audioguru

The original project has errors that can destroy the TL081 opamps and overload many other parts. That is why we fixed it and talk about the fixes in a few long threads in the forum. If the pins of Q1 are wired wrong and are mixed up then the reverse-biased emitter-base becomes a 8V zener diode that overloads the output of U2 and prevents the output from rising above about 7V. The pins on the European BC548 are CBE, but the pins on any American or Japanese little transistor are EBC.

RX-300U. The owner's manual is in Google.

I still use my 1964 HH Scott FM stereo receiver nearly every day. The only problem it had was a dial indicator light bulb burned out but it was easy to replace. My 1997 Yamaha stereo receiver has a worn out selector switch and its FM does not work anymore, even the FM display so it must be a simple power supply problem.

Simply build another power supply completely separate from the first power supply and use its "+" output as 0V and use its "ground" output as its variable negative voltage output. Even the transformer and any meters must be separate.

A transistor with the load at its emitter does not saturate and is not a switch. It is an emitter-follower. When it is turned on then its emitter voltage is a Vbe voltage drop less than its base voltage and its base voltage is probably not as high as its supply voltage. A transistor used as a saturated switch has the load connected between its collector and the power supply voltage. The base current must be at least 1/10th the collector current for most little transistors or 1/3rd the collector current for a power transistor like a 2N3055 regardless of its hFE (beta) number. The hFE or beta number is its DC current gain when it is a linear amplifier with plenty of collector to emitter voltage so it is not saturated. The definition of a saturated transistor is when its collector-base diode is forward biased and conducts so the transistor cannot turn on any more. An NPN transistor usually saturates better than a PNP.

This thread started and stopped 11 years ago. Today cheap remote controlled helicopter toys use two rotors and have a speed control for each for up, down and turning. They also have a speed and direction control at the rear for forwards and backwards.

Many ICs regulate LED current without using series resistors. Examples are the LM3914, LM3915 and LM3916.

I just lit a white LED and a red LED with a current of only 24uA. They were not very bright but were very obvious. Try it. You can see in sunlight and you can also see in moonlight that has MUCH less light. Yes.

The resistor value from the ISET pin to ground sets the LED peak current. You can also program the IC to change (maybe using PWM) the (average) current in each digit.

Texas Instruments say the 6N138 is obsolete. Its datasheet shows that its minimum transfer ratio is 300% (it is typically 1300%) so with an LED current of 1.6mA its minimum output current can be 4.8mA. But with a 5V supply and a 4.7k output resistor its output current is only 1.1mA so it will be fine. When an LED conducts ANY amount of current then it produces some amount of light. True, an LED at only 1.6mA will appear dim. But here the LED is extremely close to the very sensitive photo detector. If you want to switch when the voltage is zero then you need a more modern "zero-crossing" optocoupler.

This thread is also on at least two other websites. I think the 5V regulator is oscillating because the capacitors for the regulator should be ceramic and mounted much closer to the pins of the regulator.

The Sanyo transistor is fantastic and will work well in a new circuit to drive your LED strips but Sanyo was recently purchased by Panasonic and might not make transistors anymore. Now you are talking about a relay but forgot to tell us how much current its coil needs.

I think a little BC337 will get too hot with a current of 600mA and a saturation voltage loss of maybe 1.2V. Also its maximum allowed current is 500mA which is too low. The 12V battery might be 13.8V.

The new load will be 600mA. An ordinary transistor saturates fairly well when its base current is 60mA which is too much for an LED unless it is a big very bright one. So the receiver output must be analysed to see if it can be modified to supply 60mA.

We do not know how much current the 2V signal has. Therefore we do not know if you need a normal transistor or a darlington transistor. A darlington transistor has voltage loss that might dim the LED strips. Most Mosfets need a 10V input to turn on but some 'logic-level" ones need only 4.5V. Therefore you need a normal transistor "level-converter" to turn on a Mosfet that turns on the LED strips. Do the LED strips have built-in current limiting resistors?

A transistor is slowly damaged when operating with avalanche breakdown of a reverse-biased emitter-base junction. The 2N2222 and BT2222 are guaranteed to withstand a reverse emitter-base voltage of 6V or less. Both transistors probably have the same chip but since the BT2222 is in a tiny case then it will overheat even with a very low power dissipation.

No. An emitter resistor is not needed for only one output transistor. Are you discharging a 12V battery? My circuit with an NPN driver transistor and NPN output transistor is a battery charger, not a discharger.

You cannot change the voltage drop of R2 by changing its value. R2 quickly discharges the base-emitter capacitance of the output transistor and discharges its leakage current. 1k ohms is fine. A small thermal resistance allows a power transistor to increase its transfer of the heat to the heatsink. Then the transistor will be cooler and the heatsink will be warmer.

It will work well if the driver transistor has a fairly wide bandwidth like a BD139. In this forum there is a 0V to 30V at a few mA to 3A power supply. The improved version has two 2N3055 power transistors in parallel with emitter resistors and a driver transistor. When a slow TIP31 driver transistor was used the transient response showed bad overshoots and ringing. The BD139 fixed it. The original version used a very old maybe obsolete driver transistor that was fast. I corrected and made notes on your circuit:

Your circuits with an opamp plus common-emitter transistors have too much voltage gain so they will probably oscillate. I showed two circuits with emitter-follower driver and output transistors that will not oscillate. But if you use PNP transistors the output voltage will not be able to go near 0V unless the circuit uses an additional negative supply. I showed a circuit with NPN emitter-followers that can go down to 0V if the inputs and output of the opamp can, without an additional negative supply. Add a second output transistor and emitter resistors for both output transistors and it will work perfectly.

Your amplifier will oscillate because its transistors provide voltage gain. Then at a frequency where the opamp's phase shift would cause oscillation, its gain is less than 1 so it does not oscillate, but the total gain with the transistors is higher than 1 so the entire amplifier will oscillate when negative feedback is applied (the phase shift causes the negative feedback to become positive feedback at high frequencies). Also, common emitter transistors have more phase shift than emitter-followers due to the Miller effect (the capacitance from collector to base is amplified). Also, your PNP transistors are missing resistors that will turn them off quickly. So yours turn off slowly which adds additional phase shift.

If you connect two transistors in parallel then each needs an emitter resistor so that they share the current fairly equally. The opamp will probably not have enough output current to drive output transistors without adding a driver transistor. I sketched a circuit with a PNP driver transistor and a PNP output transistor. They have enough current gain for the opamp to be able to drive. BUT since they are both emitter-followers with a Vbe of about 0.7V and 1V then the output voltage will not go any lower than about +1.5V to 2V.

The heating of a zener diode wastes power. It is simply the voltage across it times the current in it to give the number of Watts of heating.

If it is then shorting D to S will keep it always turned on, but not cause the circuit to think there is always motion which needs an AC signal created by the Fresnel lens. The circuit will simply do whatever it does when the FIR is exposed to continuous IR light. I am surprised to see that it has a tiny solar panel inside that develops a voltage when exposed to IR light. Then the voltage turns on the Mosfet.

You could measure the one you have to find the positive supply pin, the 0V ground pin and the output pin. Then if you short the output it might become destroyed.