Jump to content
Electronics-Lab.com Community


  • Posts

  • Joined

  • Last visited

  • Days Won


Posts posted by audioguru

  1. 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.

  2. 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.

  3. 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:
    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.

  4. 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

  5. To all,
    Shall we re-engineer this dangerous (expoding capacitors) project and make it work (the author's web-site forum has many complaints about low output)?
    Well, the author recommends "more powerful" transistors, but many have tried 600V, 30A transistors and other ways, attempting to make it work, but nobody has suceeded.

  6. MP,
    Thank-you for unlocking.

    Capacitors have a "ripple current rating" which is related to their internal impedance (resistance) and physical size (to dissipate power) similar to a power resistor.
    In this project, the capacitors are charged to nearly 23V with a very high current flowing through a forward-biased base-emitter junction on one side, and the +24V (center-tapped transformer action) from the transformer on the other side. The capacitors are shown backwards.
    The capacitors are partially discharged with a very high current through a saturated collector on one side, and an avalanching (-7V) base-emitter junction on the other side. When the capacitor voltage discharges to less than 7V then it can continue a slow dicharge through the resistor, for its timing period.
    Therefore the capacitors have very high currents flowing in them which causes "ordinary" electrolytic capacitors to explode.

    But those expensive tantalums also explode (shooting bits of molten metal- VERY DANGEROUS) and there are debates about whether it is because of the high currents, or due to their reversed polarity, in the forum about this project on the author's web-site.
  7. Hotwaterwizard,
    Sorry. No, you didn't say the same thing.
    Using bigger wires going to the pump will not reduce the voltage-drop along the wires going to his castle. The light will still blink when the pump turns on.
    But I see your point that the motor will draw more current when fed a reduced voltage from smaller wires. Won't the increase in current be cancelled by the wire resistance that will limit that current?

  8. MP,
    Most LEDs will not light with only 1.5V, and worse when the battery is exhausting.
    Voltages of LEDs:

    Common Ultra-bright
    Red:1.7V 2.2-3.5V
    Yellow: 1.9V 2.5-3.8V
    Green: 2.1V 3.1-5.1V
    Blue: - 3.1-5.1V
    White: - 3.1-5.1V

    That's why the little flashlights have an inverter inside, and why National Semi uses a votage-boosting circuit in their LM3909 flasher chip here:
    You probably meant to use a common LED with 2 batteries, a current-limiting resistor and probes.

    Do you know how high is a resistance that is infinite?
    Would 1,000,000 megohms fail?
    Please tell us your maximum resistance that will pass, and how much voltage is allowed for the test. There is a commercial meter called a Megohmer, for measuring the resistance of insulation, that uses a battery of about 90V (ouch).

  9. For all,
    In another web-site's projects forum, there was a request for an inverter to provide all the power for an electricity-heated home.
    Numbers such as 12,000 Amps at 12 Volts (144,000 Watts) were discussed! Enough batteries to fill a large garage!
    But they didn't think of charging them by dragging them along the ground behind a motorcycle.

  10. MP,
    With the 8 ohm resistors gone, there is no voltage divider. The pull-down resistors turn-off the output transistors and discharge capacitances, when the driver transistors turn-off. Without pull-down resistors, the bases of the output transistors gradually (as their capacitances slowly discharge though their leakage) "float" down to a voltage level that is determined by their substantial leakage current (which increases with temperature) and so the output transistors don't completely turn-off.

    The Zapco PSU schematic is in Reply #28, and shows 4 paralleled transistors on each side, without emitter resistors.

    Zapco either matched the paralleled transistors, or had good luck.
    Think of emitter resistors as a form of negative feedback. When a high-gain transistor attempts to conduct a large collector current, then that current creates a voltage-drop across the emitter resistor, which reduces the base-emitter voltage and therefore reduces base-drive.
    A transistor with less gain will attempt to conduct less current and therefore will have less base-drive reduction. So the gains of the transistors are equalized.
    Without emitter resistors, when a high-gain transistor is paralleled with a low-gain one, then the high-gain transistor will conduct more current than the low-gain one, which results in unbalanced current sharing. Without balanced sharing, the high-gain transistor may exceed its maximum current and/or thermal rating and blow-up.

  11. Ante,
    How many inverters do we need on this site?
    I think that the other one, the Low Cost 500W (or 700W) recently revised one, should be stripped-down (and will be explained), when someone requests a reduced power output.
    As you know, I expect projects to work the first time, every time.

  12. Use bigger wire going to your house, farm or castle.
    When the motor is turned-on, it will draw many, many amps of current until it is up to speed. A capacitor won't help much unless it is huge and expensive. A high-voltage, high-value capacitor wired in parallel with the motor will lower the power-factor of that big inductance, reducing its power-draw and your electricity-bill a bit.
    The voltage is dropping along the weakest wire between the power generating station, and the point where the wires to the pump and your lights join.

  13. Hotwaterwizard,
    Your "delayed on relay" is a good circuit that will do the job, but with only a 6V supply, the input divider should be changed.
    If you use 220K for both divider resistors, and a 100 microfarad capacitor, I figure that the delay will be about 6 seconds.
    Use another 2N3904 transistor to replace the obsolete MPSU10, and use a small relay.
    The power must be turned-off for at least 30 seconds, or else the next delay time will be shortened.

    Do you understand how to build this circuit?

  14. Hi Guys,
    Why is this simple multivibrator inverter included in the Projects Section of this high-quality site? It is here:

    The author claims that it will deliver 300W as shown.
    Has anyone built it? Unfortunately, it has problems:
    1) If the parts values and transistor gains are identical, then when it is turned-on both transistors may conduct at the same time and latch-up. Then it will blow the....., well it doesn't have a fuse, so something will get mighty hot.
    2) With only 63mA of base-drive from R3 or R4 (180 ohms), if the 2N3055 transistors have their minimum-spec'd gain, then the output would be only 39W. (I know, the capacitors boost the base current, but only for a moment).
    3) The capacitors cause the reverse-bias maximum voltage rating of the base-emitter junction of the transistors to be exceeded, causing damage.
    4) The capacitors are shown backwards.
    No, I don't want to re-engineer it, I think that it should be removed from this site.

  15. MP,
    I have never heard of measuring the resistance of a tranformer's windings to determine its voltage ratio.
    The resistance of each winding is not only dependent on the number of turns (length). It is also dependent on wire size.
    For a voltage step-down transformer, thin wire is sometimes used for the primary winding and heavy wire for the secondary. They can be overloaded without much heating.
    Also, in a transformer that has one winding wound on top of the other, the top winding will have a greater length, therefore its resistance will be higher than expected.
    Just measure the voltage.

  16. Marsman,
    You could separate the CPU and PSU only if you use very heavy wire (or copper pipes), which is very impractical.
    However, some PSUs have "remote sense" wires that allow the PSU to adjust its voltage to make-up for the voltage drop of long wires that you desire. Maybe your PSU already has these wires but they are jumpered to its output. If so, then cut the jumpers and try it with each output wire (fairly heavy) and each sense wire (not heavy) joined at the CPU.

    Ante and Marsman,
    Have you seen this link? I have tuned-in since December:

  17. MP,
    Thanks for posting the article about darlingtons.
    There it is, in the schematic on the 1st page, a darlington showing pull-down resistors, and exactly the same description for their function that I gave.
    Sorry to continue this argument, but ONSemi (Motorola), in their MC34071 opamp, use pull-down current sources on the "darlington" (their wording) input transistors. See the chip's schematic. Most if not all opamps that use darlingtons have this arrangement. The MC34071 data sheet is here:

  18. Hotwaterwizard,
    Good point about spike protection!
    I was trying to figure out how a positive spike at the collectors of the output transistors can get through their turned-off collector-base junction, and how a reversed-biased diode can arrest this spike. Now I realise that the positive spike produced at the collectors of the output transistors that are turning-off, causes a negative spike at the collectors of the output transistors that have not yet turned-on, due to center-tapped transformer action. Since this negative spike forward-biases the latter's collector-base junction, then it also forward-biases the protection diode, which arrests the spike.
    Those diodes are good protection.

    Another concern: Perhaps you should change your multivibrator to the 4047 type that is used in the other post:

    1) The 4047 has a divide-by-two to ensure that its outputs have exactly a 50-50 mark-space ratio. Yours depends on resistor/capacitor values, parts tolerance and lack of drifting.
    2) I have seen it happen: If your values are the same then both transistors may "latch" during power-on, when both transistors turn-on and stay-on. Then they won't oscillate.
    3) The 4047 needs only 1 resistor and 1 capacitor, compared to all the parts for your multivibrator, and costs much less.
    But if you use a 4047, you will need to add a pre-driver transistor, like the other post, to boost its output current.
    These two projects are getting to be more and more identical!

  • Create New...