therefore produces a change in current which gives you the impedance. The output impedance could even be dependent on the load.

I think that is a shift register with a parallel ouput going to the LED's. The input to the register is pulled high. Each pulse will send a high that migrates through the shift register. At the end, all the LED's are lit and the device needs to be reset. This is a way of counting clock pulses where the clock shifts a one instead of shifting the sequence.

You might consider that this is a signal that is biased around 12 volts. Use a transistor which will give you the voltage you need under higher current. I think you need a high current transistor because it will allow you to get a reasonable amount of gain. In other words, let's say the load resistance is 12 ohms with a common emitter configuration. The high current transistor will have a low value emitter resistor because of the high current. This gives you a reasonable gain. A low current transistor would have a large emitter resistor and the gain will be low. Also, with a higher current transistor you can have a greater change in VCE and still stay within the power rating.

Test it the way you would with any other component; with a bridge circuit. You have to know the frequency it was designed for. You might find that the device is so accurate that you would be looking at very small variation. But this is what a bridge is designed for. You might start with the fact that the transformer can be broken down into an inductance value for each winding. Also, this is going to have to be subjective, since any real equation dealing with a transformer is not going to be accurate enough. What you might look for in the final result is the amount of similarity between the input and output signal.

A modulus 10 counter uses flipflops and some associated logic to give you 10 stabile states. Each clock transistion will give you a change in output. There is a guy in the microcontroller section that is going to use a johnson counter to control the triggering of a one shot. With a latch on the last pin, he can trigger the one shot periodically to give to give him a pulse width modulated output. This is going to be summed with a clock to produce a special clock. Recall that a 555 will give you astable or one shot capability. The one shot has a variable duty cycle, but it will only produce it one time with a triggered input. You have to keep triggering the one shot if you want a continuous waveform.