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# Cabwood

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1. ## Transistor combinations questions

Not much actually - when the switch is pressed, the pair of inverters will change state and settled into a condition in which the switch is no longer fighting the collector of the first transistor. If the first transistor is switched fully on, though, it is effectively a short circuit from its collector to ground, and pressing the top switch (in the absence of R) will connect the +ve supply to the same collector. This will cause oodles of current to flow through the transistor, until the state change induced by pressing the switch propogates around to switch this transistor off again. So it's a momentary current surge. The current pulse is very short, but large, undesired and possibly dangerous to the transistor none-the-less. R simply makes sure that the transistor's collector is never subjected to the full clout of the +ve supply.
2. ## Transistor combinations questions

Very true, well spotted. I am tempted to call this a "hard reset feature" of the circuit.
3. ## Beginner Question about Ohm's Law Thanks!

Zeppelin pointed out that the battery in my little circuit is drawn upside down. He's right - the positive side should be at the top, so that the upwards Vs arrow is in the right direction. Kirchoff must be turning in his grave. My apologies. To hide my shameful error, I've edited the post wth a tweaked diagram.
4. ## Beginner Question about Ohm's Law Thanks!

The picture is a simple battery connected across two resistors in series, as you describe. The first thing to understand is that the current is the same everywhere in the loop. There's nowhere for current to leave the circuit, so the current is the same at every point around it. I've labelled it I. The second thing to know is that the total resistance of the two resistors in series is the sum of their individual resitances. That is, R1 + R2. We can use Ohm's law to find out I. Since we know the battery's voltage, and we know the total resistance that this battery is pushing current through, we can work out the current I through those resistances: I = Vs / (R1 + R2) Now that we know the current I, we can work out what voltages will appear across the two resistances, again with Ohm's law: VR1 = I * R1 VR2 = I * R2 Play with some numbers: Let's say Vs (battery voltage) is 9V, R1 = 1kOhm and R2 = 2kOhm. I = 9V / (1000 + 2000) = 3mA VR1 = 3mA * 1000 = 3V VR2 = 3mA * 2000 = 6V The sum of the two voltages across the resistors VR1 + VR2 (that you can measure to prove this empirically) will always be 9V, regardless of the resitances used. In other words, the resistances can be chosen to divide the voltage across them into two separate voltages whose sum is 9V. The ratio of those two voltages is the ratio of the resistances. Thus, to divide your 9V by exactly 2 (to create 4.5V and 4.5V), choose two resistors of the same value. You can use as many resistances in series as you like, to divide a voltage into any number of smaller voltages. The maths above reduces to this: VR1 = Vs * R1 / (R1 + R2) VR2 = Vs * R2 / (R1 + R2) Note that you don't need to know the current at all, but understanding the principle of the current being the same through both resistances allows you to see how the resistances each develop a share of the total voltage across them. One last observation; if you are thinking you can now power a small 3V lamp by connecting it across the 1kOhm resistor in our example, think again. The lamp's low low resistance (just a few Ohms) in parallel with R1 will change the whole scenario.
5. ## Transistor combinations questions

Just a couple of small diagrams of how inverters can be used to make a bistable multivibrator. The resistors 'R' are present to avoid connecting a power rail directly to the output of a gate, which is bad. The switches are pressed momentarily to force an input to a specific level (high or low depending on which switch), and cause the multivibrator to flip into one of its two stable states.
6. ## Transistor combinations questions

Astable means NO stable condition. That is, an astable multivibrator will never settle into a fixed state - it will always switch from one state to another and back, all on its own, without any external input. Lamp flashers are astable, needing no external input to switch from off to on to off to on etc, all on their own. Bistable mean TWO stable states. A bistable multivibrator will settle into either of two states and stay there until unsettled by some external input - then they switch to the other state, until it is externally signalled to switch back again. They do nothing on their own. Static RAM is implemented using bistable multivibrators to store binary 1s and 0s, and keep them unchanged until told to change. Bistable multivibrators do not change state AT ALL unless externally signalled to do so. Monostable means ONE stable state. These multivibrators settle into some state and stay there until signalled to change. When so triggered, they switch to the other (unstable) state, stay there for a while and return automatically to their previous, stable state, where they remain until the next external trigger signal. Sometimes monostable is called "one shot". By 'state' we normally mean "on' or 'off'.
7. ## A simple transistor switch for an LED light.

That was indeed an awsome reply!
8. ## Beginner's Question 2

It's one of the most misunderstood things in electronics - voltage. The problem is that people think 9V is 9V and that's that, but really it means 9V higher than somewhere else. On a 9V battery, the '+' terminal is 9V higher than the '-' terminal. In other words, the battery is generating a difference of 9V between its two terminals. It might help if the negative terminal on a battery was labelled "some voltage or other" and the positive labelled "9V more than the other terminal". Now ask yourself this - what is the difference between the voltage on the - terminal of that battery, and the top of the Eiffel tower? It could be 10 million volts, for all you know. In that case, the positive terminal of the battery is ten million and nine volts higher than the top of the Eiffel tower. All relative. A floating ground is some point in a circuit that is not connected in any way to anything outside of the circuit, but acts as ground for the circuit itself. It could be thousands of volts different from the ground of your home, because it's completely isolated from it. If you then connected that floating ground to the earth/ground pin in a wall socket, then it wouldn't be floating any more. The idea is that you don't always want the 'ground' supply rail of your circuit/appliance to be held at the same voltage as the ground of your building. A nice example is when you have two bench supplies in your electronics workshop, and you want to combine them to form a +15V:0V:-15V dual rail supply. In that case you connect the two in series, and you most definitely do not want the negative rails of the two supplies to share the same ground. When you connect them in this way they need to have floating grounds (completely isolated from the building's ground) to avoid a short circuit. And anyway, your circuit probably doesn't really care what the rest of the world's voltage is, and since it's not connected in any way to the rest of the world, why should it? If you do connect equipment together (hifi to TV for example), the ground connections in your home provide a neat way of making sure all the equipment is held at the same voltage level, and you don't end up connecting something floating around 100V to something floating at -50V - sensitive and fragile signal inputs and outputs will be exposed to the difference of 150V! With these crap "double insulated" appliances which claim to not need a ground connection (and use only two core mains cable - live and neutral, but no ground), you'll often feel the buzz of a hundred volts AC or so as you connect them together, because their internal "floating" grounds are different. This is a great way to explode PCs - connect together two serial ports on different PCs which aren't grounded via the mains socket, and see what happens. No, seriously, don't do that.

10. ## free energy motor

Indulis - great strides have been made, but the greatest ones are always those that disprove previous assertions of law, or at least of commonly accepted limits. The copper is proably the same, but they twisted it. I mean, somebody had to think "hey, I wonder what would happen to latent capacitance and inuctance of these parallel conductors if I twisted them together", (I don't really know what happened, of course, or even if that is the solution) and in that blinding flash of creativity, previously accepted limits regarding noise rejection and bandwidth of two copper wires were consigned to the scrap heap. I rather hope that the same kind of creativity will soon blast to smithereens a few really really popular physical laws, like the one that says there's no such thing as a free lunch.
11. ## 6V 5A PSU problem.

The biggest source of ripple is from an inadequate reservoir capacitor between the bridge recifier and the regulator. Make that bigger, and you reduce the ripple at the regulator's input, and the regulator's ripple rejection isn't stretched so far. Assuming you are drawing 1A from the supply's output, A 1000uF capacitor will experience a ripple of roughly 8V, which must be rejected by the regulator - hard work by any account. Double the capacitance, and you halve the ripple. Under those same conditions, 10000uF will therefore reduce the ripple down to less than 1V, which will result in less ripple at the regulator's output. Keep in mind that the drop-out voltage of the regulator will determine the minimum input voltage that it can properly regulate. The LM317 for example needs an input which is at least 2.5V higher than it's output. So, to get it to reliably output 6V it must have an input voltage which never drops below 8.5V. If the input ripple (across the reservoir capacitor) is too large, that voltage may drop below 8.5V, and since the regulator may then fail to regulate, it will appear at the output as 100Hz (UK) or 120Hz (US) ripple. If this is not the problem, then how about passing the existing output through another regulator (two stage regulation, I suppose you would call it). Since the ripple at the input of the second regulator is so small, the output should be pretty well flat. The problem with this is again the drop-out voltage of the regulators. You'd need to ensure that each regulator's input is at least a couple of volts higher than its output, typically. It doesn't make sense to put large capacitors at the output of a linear regulator, for the reason you have observed, and because most circuits don't sink current in huge spikes. When circuits do do that it's better to position the capactior electrically close to the current sinking component itself.
12. ## free energy motor

At university in 1988 I was taught some facts and figures about networking and electronic communications. Some of it was very compelling matthematical proofs of what can and can't be sent down a wire link, such as telephone wire. At that time it was deemed impossible to carry anything over 100kHz more than a few tens of meters without irecuperable losses. Well, that was a load of crap, considering DSL signals in the tens of megahertz are routinely connecting people over a single copper twisted pair to exchanges hundreds of metres away. Maybe we employed those laws of physics in an innovative way, or maybe these "known" laws broke down under certain conditions. Or maybe they were just hokey laws from the outset. Whatever the way, I can't take "can't" seriously at all. Over-unity systems are currently debunked with as much enthusiasm as they are dreamed up, but some day I'm sure someone will dream one up that works. Then the laws will magically adapt themselves to cater for this anomaly, and everyone will declare nonchalantly "well, yes, if you do it like that, of course it will work", while at the same time wishing the idea had been their's, and secretly banging their collectively expert heads against the bedroom wall at night.
13. ## free energy motor

The concept is called "over-unity systems". Just Google for over-unity, and you'll find an overwhelming number of web sites that claim to know all about them. And probably don't. Maybe somebody succeeded with this, but nobody's ever managed to demonstrate it to the public at large. It's a myth, that might be true, but right now it's just a myth. And I'm willing to bet that even if it could be done (which would defy current laws of physics), with current technology the best you could do is light an LED with the power generated. Having said that, play with it. You might be the one to bring this wonderful dream to fruition for us all.
14. ## HELP NEEDED

I'm afraid you are going nowhere with this. Any design needs numbers. Get numbers. Get details. Without them your project is only a little less vague than "make a kind of thing that sometimes does some stuff, perhaps, and some other stuff." It is just not feasible to start off with "any range" or "as high as possible". You might end up with something perfectly good for measuring microwave frequencies in the millivolts, but completely useless for tuning a guitar. Or vice versa. If you want a frequency counter for all occasions you'll be needing to visit your local Fluke rep.
15. ## HELP NEEDED

What range of frequencies, approximately are you measuring? And what amplitude (voltage) range is the input likely to be?
16. ## HELP NEEDED

Count the cycles of a precise 1Mhz oscillator during one cycle of your input waveform. Are you sure you don't want to use a microcontroller?
17. ## HELP NEEDED

One way to ignore higher-than-fundamental harmonics is with hysteresis. In the picture you can see an input waveform with multiple peaks, and two levels A and B. When determining a rising edge in the input, we compare with level A - we assume the cycle starts when the waveform is determined to be greater than A. Then we switch to comparing with level B, and assume the mid-point of the cycle to occur when the input drops below B. The cycle is complete when the waveform rises again above A. This hysteresis can be achieved with positive feedback as shown using an opamp as a comparator. The comparator's output is always high or low, +Vs or -Vs (or nearly so, as many opamps' outputs don't swing all the way to the supply rails). A known fraction of that output is fed back to become a reference voltage that the comparator will compare the input signal to. For example, assume the amplifier is supplied with +12V and -12V. Also assume R1 and R2 equal, such that 50% of the output is fed back to the non-inverting input. When the output is high (+12V), the non-inverting input (+) sees 50% of it, or +6V. When the input signal on the inverting input (-) exceeds this +6V, the output will swing low. Now of course the non-inverting input sees half of this new ouput, or -6V. When the input signal drops below -6V, then the output swings high again. And so on. You end up with a rectangular wave of the frequency of the fundamental of your input signal, and from there you can easily measure its frequency. You can change R1 and R2 to change the levels A and B that determine the rising and falling edges, but ultimately this simple circuit will only work if the input waveform is known to slightly exceed A, and only slightly drop below B. The two obvious ways to achieve this condition are: (a) Build an AGC (automatic gain control) which amplifies or attenuates the signal to have the approximate amplitude required by the hysteresis levels, or b) use peak and trough detectors to determine the levels A and B, and feed the non-inverting input of the comparator with these derived values. You would normally want more than 50% feedback in a frequency measurement application, to avoid erroneous transistions due to input noise. Perhaps 90% would be appropriate, requiring R2 = 9 x R1
18. ## LED dimmer

Here's the last version of this that I'm going to do. At the top left is a nearly-trianglular waveform generator. The peak and trough voltages are carefully chosen. The bottom left part is the ramp generator. I decided to be careful here because the level limits are important, given the range of the triangle wave it will be compared to - hence the use of two opamps. The last opamp of the four available is the comparator, which switches TR1. R12 is there to compensate for the opamp output not swinging low enough. The ramp output swings through the whole range (3V - 9V) in about 5 seconds. Change R8 or C2 to speed it up or slow it down. It turns out that this was a waste fo time, because it doesn't do any better than the simple two-transistor dimmer I posted first. The perceived brightness of an LED seems to be proportional to the log of the power it dissipates, which means both circuits seem to change brightness quickly when dim, and slowly when bright. So, I played with a log amplifier to do this, which worked OK, but it's severe overkill for a simple dimmer circuit.
19. ## Paralell resistors

No. I said sorry because I had just realised I repeated exactly what you said.
20. ## LED dimmer

Ugly, and incomplete, but hey, it's a work in progress! This was the test circuit I messed around with. The biggest trouble with it is my opamp's limited output swing, making levels difficult to predict, and making it complicated to switch the transistor. Amongst other things.
21. ## LED dimmer

Well, I've made the PWM version, and guess what - the apparent rate of brightening and dimming is very similar to the circuit I posted before, which makes me wonder - does the eye perceive brightness in the same way the ear perceives volume? If so the relationship would be non-linear (perhaps logarithmic, like the ear) - to achieve an apparent unit increase in brightness, I need to double the current. Something like that. Instead of a linear ramp, I'll need a log curve, rising slowly at first and getting faster. A capacitor charging has this behaviour, which is perfect for dimming, but for lighting up the curve must be initially slow-changing. Ideas anyone?
22. ## LED dimmer

In the picture, the blue sawtooth waveform is generated by an oscillator. 1kHz should be fine. The red signal is a ramp generated when you press the switch. When the switch is released, the ramp would rise instead of fall. The green waveform is generated by comparing the blue and red - when the blue voltage is greater than the red voltage, we derive a high signal, and low at all other times. This resulting green square signal would be used to switch on and off the LEDs. As you can see, the lower the red ramp voltage, the longer the LEDs would be illuminated by the green pulse during each cycle. This is the principle of Pulse Width Modulation - PWM. In reality we would have the ramp rise and fall slowly, over hundrerds or thousands of sawtooth cycles, not just the 7 shown here.
23. ## LED dimmer

That transistor is a limiting factor, for sure, but what Theatronics and I proposed was a PWM design which would not only make the on/off change smoother visually, but would also vastly improve the power efficiency, so heat (brute force) wouldn't be such an issue. The trick is to have that transistor permanently either fully on, or fully off, so that its I x V is always near zero. So you make the LEDs flash (too quickly to see) alternately fully bright and dark, and change the ratio of light period/dark period to give the illusion of gradual dimming. With such a design, the only limit on the transistor used would be maximum collector current. For the BD135, that limit is 1.5A! The design is more complex, of course. Does this interest you enough to have me draw one up, test it and post it?
24. ## LED dimmer

PDMAX is the maximum power it can dissipate when cold. Since it's likely to heat up for a few seconds as the lights come on and off, You are more interested in the maximum it can dissipate without a heatsink, in air at 25 degrees C. This figure is quoted as "PC(Ta=25)". A good transistor for up to about 200mA would be the BD135, with a PC of 1.25W. With a small heatsink, you could increase this to about 350mA. The voltage across the transistor traverses a range of 0-12V, during which the current rises almost linearly. The power dissipated in the transistor is the product of the current through it and the voltage across it (P = I x V). If you plotted a graph of that power against voltage you would see that it peaks at about 6V. It is clear that when the transistor has all 12V across it, and no current is flowing, P = I x V = 0. When the transistor is fully on, and maximum current is flowing, it has almost 0V across it, so P = I x V = 0. The peak power must therefore be somewhere between these extremes.
25. ## Scopes

I used a Goldstar CRT scope for years, and then changed to an Tektronix LCD. The CRT is clearer, and there's something about analogue that's just plain sweet. However, now I'm used to the digital scope, I wouldn't go back for all the money in the world. It's just great. Of course, if you really did offer me all the money in the world, I would ditch my Tektronix in the blink of an eye.
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