audioguru

I have never seen a bad ALPS pot and have never ordered one. If the code had an "A" in it then it would have an Audio logarithmic taper. It has an "M" so it is a linear pot.

Without using thermal compound, the transistors are in one city and the heatsink is in another city. Then the heatsink will not be very effective. If the value for the emitter resistors is too low or the resistors are missing then one transistor will probably heat more than the other.

The 2SC5200 power transistors in a plastic case are much better than 2N3055 transistors in a TO-3 metal case. They should transfer more heat to the heatsink than the 2N3055 transistors. The heatsinks in your photos are tiny so no wonder the output transistors get hot. I do not know if you use thermal compound between the transistor and the heatsink to fill in the tiny imperfections and allow much better heat transfer. Our 5A version of this 30V power supply uses three 2N3055 output transistors in parallel, each with a 0.33 ohm emitter resistor. Each transistor is different. The Vbe of one transistor might be 0.9V and the Vbe of another might be 0.8V. If they are in parallel without emitter resistors then the one with the lowest Vbe conducts all of the current and burns out while the other transistor does nothing. The value of the emitter resistors (0.33 ohms works well) must be selected to balance worst-case unmatched transistors. Get rid of the 1.3W zener diodes because they might not even regulate with the low current given to them. I do not know why you are not using the TLE2141 opamps that we use. Their maximum supply voltage is 44V so the do not need your LM317 low voltage regulator.

you reduced the values of the resistors so that the low supply voltage for your voltage amplifier opamp does not cause poor voltage regulation. then if the output transistors are a little different then one will conduct much more current than the other and get too hot. Also the input offset voltage for the current regulating opamp will affect its regulation. The regulation should be much better than a 60mV drop. Maybe your wiring has a resistance of 60mV/3.4A= 0.0017 ohms. Use thicker wires. Probably because the low supply voltage for the voltage amplifier opamp causes it to saturate when the output voltage and current are high. It is simple to calculate the dissipation (heating power) in the output transistors then select a heatsink that will cope with it. The heatsink must not be inside an enclosure.

It is very confusing when each new schematic has different parts designation numbers. Your new D10 has backwards polarity then the current regulation will not work. When the new U1 detects over-current then its output goes low enough to pull down the voltage feeding the voltage amplifier opamp so that the current in the load is reduced.

The BZX55 zener diodes have their voltage spec'd when their current is only 5mA. Then the 6.2V zener diode has a power of only 5mA x 6.2V= 31mW, not anywhere near its maximum of 500mW. The 3.3V zener diode has a power of 16.5mW which is also nowhere near 500mW. Why don't you look at the datasheets yourself? the BZX85C3V3 zener diode is 3.3V only when its current is 80mA! In this circuit the current is only about 7.8mA. The NE5xxx opamp has a slightly higher maximum output voltage than the TL081 but its 33V positive supply voltage from the LM317 is much too low so it will not make much difference in that circuit.

Your project has poor voltage regulation when its voltage and current are high because the voltage regulating opamp has a low supply voltage of only 33V. Its maximum output is about 31.5V if you are lucky, the base-emitter voltage of the BD139 could be 1V and the Vbe of the output transistors and their emitter resistor voltage drops could be 2V. That is why we use opamps that have a maximum allowed supply of 44V so that they do not need a voltage regulator.

Liquibyte, You have the transistor shorted with a wire from its collector to its emitter instead of a 10M voltmeter.

I do not know why so many calculations were made for the very simple circuit, most of the calculations are not needed to simply find the RMS power in the 5K ohms load. I simulated the circuit with LTspiceIV and it shows an output power of 9.0uW in the 5k ohms load.

The current regulator opamp is not oscillating. Its output is saturated as high as it can go which is about +28V because it is not regulating the current so of course it shows the 120Hz ripple from the unregulated +39V, which is reduced to +29V by the 10V zener diode. The moment the current regulator opamp begins to work then its output will have no ripple.

We cannot teach the basics of opamps in one post but here goes: 1) Everyone who modifies this circuit uses a different parts numbering system so I will describe the three opamps as the voltage reference circuit, the voltage amplifier and the current regulator. 1) The voltage reference opamp provides a constant current to the 5.6V reference zener diode and has a gain of 2 times so the reference voltage is 11.2V. 2) The voltage amplifier opamp drives the BD139 driver transistor which drives the two 2N3055 output transistors. Two resistors in the amplifier allow the amplifier to have a gain of 2.68 times so that the 11.2V reference is amplified up to 30.0V at a high current. 3) The current regulator opamp compares the voltage produced by the load current in the 0.47 ohms current sensing resistor with the voiltage of the current setting potentiometer. If the sensed voltage is too high then this opamp reduces the voltage from the voltage setting potentiometer through a diode until the load current is the same as is set. If the output is shorted then the current regulator opamp causes the output voltage setting to drop to almost zero so the current is not higher than the setting of the current setting pot.

It slows down changes in the voltage regulation. Add a load then the voltage suddenly drops and when this capacitor charges then the voltage slowly comes up to normal. Disconnect a load and the voltage suddenly increases and when this capacitor charges then the voltage slowly comes down to normal. It prevents high frequency oscillation, maybe because the output transistors are very slow.

C6 in the latest SIM circuit prevents the current regulator from quickly cutting back or quickly allowing more current. Then if the output is suddenly shorted the current in the circuit will skyrocket until the slowly ramping opamp can catch up which is very bad. Maybe the current sensing resistor should be non-inductive (not wire-wound).

Since you bought it from Amazon then maybe the IC is a fake or maybe the circuit is designed wrong. Please provide a link to it. The output pulses? When it is playing loudly? Does the heatsink get hot? What impedance are the speakers? What is the maximum continuous current from the 18V power supply? If two 8 ohm speakers are playing at 14W each with a little clipping distortion then the power supply must provide 18VDC at 2.5A (45W).

The very old LT1001A opamp is very slow like a lousy old 741 opamp. The TLE2141 is MUCH faster. You have C6 as only 33pF but it should be 330pF. Why does the circuit have C11? It is causing the current spike and maybe is causing the oscillation.

A 28V transformer might produce 29V when the project has no load. Then its 41.0V peak voltage is reduced to 39.6V by the rectifier bridge. It will be fine if the mains voltage is very stable.

Good point. Eliminate Q3, turn the polarity of the LED around so its cathode is at the collector of the new lower NPN transistor and its current-limiting resistor is between the anode of the LED and the +27.6V. Another good point but there is no problem. If the output is shorted then we want the output voltage to go to 0.00V. The new upper NPN transistor has its emitter connected to the circuit ground. Its collector is connected to the input of the output amplifier that must be +1.41V higher (the voltage drop for 3A in the current sensing resistor) for an output voltage of 0.00V when the output is shorted and the current is set to 3.0A. So the transistor does not need to saturate. The output amplifier does not need a negative supply for its output voltage to be 0.00V because the output of the opamp drives the darlington-connected driver and output transistors so its voltage is two diode drops higher.

Then you must replace the 10V zener diode with a resistor. I do not think U1 needs its power supply voltage regulated because a TLE2141 opamp has a minimum Supply-Voltage Rejection Ratio of better than 90dB so 2V of ripple causes an output of less than 62.5uV (typically less than 2uV) which is very close to zero.

When the load current is higher than the current regulator setting and the voltage is set high, the "on" voltage spike is caused by the time it takes for the negative supply to reach a low enough voltage (-1V?) for the current regulating opamp to pull down the voltage feeding the voltage output amplifier. Here is my proposed fix:

In Canada we also have three wires, live, neutral and earth ground. I have seen electrical outlets wired wrong by drunk electricians or by homeowners who do not know how to do it correctly so the "earth" terminal is actually "live". Then if a person touched the metal chassis and an earth ground, the person would be electrocuted if the outlet was wired wrong and if the chassis was connected to the earth wire on the plug. If the chassis is not connected to the earth wire then there will not be a problem unless the transformer live wire or live wire in the cord shorted to the chassis.

RV1 nulls the input offset voltage of the voltage setting opamp U2 so that the output is exactly 0V when the voltage pot is set to zero. Without RV1 then the input offset voltage of the opamp (if there is any) will cause a small positive or negative voltage output instead of 0V. But some people say RV1 does not do anything. 2) when mounting in a metal box with the chassis earthed to mains earth (green/yellow), should the -ve on the electronics also be connected to mains earth?

Did I calculate it wrong? 28VAC has a peak voltage of 39.6V and the rectifier bridge drops 2V to 37.6V and the capacitor ripple drops it to maybe 35.6V. if the current sensing resistor is 0.47 ohms then it drops 1.4V to 34.2V. Then the 0.33 ohm emitter resistor drop 0.5V to 33.7V so the output transistors have a total dissipation of 101.1W. I DID calculate it wrong. Sorry.

The MC34071 is no longer available in a through-hole package (it is only surface-mount now) so the TLE2141 opamp from Texas Instruments is used. I think a 24V transformer produces peak rectified voltages of 32V at 3A and ripple reduces the minimum voltage to about 30V. That is why the circuit shown with the 24VAC transformer is rated for an output voltage of only 25V. The original project also used a 24VAC transformer and also could not produce 30V at 3A. An old CA3140 opamp is very noisy and has a high maximum input offset voltage. Its supply voltage might be higher than its maximum allowed voltage of only 36V when there is low load current or no load.

I agree that the battery might be bad. If it is lead-acid or Lithium then its life might be finished. Replace it.

Its datasheet says the relay is a Chinese "Songle" relay. What is that?? They cannot spell "single"? Its coil current is 120mA at 3V. I betcha the sensor cannot provide an output current as high as 120mA at 3V. Then the relay needs a transistor to drive it and the sensor drives the transistor. Instead of a transistor and a high current relay why don't you use a low input current darlington transistor to drive the sprinkler?