audioguru

Ante, I was agreeing with you when I mentioned computer wiring, since you mentioned wiring then I knew that you were also considering PCB trace sizes: In this project do you believe that the capacitors are shown with the correct polarity? Do you have any comments about its high capacitor currents? What are your comments about many amps of current going into avalanching base-emitter junctions?

MP and ANTE, Maybe they use more transistors than is necessary in order to reduce the gamble of a high-gain transistor "hogging" the current. Also, extra transistors are probably cheaper than having many emitter resistors. You didn't comment about my explanation for emitter resistors that I posted previously:

Ldanielrosa, Input capacitors will not provide isolation from the mains, since they pass the AC. If you use their capacitive-reactance like a series resistor, except without heating, then they will cause the rectified voltage to fluctuate with load changes (very high with no load and quite low with a load).

The exponential charge/discharge of a 555's timing capacitor IS like a pendulum! But the effect is reduced by the 555's lack of full charge and discharge (1/3 of +V to 2/3 of +V). So maybe it would be best to low-pass filter a square-wave in order to obtain an alternating voltage that changes very quickly at the beginning and very slow at the end. What do you think about that? How many LEDs?

MP and Ante, You made very good points about the transformer. Although the author recommended an over-rated 15A transformer (15A X 24V = 360W), he did mention re-winding a powerful microwave transformer. Maybe it is the transformer that produces the power, not the transistor circuit.

Ante, I agree that a power circuit will have a variation in its performance if incorrect wiring techniques are used, such as moving a computer's power-supply into the next room without considering the the voltage-drop of the long wires. A power circuit will also have varying results if the circuit-designer does not consider the sample-to-sample huge variation of a transistor's current-gain. The circuit should operate well with ANY guaranteed sample. But let us not discuss "yield" nor luck. Circuit performance and safety also depend on assembly details: The instructions for this project do not mention heat-sinks.

Kasamiko. Yes, the 2N3055s will be fine for this project since they must conduct only 10A. Please let us know how well it works.

Hotwaterwizard, If the capacitor is discharged, and you apply power at the time that the AC voltage is at its peak of 170V, the LED must pass 1.7A for a moment. If you turn-off the power when the AC voltage is at its peak of one polarity, then while the capacitor is charged, apply the power when the AC voltage is at the other polarity, 3.4A! Can an LED withstand such abuse? Doesn't the LED flicker? Kevin, The 12V and 24V circuits also indicate on many other voltages. The resistor limits the current, not voltage.

Stuee, 1) Ammeter: Meter manufacturers should have a current-shunt that you connect in series between the amplifier and battery. To allow for a voltage-drop of 0.1V at 100A, the shunt must be 0.001 ohms and be able to dissipate 10W. Then you connect a digital voltmeter having a full-scale sensitivity of 0.1V across the shunt. In order to keep the readings stable, use a peak-detector circuit ahead of the voltmeter. Our LED digital voltmeter project is here: http://www.electronics-lab.com/projects/test/014/ Our voltmeter can be modified to give 0.1V full-scale. 2) Digital volume control with display, channel selection and memory. This sounds like it will be a complicated circuit, maybe another forum member has an idea for a manufactured system.

Nettron, Have you considered distorting the ramp so the LEDs slow down at each end like a pendulum?

Nettron, Don't use the ramping voltage of a 555's timing capacitor to drive an LM3914 because it is not linear with time. It would start counting the LEDs quickly, then slow down when nearing each end. The integrator output of a dual-opamp triangle-wave generator is perfectly linear.

Uman, A very low-power CD4060 oscillator/counter chip can give a short pulse after 60 minutes if its oscillator is set to 273Hz. This pulse can trip a latch that controls a transistor that feeds power to your circuit. A simple 2-diode OR gate can be connected to the CD4060's reset pin, one diode for power-up reset and the other diode for reset after an event. The data-sheet for the CD4060 is here: http://www.fairchildsemi.com/ds/CD/CD4060BC.pdf The number 2240 sticks in my mind, it is a chip that combines a 555 with a counter/timer but I couldn't find it. Maybe you can find it.

Kasamiko, Your 2N6124 and MA21 transistors should be OK. I couldn't find a data-sheet for a DM74HC74, but all "HC" chips that I have seen have a maximum operating voltage of only 6V. "LS" and "HC" chips are made for extremely high speed so why use them here? An ordinary old 74C74 or 4013 will work fine and run with up to 15V. But it is best to use the 4047. Don't forget to change R9 and R10. Are you going to replace the MJ15015s with 2N3055s?

Nettron, We have a 6-LED knightrider circuit in our projects section. The 4017 counts forward lighting LEDs 0-5, and diode gates are used to light those LEDs in reverse order while the 4017 continues its count of 6-9. From a 9V battery, the 4017's outputs are current-limited to about 11mA, so a current-limiting resistor is not needed. The project is here: http://www.electronics-lab.com/projects/motor_light/035/index.html An LM3914 dot/bar-graph chip lights 10 LEDs sequentially in both directions when fed with a dual-opamp triangle-wave generator. The LM3914 can be cascaded to allow umpteen-dozen LEDs, but the set-up adjustment will be tricky after maybe 50 LEDs. A D-A converter would feed cascaded LM3914s beautifully.

Hotwaterwizard, Your very nice circuit actually is RTL, not TTL. I hope that Telecombug doesn't lose marks for that oversight. Kevin, Do you see how this OR gate works? It is simply that one transistor OR the other transistor OR both transistors turn-on the LED. There is no phase-splitter.

OOOOps Kevin, The opamp roll-off is 6dB per octave (each doubling of frequency) or 20dB per decade (ten times the frequency).

Kevin, The input capacitor, if used, to an opamp circuit is external to the feedback and therefore has no effect on the high frequency roll-off. The input capacitor, souce impedance and opamp's bias resistor determine the low frequency phase-shift and roll-off. Where did you get 40dB from? Most internally-compensated opamps use a single capacitor for their 20dB/octave high frequency roll-off. The flat frequency-response part of the graph gets extended due to your gain-determining negative feedback. This feedback exchanges the enormous open-loop gain, for bandwidth.

gEcky, Did your switch melt? Perhaps it was poor quality. The transformer may have a short from primary to secondary which would cause your problem, and is VERY DANGEROUS. Check that transformer and use a quality switch.

Nettron, Nice circuit. But 5 chips (the 5th is a 5V regulator) just to flash leds? The original knightrider car probably used light-bulbs. I would guess 8 to 10 of them.

Kasamiko, Oh, oh, I see a problem: 1) When one transformer wire is pulled up to 11V by its conducting output transistor, then the other transformer wire is forced to -11V by center-tapped transformer action. 2) The output transistor that has its emitter forced negative, should be turned-off, but current will flow into its base from R9 or R10, which will cause it to turn-on. 3) This will waste power and create unnecessary heat. If you increase the value of R9 and R10 to 1.5K, then the above problem will not occur, with no other consequences. It looks like the output transistors will protect all devices from spikes, since a positive spike at one transformer wire will create a negative spike at the other transformer wire, which will turn-on that transistor (as above), arresting the spike. But the output transistor may not turn-on fast enough. Try it and see.

MP, In this series circuit that will have extra wire resistance, where is your "more" current going to go? See my example #1 to see that the current is LESS when wire resistance is added to the loop. Why is a more powerful PSU needed? With more wire resistance, the current and power will be LESS. If you use a larger wire guage or copper pipe, then the current and power from the PSU will NOT be more than it was with original short wires. My example #2 shows that IF you could raise (maybe not possible) the voltage of the PSU, then the voltage and current at the CPU will be normal, but the PSU supplies a little more power.

Hi Kasamiko, That's a nice design. But the oscillator/flip-flop is rather complex and U2 is operating at a supply voltage that far exceeds its rating. The 4047 from the other project is better suited here. The MJ15015s are expensive and your circuit doesn't need its high ratings, except it can use a smaller heat-sink than its 2N3055 cousin. 2N3055s should work fine here since it has the same gain and current ratings. I couldn't find data sheets for 2N6124 nor MA21 so I don't know if they are OK. I haven't fully analysed spike protection, but the low values for R12 and R13 is a clever way to protct the output transistors against base-emitter avalanching.

EDY, Your problem is not the cables in your home, it is the cables or connections that FEED your home. The light-blink may be the 1st symptom of many more problems to come if those cables or connections are bad. The undergound wiring for my neighbourhood had to be replaced because it became unreliable. If you want to procede with adding a capacitor to your motor, you will be working on the mains, so be careful. A good link about power factor correction is here: http://www.lmphotonics.com/pwrfact.htm They say that tables cannot be used since motors vary very much. So contact your pump and motor manufacturers and ask for their recommended capacitor for power factor correction. The link above also says to put the capacitor on its own pole of the relay, since if it is still connected to the motor when the power is cut-off then it will resonate with the motor's inductance and create huge currents and damage. Good luck and please let us know about your results.

I don't see anything. I'll wait for it to be converted. Kamasiko, Have you measured its output at full load? At full load, does anything get hot?

MP, I am sorry to continue to disagree with you. The current will NOT increase due to the added resistance of the wires, if the supply voltage remains the same. The resistance of the wires is in series with the load, not in parallel. Therefore the load will receive less voltage due to the voltage-drop across the wire resistance. And with the load receiving less voltage, the current will also be less. Example #1 (5.0V supply, 0.5 ohm load): Original With 0.1 ohm wire resistance Current: 10A 8.3A Voltage at the load: 5.0V 4.2V Power from supply: 50W 41.7W Power at load: 50W 34.9W Example #2 (6.0V supply, 0.5 ohm load): With 0.1 ohm wire resistance Current: 10A Voltage at load: 5.0V Power from supply: 60W Power at load: 50W