audioguru

Mikail, Welcome to this web-site. With an unknown transformer you must take precautions to ensure your safety: 1) Make certain that it is a mains-power transformer, not an audio one. 2) Make certain that you know which wires are the primary that connect to the mains. 3) Make certain that the primary voltage matches the supply of your country (115V or 230V). 4) Connect the primary to the mains and measure its secondary voltage. Subtract about 5% to allow for its voltage to drop when loaded. This will approximate its rated secondary voltage. Then you should visit your electronic parts supplier and find their transformer that is the same voltage and size. If its current is marked then you have a match. Sometimes transformers are rated in voltage and power (VA). You can calculate their current by dividing the VA rating by the secondary voltage rating.

Here is a programmable digital sine-wave chip developed by Fairchild awhile ago for telcos. They say that it uses "no external components". I haven't checked its availability. They talk about the phase-accumulators having interference causing distortion readings (0.1%) to be higher than it actually is, so I suspect that is has some phase-jitter. The data sheet is here: http://www.fairchildsemi.com/ds/ML/ML2035C.pdf

Groovy, Here is a project that uses a mini-micro to digitally bandpass-filter its internally-divided clock and over-sample it. Although it was limited to a max freq of 20kHz it may give you ideas for use with faster chips. The project is here: http://www.web-ee.com/Schematics/Digital_Sine/DigitalSine.htm

I disagree with you, Serge: In all of your circuits, of course the IR diode is foward-biased, with its anode shown connected to +9V, +4.5V or +5V, and the cathode connected through a resistor to negative or ground. The diode is shown conducting forward current, all the time. Please check your working circuits and you will find that the IR diode is connected in reverse. For example, in your 3rd circuit, the anode of the IR diode is shown connected to +5V. Since its cathode connects to ground through a resistor, then you must measure about +4.3V at the inputs of the logic gates without IR radiation, since the diode will be forward-biased. But when you measure it, I think that you will actually measure about 0V, since the diode is actually reversed and will not be conducting. I do agree that the 9V battery and resistor in the 1st and 2nd circuits and the +5V and resistor in the 3rd circuit will convert DC to AC during modulated IR radiation, but only if the IR diode is reverse-biased. Again, I ask what is the function of the rectifier diode and why is it forward biased? Are you using the voltage divider that includes the rectifier simply to bias-up the input of the schmitt trigger and therefore make it more sensitive to low-level signals? If so, then the rectifier is not needed.

Hi Guys, Now that the American high-definition TV and progressive-scan DVD players are available here (Canada), I need to clear up some misunderstandings: 1) The TVs are still too expensive. Aren't many being purchased? Many DVDs are being rented/sold and I don't think that most are viewed on computers. Are the manufacturers just getting a huge profit? 2) Very few shows are broadcast in high-definition. Why not? 3) The salesmen in the TV/stereo stores tell me that DVD players with progressive-scan will work and show clearer pictures on my older ordinary TV that doesn't have progressive-scan. Are they lieing to me?

Ante, A zener begins losing control when it is conducting its "knee" current. A zener operating at its knee makes a lousy regulator. Did you know that the reverse-biased base-emitter junction of an NPN silicon transistor makes a pretty good zener at about 7V. But they say that the transistor later should not be used normally, because the avalanching causes damage to its hFE spec. I think that is why there are input-clamp diodes in 5532-5534 opamps, and that protection diodes are recommended in 2-transistor multivibrators.

Hotwaterwizard, Mosfets beat bipolars because they are much more efficient when switching high currents. A mosfet has an extremely low on-resistance so that the load gets the most power, and they switch very quickly so that they spend very little amount of time dissipating power. There is a very high power (2kW) inverter project (I think for a telco) where the author recommends testing all (there are many) of the mosfets so that the project won't explode if it had a bad one: 1) A 12VDC source and a lightbulb load. Without any connection to the mosfet's gate, the lamp should not light. 2) With one hand, hold the voltage source. With the other hand, touch the gate. The lamp should light. Never mind using a resistor! Many power mosfets have static protection built-in. Ante, You can make a solid-state amp produce "tube sound" by adding the same amount of 2nd harmonic distortion and soft-limiting.

Kevin, A compensated (with built-in roll-off) opamp that is using negative-feedback IS stable because the roll-off makes certain that at high frequencies, the gain is reduced so that the circuit will not ring nor oscillate. The compensation ensures a good phase-margin. Stability of an opamp is referring to its lack of ringing or oscillation at a high frequency. The only times that a compensated opamp with negative feedback cannot drive its output to a certain voltage level is: 1) When the load is drawing more current than the opamp's rating then the opamp will current-limit the output swing. 2) When the frequency is high enough (an old 741 at 8kHz or a more modern TL071 at 100kHz) so that the opamp is slew-rate-limited where its internal circuitry cannot charge and discharge its compensation capacitor quickly enough. When the opamp is slew-rate-limited then a sine-wave input results in a reduced-level triangle-wave output. Spec sheets show the maximum frequency where slew-rate does not limit the output swing in a graph called POWER BANDWIDTH.

A comparison between zener vs series voltage regulator: Example: 20V power source, 10V regulated voltage, 100 ma load current: Zener S.V. regulator Resistor voltage = 10V - Resistor current = 200mA - Resistor power = 2W - Zener voltage = 10V - Zener current = 100mA - Zener power = 1W - Total of resistor plus zener power = 3W - S.V. regulator voltage = 10V S.V. regulator current = 106mA S.V. regulator power = 1.06W Therefore in this case, the zener plus resistor power loss is almost TRIPLE that of the S.V. regulator. The only time that a zener has a low power loss is when it doesn't conduct: Resistor voltage = 10V resistor current = 100mA resistor power = 1W Therefore in this case, the resistor power loss is almost the same as the S.V. regulator, except there is no regulation.

I agree with you Ante, A Schmitt trigger is better than a simple transistor since it will stay after switching, while a simple transistor may cause the lamps to flicker if a shadow falls on the LDR or a if cloud goes by. We still don't know how much current is required. Recall that an incandescent light bulb draws 10-12 times more current when cold, so the transistor must handle the power-up surge and be driven hard. The data sheet circuit certainly isn't suiteable nor practical for switching.

Juan, If you are referring to C7 in this project, check its polarity. Polarized capacitors make spectacular explosions if connected backwards.

Hotwaterwizard, Do you remember when transistor AM radios were invented and most had 6 transistors? Well one brand had engraved on its case, "14 TRANSISTORS". I opened one up and counted, 7 for the radio (the 7th was used as a detector diode) and 7 connected together in a neat circle doing absolutely nothing! Today, many portable boom-box radio and tape or CD units claim 3-way speakers. A woofer, a tweeter and a label instead of a super-tweeter. Sometimes they use a peizo whistle for a tweeter (sounds real bad). The new class D amp chips put out a lot of power from a surface-mount chip, without a heatsink. They don't sound very good yet so I will use one for DC motor speed control since they use PWM and the high-current Mosfets are built-in.

MP, A zener regulator is always conducting current while regulating, regardless of the amount of load current. The zener conducts its wasted current to ground, not to the load. The resistor feeding the zener must supply both the zener's and the load's currents. A series voltage regulator IC conducts only load current, without wasting any current to ground (except for its tiny operating current). So the IC will heat much less than the zener plus the resistor, above, and therefore the IC will have much less power loss.

Siddharth, 1) What are you going to use this voltage-changing circuit for? 2) How many volts will feed the regulator? 3) How much output current when its output voltage is high? 4) What voltage output do you need when it is dark? 5) What voltage output when it is in bright light? The transistor doesn't have to be a switching type since an LDR does not change resistance super-fast. A 2N2222 switching or 2N3904 audio will be fine.

Its a good idea to use a car remote. Simply use a small piece of wire to replace the clutch or brake switches so that it will think that is safe to "start" your computer. I'm sorry if this is too complicated.

Thanks, MP, But we are talking about using LED-driven, solid-state triacs here, not mechanical relays.

I agree with you, Kevin: 1) Why is the IR photodiode FORWARD biased? Most photodiodes are REVERSE biased, so that there is a very small leakage current when dark, but IR radiation causes much more leakage current. An IR photodiode project is here: http://www.comdev.ca/esq/Lightwavecommunicator.pdf The photodiode leakage current modulation causes a voltage signal across it. The amplifier following the photodiode is used only to make the circuit extremely sensitive for long-range pickup. Our project doesn't need long-range and therefore doesn't need that amplifier. 2) Why is the function of the rectifier diode and why is it forward biased? Since it is in the middle of a voltage divider then it will not rectify.

Kevin, A zener diode regulates voltage well because it has a LOW dynamic impedance. Therefore if you vary the current through it, the voltage across it won't change much. See the spec sheet for a 1N5231 (I couldn't find one for a 1N5338) 5.1V zener diode here: http://www.microsemi.com/datasheets/SA5-44.PDF Notice that with a current of 20mA, its dynamic impedance is only 17 ohms. A resistor with 5.1V across it and 20mA through it would be 255 ohms. A good description of zener diodes is in our web-site, see Articles-General-Zener. Patabus, Why throw power away in a zener diode? We don't know your application but maybe a 7805 voltage regulator would be a better choice.

Ante, Good idea! And its not complicated .

Hi and welcome, Your attachment doesn't show up, I tried once and it didn't work either. It sounds like your project needs a circuit-breaker (disconnects when current is exceeded) with its own reset button, for each outlet. It also sounds like you need these outlets to be powered sequentially when the power is first applied, in order to avoid a huge power-up current surge. (Do your lights dim when your computer is turned on? What would happen if you turned on 8 computers at the same time? Poof.). So you need an electronic power switch and a current-measuring circuit, for each outlet, controlled by a micro-p. A Triac is a good power switch device. Check the Articles section of this site for details. The gate signal for the triac can come from the micro-p. The current-measuring is done by simply wiring a low-value resistor in series with each outlet (ohms law, 10 amps through a 0.1 ohm resistor gives 1 volt). The voltage across each resistor feeds the micro-p. On each measurement when its power is first switched on or the outlet is reset, allow for a delay time in the micro-p. routine for the power-up surge to pass. When power is first applied to the whole system then the micro-p. sequentially turns on each triac. If the measured current exceeds a certain amount then the micro-p. turns off the outlet's triac. Each reset button feeds the micro-p. Note that the triac and current-measuring resistor are directly connected to the mains, so for safety the micro-p. must be completely insulated in a plastic box (don't hook-up your computer to the triacs nor the resistors) unless you use a complicated isolation method for the signals. I hope this helps you to get started.

Congratulations for your win in the contest! And thanks for listening.

Wath out guys, I'm gaining quickly.

Your 4017 pin numbers do not seem to be in the proper order for the lights to be sequential. For example, 4017 pin3 should feed the 1st light, but you have pin1, pin 4 should feed the 3rd light, but you have pin 2 etc. Don't you have them mixed-up? And the diodes? You don't talk about the switch but I think that it is used to shorten the sequence, but position 4 seems to light only one light continuously instead of 5 lights. All other switch positions seem OK. Please check your switch position 4. What is the regulator used for? With your well-filtered 12V unregulated, the parts don't need it.

On your schematic, I can't see where the diodes connect to. I built a similar Knightrider circuit that uses only LEDs and it clearly shows the diodes: http://www.electronics-lab.com/projects/motor_light/035/index.html Please compare its diodes to yours. Thanks

On the schematic, R3 should connect to the LED anodes, but not to T1 as shown.