audioguru

Will you teach your processor that the intensity of the 3 colors and white from the sensor are a "..." light source at a certain distance with a certain brightness? I think your processor will be frequently confused because for example the sun and a close candle both produce fire colors. The frequency scaling has something to do with the resolution of the outputs.

Your photodiodes will produce an output current if they have no bias voltage because they perform like a tiny solar cell. Then a photodiode directly feeds the (-) input of an opamp. The opamp will have an output voltage in the dark caused by its input offset voltage and current. Another way is to reverse-bias the photodiode Then it leaks a tiny dark current and a current dependant on the intensity of the light. Since the photodiode simply produces an output level dependant on the brightness of the light then you need to guess what is the source.

Your new circuit should work fine.

Sonos did not put headphones jacks maybe because if you use their Trueplay, then the Sonos is equalized for the speakers you are using, not for headphones. Then headphones might sound awful unless you can turn off Trueplay (but of course she will forget to turn off Trueplay). You did not say and we do not know what is a Sonos ZP-100 and we do not know what are its output signals. The Sonos website shows no details and no spec's. Their website does not list a ZP-100. Are L- and R- outputs actually grounds, or are they the other half of differential outputs from a bridged amplifier? Looking at the attenuator resistors I think they are grounds.

The circuit might work when the parts are soldered. Do not use a solderless breadboard that cannot handle the high currents and its contacts might be intermittent. Some projects on that website have errors.

We speak English in this forum, you don't. Maybe you are a SPAMMER?

A transformerless circuit powered from the mains electricity is Extremely dangerous. All wired parts must be completely covered so they cannot be touched. A fuse should be used to prevent a fire if a voltage spike shorts the capacitor. The reactance of the capacitor limits the operating current but its charging current is extremely high when the circuit is turned on at the peak voltage of the electricity sinewave so a current-limiting resistor is needed. The spec's of your LEDs will show a range of forward voltage not simply 1.5V and might show the maximum peak current for a short duration while the capacitor charges. Do you have the LEDs back-to-back or do you have a rectifier?

The 10V zener diode reduces the total supply voltage of U3 from its absolute maximum of about 44 V to about 34V. Then opamp U3 will be cooler and it will live longer. EDIT: But it is not needed in your circuit since your input and output voltages are lower.

I use AA alkaline cells only for very low current products where the idle current of The Batterizer is higher. Then it wastes more power than it saves. I use AA Ni-MH rechargeable cells for high current products that The Batterizer cannot supply. The marketing people know that many people are gullible and believe their lies.

You want a power of 5V x 500mA= 2.5W. An AAA cell voltage is 1.5V when new and drops to about 1.0V so its average voltage is 1.25V. To produce 2.5W its current must be 2.5W/1.25V= 2.0A. The voltage stepup circuit will get hot with 2A so will use maybe 500mA extra current making the heat. Then the battery current will be 2.5A which is too much for a little AAA alkaline battery. The datasheet of an Energizer AAA alkaline cell shows that its internal resistance is a maximum of 300 milli-ohms so its voltage loss will be 2.5A x 0.3 ohms= 0.75 V so its voltage will be 1.5V - 0.75V= 0.75V when brand new and its voltage will drop as it is used. But since the voltage to the stepup circuit is only 0.75V then its current must be 5A which is impossible from a little AAA cell.

The datasheets for the BD139 and BD140 show that the chip will be at its maximum allowed temperature if it dissipates 12.5W and its case is cooled down to 25 degrees C somehow (with a huge heatsink in liquid nitrogen?). In free air a pretty big heatsink will get hot then the chip will be too hot. EDIT: I forgot. I do not know of a power transistor that can dissipate more heat and is as fast as the BD139. The problem with using two BD139 transistors in parallel, each with an emitter resistor, is that they will overload the opamp that drives them.

Your main filter capacitor value is much too low for an output of 10A. use at least 20,000uF. The transformer voltage is too high for a maximum output voltage of only 20V and its current rating is much too low. The poor little BD139 driver transistor will smoke and die if you are unlucky to find those very old output transistors with low current gain. Q1 is completely wrong. Its maximum allowed reverse bias on its base-emitter is only 5V.

I am glad that the value of the R7 heater has been reduced to 0.1 ohm. The original was 0.47 ohm which heated with (3A squared x 0.47 ohm=) 4.23W. The TL081 opamps that were used had a maximum input offset voltage of 15mV (!) so the minimum regulated current with an accuracy of 20% was 15mV/0.47 ohm x 1/5= 6mA but they claimed 2mA. I do not think anybody needs current regulation less than 20mA (for an LED?).

The two series diodes at the outputs of the regulators RUINS their excellent voltage regulation. Why two diodes, why not just one diode, it will also ruin the voltage regulation but not as much as two diodes? The sliders on your 5k pots are shown disconnected. The 240 ohm resistors are for the more expensive LM138 and LM137. LM338 AND LM337 need 120 ohms (and the pot value also halved) to prevent the output voltages from rising if the load current is low.

C11 and C12 slow down the outputs too much, causing poor voltage regulation if the load current changes of if the load current is modulated. The circuit will have serious overshoots and might oscillate.

You forgot that the base of the NPN transistor cannot go higher than about 0.65V so in the light the base has some current. The 7k resistor has a current of (6V - 0.65V)/7k= 764uA. The 4k LDR has a current of 0.65V/4k= 163uA. Then the base current is 764uA - 163uA= 601uA so it is turned on a little and the LED lights a little.

The datasheet for the BC548 shows that it saturates properly when its base current is 1/20th its collector current. When the transistor is saturated its collector current is 15mA then its base current should be 15mA/20= 750uA. Then R1 will be 5.3V/750uA= about 7k. In room light the base voltage will be about 2.18V so the transistor will not completely turn off causing the LED to look dim. You might be lucky to find a transistor with high gain (they are all different) then R1 can have a much higher value. You should be using a darlington transistor that is guaranteed to have very high gain. R2 simply prevents the base current from being too high when R1 is reduced too low.

Your frequencies are completely wrong and what you say is also completely wrong. Microwaves do not end at 75mHz which is a VERY low frequency. Microwaves are from about 3GHz to about 300GHz. FM radio is around 100MHz which is VHF. UHF is from 300MHz to about 300GHz, IR is from 300GHz to about 320THz, visible light is near 320THz and UV is from 325THz to about 30PHz. Microwaves do not cause cancer, they cause heating. Ultra violet radiation (mostly from sunshine) causes skin cancer. Why do you talk about aliens?? Are you an alien?

When a capacitor is discharged by a high load current then its voltage begins to drop immediately since it has no voltage regulation. When you want the output capacitor discharged then you are using a transistor that will vaporize. The only thing limiting the current in the discharge transistor is its current gain since the opamp can supply its base with only about 20mA .

If there is a positive input voltage to the opamp (+) input, the current flows from the emitter of the BD139 through the base-emitter diodes of the output transistors and through the low value resistors to ground. The 1k emitter to ground resistor of the BD139 also has current in it. When there is 0V on the discharge power resistor then there will be no current in it and no current in any other part.

The opamps will be cooler if they are TLE2141 singles.

You do not need a huge output capacitor because the circuit has voltage regulation. Actually this original old circuit from a Greek kit has many overloaded parts and is not reliable. We fixed in and you can see the improved circuit in the forum. A Chinese kit copy (from Banggood or something) is sold at a very low cost but it also has some overloaded parts.

It is not smart to zap and burn a person with high power RF energy. It is not a medical device and will not cure anything, It might cause cancer. It is absolutely normal for an old person to age. Let it happen. I recently turned 70 years old but I take good care of myself so I am very healthy and I feel like 40 or 50. Many people today do not take good care of themselves and at 40 or 50 they can barely walk. Today I can run fast and I can almost fly. High power RF energy is never going to go into me.

Kevin, we cannot hear you.

I do not know which planet you live on but here in Canada the ringing is a 90VAC/20Hz signal on the telephone line. The 52VDC is still there during ringing and when a phone goes off hook it draws current from the 52VDC which is sensed and stops the ringing and connects the audio from two phones.