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Posts posted by audioguru

  1. In order to avoid continuing confusion about knightrider/chaser, for those who have not seen the American TV show, is it possible to include the animated GIF:

    into the project article, above?

    Is it also possible to remove the phrase,"Knightrider" from the Projects Index for the chaser circuit:

  2. Hi Mike,
    Welcome to this community.
    So far, you have gathered most of the information that you need:
    1) You have a power supply with a full description and its ratings.
    2) You described a purpose: a certain time delay before energizing the load.
    3) You know that a capacitor takes time to charge.
    But the capacitor needs to be part of a timer circuit, which activates a transistor switch. Please look in our projects section for a timer and I'll look too.

  3. Mastrila,
    Your variac, isolation transformer and rectifier diodes probably are not made to have a low voltage drop during the brief but HUGE current surge when the filter capacitor, C1, is re-charged. Therefore the voltage across C1 (at the lowest point of its ripple) probably is too low. Let's call this voltage, V+. V+ must be more than the project's output voltage by at least the total amount of voltage-drops that follows:
    1) The base-emitter voltage of Q4. (0.9V)
    2) The base-emitter voltage of Q2. (0.8V)
    3) The voltage-drop across R15. (1.0V)
    4) The output-high voltage of U2. (1.2V)
    With the project providing 3A, voltage drops above add up to 3.9V.
    Therefore if you want 30V output, V+ must be at least 33.9V.
    A 24VAC transformer, low-current rectifiers and a low-value for C1, supplying 3A, won't give you this. What do you measure across C1?

  4. MP,
    I did not see any function for diodes D3 and D4 since they are always reversed-biased. Pull-down resistors having only about 20mA were suggested to replace them for reasons as follows:

    1) To turn-off the output transistors, fast. Without pull-down resistors, the bases of the output transistors just float down.
    2) To bypass the leakage current of the output transistors and ensure that they actually turn off. The leakage current increases with temperature (thermal runaway).
    3) To discharge the capacitance (the miller effect multiplies it) at the bases of the output transistors quickly. Without the pull-down resistors then the output transistors become integraters with their own substantial capacitance.

    Darlingtons are emitter-followers with their collectors tied together. The miniscule 20mA of current into the pull-down resistors won't affect the amps of current that are fed to the bases of the output transistors.

    Darlingtons are not used for high-speed-switching nor high quality audio because they are too darn slow.

  5. Hotwaterwizard,
    I am sorry, but I made errors when I suggested modifications to your project:
    1) DO NOT add diodes across the transformer's primary.
    2) DO NOT add 8 ohm resistors in series with the bases of the output transistors.
    My reasons are explained in the other inverter post here:

  6. Mixos,
    Steven's circuit is not a knightrider, it is a chaser that misses a few steps. We have discussed knightriders and chasers recently in this site at: Electronic Projects Design/Ideas, "LED Flasher (very interesting)".
    In Steven's circuit, R5, the LED current limiting resistor, is not needed because the 4017 outputs have built-in current limiting at about 9mA with an 8.3V supply.
    In Steven's circuit, the transistors in the multivibrator have their base-emitter junction driven beyond their 7V maximum rating which causes damage to them.
    A good knightrider project is in our site's project section here:

  7. Hotwaterwizard,
    Your inverter design is coming along, but I have some suggestions:
    1) D3 and D4 don't appear to do anything. I suggest replacing them with 100 ohm, 1/4W resistors so that the bases of the output transistors have a pull-down instead of a float-down.
    2) Add diodes in series with the multivibrator transistors' bases with the cathode towards the base, and add 1K ohm resistors from the bases to ground. Most silicon transistors have a maximum reverse-bias voltage of only about 7V for the base-emitter junction due to avalanching. But the avalanching causes damage to the transistor's hFE. In your circuit, the bases are driven to almost -12V.
    3) Add 2A diodes across the transformer windings with the cathodes to the switched +12V center-tap and the anode to the output transistors' collector. These diodes protect the output transistors from a flyback voltage spike that would occur if the load was removed with the circuit operating.
    4) Add 8 ohm, 10W resistors in series between the emitter of the driver transistors and the bases of the ouput transistors. These resistors will reduce the voltage across the driver transistors when conducting, and reduce their unlimited current (less E times less I = less heat).

  8. In my plane-days, I used a #6 dry cell. Have you seen one? 1.5V and it weighed about 4 pounds. When the engine wouldn't start then it was time for a new battery. I might try an electric plane this summer, but instead of Radio Control, I'll use a timer to control PWM motor speed control, and let it go free-flight:
    1) Full power for takeoff, for about 15 seconds.
    2) 2/3 power for climb, for about 30 seconds, then it crashes.
    3) If it's still flying, then 1/2 power for cruise, for 1 minute.
    4) Then cut the motor and let it land in a tree.
    But if it gets into a strong thermal then i'll never see it again (been there, done that).
    Maybe during #1, it will loop and hit my back.
    Good luck keeping your glow plugs hot.

  9. Siddharth,
    I am glad that you figured how to calculate the resistors.
    But a 7805 won't work with only 3V because it needs at least 6.5V. I suggest powering it with 6 cells which give 9V when new, and 6.6V when exhausted. Use a 9V torch bulb. This circuit is going to eat batteries quickly.

  10. Ante,
    Your experiment inspired me to try it too, and then I continued experimenting:
    1) New 9V alcaline battery feeding a 7805 with no load. Current drain = 6mA. Its been going a day-and-a-half now, and its battery measures 7.8V. The regulator feels cold and its output measures 5.05V.
    2) I made another one but put a 51 ohm load on it. That's 98mA into the load and the regulator was using only 6mA. The regulator was dissipating 406mW and felt slightly warm. At 4 hours, the battery measured 6.6V, and the regulator output still measured 5.05V.
    3) I built a zener circuit with a 1N4733, 5.1V/1W zener, and calculated a series resistor to feed it and feed the load. With a 51 ohm 100mA load, I figured that with a 6.6V battery then a 15 ohm resistor would feed the load 5.1V. With a new 9V alcaline battery, the 15 ohm feed resistor got VERY hot since it was dissipating 1W. The Zener also got hot and was dissipating 816mW. At first, the total drain on the battery was a whopping 260mA. The battery voltage dropped to 6.6V in about 2 hours, when the zener felt cold, and the load voltage measured 4.4V. Lousy regulation.
    4) With a new 9V alcaline battery, I tried the zener circuit without a load. Again the drain on the battery measured 260mA, but the zener was smoking a bit since it was dissipating more than 1.3W. The battery voltage measured 6.6V in about 3 hours, when the zener felt warm, and the output voltage measured 5.0V.
    So with a new battery and 98-100mA load, the regulator dissipated 406mW, and the zener/feed resistor dissipated 1.8W.
    And with a new battery and no load, the regulator dissipated 24mW and the zener/feed resistor dissipated 2.3W.
    The zener regulation can be improved by feeding it even more current.
    The zener/feed resistor costs more than a 7805 and eats power, so why use a zener? Maybe it's better for low current applications.

  11. Testing cheap opamps?
    I designed an audio equalizer that adjusted the levels at 7 frequencies, to flatten the response of a new loudspeaker. I also designed a complicated tester for the circuit which automatically stepped through many specs such as current drain, output noise, harmonic distortion (0.05% max limit), and the level at all the 7 frequencies, using green/red LEDs to show pass/fail and the test's progress. Using TL074 quad opamps, 15,000 units were built locally and I was responsible to fix the failures:
    1) One unit had the IC installed backwards.
    2) Two units had shorted power supply filter capacitors.
    15,000 units with quad opamps = 60,000 opamps were tested and not a single one failed! Within 1 year of the sale, and they sold quickly, not a single unit was returned. These tests were much more stringent than the simple test that is posted.
    Why bother to test opamps? Opamps that I have seen work VERY well.

  12. Siddharth,
    Thanks for telling to us the application and all of your requirements. Your original circuit will not do what is required if you connect an unknown quantity of incandescent lights between the output of the regulator and ground:
    1) The regulator will shut-down if you try to power too many lights because it has a maximum output current of only 1.5A.
    2) With daylight, the lights will be bright. For the lights to be bright at night, swap the LDR and R1.
    3) With daylight, the lights will dim a bit but will not shut off, because the output voltage of the regulator cannot be less than its label.
    4) Without having hysteresis, the lights may flicker during shadows or clouds.
    Replacing the regulator with a power transistor will avoid these problems.

  13. Kevin,
    A compensated opamp works VERY well when used within its simple rules and specs. Don't try to build your own opamp from scrach, just use the ones that are so popular.
    The only problem that I have ever had is that some opamps oscillate at a very high frequency when driving a shielded audio cable and/or the shielded cable to my 'scope (capacitance). But that problem is simply solved by adding a 100 ohm resistor in series with the opamp's output.

  14. Kevin,
    In the 1st and 2nd circuits that use a 9V battery, the IR signal is taken across the 1K resistor. The +5V is usually bypassed with a capacitor and therefore is at signal ground.
    It doesn't matter what the voltage is for the - terminal of the 9V battery, since it only affects the 1pF capacitor, which surely can withstand a few volts. If the IR diode is forward-biased as shown and has about 1.5V across it, then there will be about 7.5V across the 1K resistor. So the - terminal of the battery will be negative 7.5V from +5V which is -2.5V with respect to ground. If the IR diode is reverse-biased then the 1K resistor will have 9V across it when dark, but signal with IR radiation. So when dark, the - terminal of the battery will be -4V with respect to ground.

  15. Good news, Guys,
    I found a high-voltage, readily available and inexpensive replacement opamp for U2 and U3, which are operating in this project with a supply voltage beyond the maximum specified ratings of the original TL081 opamp. It is ONSemi (formerly Motorola) part number MC34071AP. Its maximum supply voltage rating is 44V (this project gives 38V plus).

    Do you recall that a couple of years ago, Chinese electrolytic capacitors were very unreliable? Maybe Juan got some. I know some techs who made a fortune relacing defective computer capacitors.
    ONSemi is not recommending the TL081 for new designs (will soon be obsolete).
    Please see ONSemi's spec sheet for Juan's MC34063, where they say that maximum ratings are not to be used continuously nor simultaneously. Their wording for the TL081 is the same as TI's (just a copy) which is very confusing.

    A-HA, I saw your other post about the 24V to 12V switching power supply that is heating too much when powered by this project. My suspicions are correct that C7 is too small physically (low ripple-current rating) and value (not enough storage to provide high-current pulses that are drawn by the switcher). Hopefully, a big C7 will solve both problems.

  16. Ante,
    Yes, let's get that transistor switching fast. The schmitt trigger, because of its positive feedback, will do that.

    That's a nice circuit that you have found but it can be improved:
    1) The opamp is not a schmitt trigger because positive feedback is missing. Add a 100K ohm resistor between pins 3 and 6. That should give about 2V of hysteresis.
    2) The expensive and failure-prone relay might be able to be eliminated if you tell us what your maximum load current will be.
    3) The expensive "replacement series" ECG128 transistor can be replaced by a much cheaper one if we know what your maximum load current will be.
    Where are those bubbles? What are you going to use this light switch for? Maybe the light heats a soap-solution that makes the bubbles but only at night?

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