audioguru

Siddharth, Actually, the opposite is true. When a digital circuit changes its output state from a "0" to a "1" and back, a current pulse is created in the ground wiring which causes voltage spikes (digital hash) along it. The digital circuits are not affected by reasonable (10% to 30% of the supply voltage) voltage spikes in the ground wiring due to their noise-immunity specification. However, analog circuits (audio or video) do not have noise-immunity. If you supply an analog circuit with a digital ground wire then you will hear or see digital hash. By separating the analog ground wiring from the digital, this keeps the analog circuitry clean of the transients introduced by the digital circuitry.

The site's clock and my time-zone shift worked fine before the Apache software upgrade. Now it is exactly 10 hours behind. Am I the only one affected?

FlipFlop, In addition to stopping corrosion of traces, the solder-mask saves manufacturers the cost of extra solder and its weight in the product.

Cmer, What kind of LEDs and why must they be in series? If they are common red LEDs, and they have a voltage of 1.8V each, then you must use a DC to DC converter to step-up the 5V so that they will have 5.4V across them plus more voltage for a transistor or IC current-limited switch. Other colours and types of LEDs will need even more voltage. However if they are in parallel, each having a current-limiting resistor, then they will flash when driven by a 5V oscillator such as a 555. Even white LEDs can be driven this way.

Now that the server has its Apache update, I frequently get new errors from the site or server: "Fatal Error. Cannot redeclare phpadment....." and ".....advertisement....." so I must go back in my browser, then forward and it is OK. The clock is showing the wrong time.

Jacob, Do you want to measure the time that the object stays on the platform, after landing, before it bounces or falls off?

Kevin, I think that you "roasted" the little zener if you put 1.5A through it. Next time use a current-limiting resistor in series.

MP, Thanks. I didn't have Paint installed, but I do now. Now I can scratch-out simple schematics quicker than Eagle. I just now converted a 4MB BMP schematic to a 240KB JPG and they look the same. I'm saying goodbye to all my BMPs.

Ante, I am on cable and are running at 350KB/s to 1MB/s and on most other sites a screen snaps on display only a moment after I click. Mixos, Thanks for trying. It must be difficult for you.

MP, Back to the "darlington transistors" argument. Power darlingtons also use base-emitter resistors too: http://www.fairchildsemi.com/ds/MJ/MJE803.pdf

Sorry for my repeat of GPG's formula. I didn't see page 2 when I replied.

That circuit provides a regulated output voltage by this calculation: Zener voltage plus 0.65V (base-emitter) multiplied by (R1 divided by R2) plus 1. The zener is the voltage reference that the divided output voltage is compared with. Since a zener is low impedance, changes in current through it does not produce much voltage change across it. If a resistor was used instead, then there would be no reference voltage (it would change with any output change) and the circuit would not regulate.

Kevin, What opamp circuit has its input capacitor connected to ground? Input capacitors usually couple in series an AC signal source to the opamp input and bias resistor for noninverting, or opamp input resistor for inverting circuits. The low-frequency rolloff of an input coupling capacitor is like a differentiator. What is an emitter element? If you are talking about an inverting opamp circuit, then its gain is simply R2 (feedback) divided by R1 (input) plus the source impedance.

Kevin, That's correct, when the capacitor voltage reaches about 11.3V, then the timing period is over and the transistor turns off. The switch resets the circuit by discharging the capacitor (and turning-on the transistor). The transistor stays-on during the timing period when base current flows through the resistor to charge the capacitor. When the capacitor charges to about 11.3V, then there will no longer be base current and the transistor turns-off. But this simple circuit relies on having very low leakage currents in the capacitor and transistor, otherwise the transistor will never turn-off. I believe that ALL transistors should have a resistor between the base and emitter, to bypass leakage currents.

Today the server is worse than it was before the long update the other day. I had to make many tries to log-in, going from page-to-page and trying to post gave errors. Once, the server gave the message, "too many connections".

Kevin, The input capacitor does NOT cause the gain to increase. It simply rolls-off the low frequencies, below its "corner" frequency at 20dB/decade. The corner frequency is determined by the capacitive reactance and the source and bias resistors. Below the corner frequency, the capacitor passes less signal because its reactance is high (making an attenuator). Above the corner frequency, the capacitor passes the full signal because its reactance is low (like a piece of wire).

Uman, A project that uses the 4060 that I recommend is here: http://www.mitedu.freeserve.co.uk/Circuits/Timing/24hour.htm

gEcky, 1) A quality switch should not blow-out. 2) I am still confused by your "+/- 24V" transformer. Is it 24V center-tapped, which is also 12-0-12? Then its rectified and filtered output is about 16.2V. 3) Your LED has about 1.8V (for red) across it. If you use a 220 ohm resistor in series on 5V, then Ohm's Law calculates the current to be 5-1.8 divided by 220 = 0.0146A. A resistor to provide the same current from 16.2V is calculated: 16.2-1.8 divided by 0.0146 = 986 ohms. Use 1K ohms. The power in the resistor is calculated: voltage x current, so 16.2-1.8 times 0.0146 = 0.21W. A 1/4W resistor will get quite hot, so use a 1/2W resistor.

Output Transistors in Square-wave Inverter Circuits: 1) How much current must be drawn from a 12V battery to produce 300W output from a square-wave inverter circuit? Well, you need to draw at least 300W, plus some more to account for losses such as transistor saturation voltage-loss (which produces heat) and transformer wire resistance. So let's assume that the losses are 10%. But since the 12V battery actually measures 13.2V, which is also 10% higher, then we can ignor the losses since they cancel the extra battery voltage. Since Power = Voltage X Current, then the current from the 12V battery must be 25A. 2) How much current must the transistors conduct? Each output transistor must conduct that 25A, alternating back and forth into the center-tapped transformer winding. 3) Can 2N3055 transistors be used? No, unless they are paralleled (matched, or with emitter resistors to equalize variations of their current gain). The 2N3055 transistor has a maximum collecter current rating of 15A. But it has a guaranteed minimum current gain of only 5 with 10A of collector current, and even less current gain (not specified) at a higher collector current. If we parallel two 2N3055 transistors and operate them with 10A of collector current each, then the current sum is only 20A, which is still not enough. But since the current gain could be 5, the bases current will be 4A, which we can add to the transformer winding if we use a darlington configuration for driver transistors. Now the total current will be 24A, which is close to the 25A required. So the original 2-transistor inverter must have at least 6 transistors. (to be continued)

Hey! Who started this topic here and put my name on it? But let's continue without locking nor deletions: 1) The author's web-site's forum has 5 pages of complaints about this project with all-kinds of capacitors (tantalum, elecrolytic, non-polar) blowing-up. 2) One that didn't blow-up complained that its output was only 37W. 3) There was only one who claims to have success with this project: __________________________________________________ Re:"Power?" Inverter

MP, Credit? Of course I'm proud of myself. Philips? While they were showing to me their top-secret audio-cassette and LED, I was teaching them about transistors. Been there, done that: Job? I'm playing with "Freedom 55" now, plus a little consulting on the side.

Ante, That's a good comparison, they both look almost the same, even with nearly 3:1 compression. What program do you use to convert GIF to JPG? Will it also convert BMP (huge file) to JPG?

Equinox, Isn't the contol circuit just a thermostat that is designed for use with your heater? Isn't the heatexchanger just a fan that blows air through fins attached to your heater? Didn't they come with connections instructions?

Nettron, I mentioned CMOS because Eki's kit doesn't appear to have LED current-limiting resistors. CMOS outputs current-limit at about 11mA from a 9V battery. I agree that linear will look better than logic, because the LM3914 "blends" between adjacent outputs, while logic will go: jerk, jerk, jerk, etc. Ldanielrosa, Microcontroller use also is not very original. I can purchase this circuit from my local car-accessory store and have a choice of LED colours. The product has a microcontroller chip under a blob of black epoxy, and the LEDs are surface-mount. It is supposed to make your car appear to have an expensive (and noisy) security system. Cell-phone covers also have many blinking LEDs.

Ante, In those days, few people except you and me knew much about transistor circuits! Those Philips engineers just didn't know, so uni-polar was chosen. For the capacitors in this project, most of the time the collector-side of the capacitor is positive: 1) During its charging when the collector is near +24V (center-tapped transformer action) and the other end of the capacitor is at about +1V (forward-biased base of the other transistor). Notice that there is little current-limiting here. 2) During most of its discharging, when the collector is at about +0.5V, and the other end of the capacitor is at about -7V (avalanching base-emitter junction of the other transistor) (again, little current-limiting), then the capacitor continues discharging into its resistor. 3) With the collector still at about +0.5V, the other end of the capacitor rises to about +0.7V when the base of the other transistor becomes forward-biased and that transistor turns-on. So the capacitor gets only about 0.2V of reverse-polarity, for only a moment. Any capacitor can withstand a reverse-polarity of only 0.2V for a moment. Few, if any, capacitors can withstand a reverse-polarity of 23V with a very high charge current, nor continuing charge/discharge high currents.