5 ms·

I don't think any circuits will survive 110 / 230V.

oofabz 12y ago
There are circuits that can survive high voltages, but they are rare. You can buy power transistors (BJTs) which can handle 400V (VCE). The base can't take high voltages but it's simple to protect with a resistor. Here is a power supply I built that works directly with mains voltage, no transformer: https://frammish.org/realized.png https://frammish.org/realized.png https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-positive.jpg https://frammish.org/protoboard-positive.jpg
- deutronium 12y ago
  Very interesting, do you mind explaining roughly how it works? And what the output voltage is.
  - oofabz 12y ago
    I'll try. This is a constant-current power supply, so the voltage depends on the resistance of the load. The LED on the left is the load. The circuit does not work well and I do not advise building it. It had extremely noisy output and tended to destroy the expensive LEDs I hooked up to it. The four diodes and large capacitor on the right are a full-bridge rectifier to provide 170V DC. The top left PNP transistors are a current mirror which reflects the current through the LED to the right. The 1k potentiometer on the base reduces the mirrored current in order for the logic to consume less power than the load. The top left transistor is mostly responsible for the total inefficiency of the power supply, because it adds 0.7V (VBE) on top of the voltage of the load. One of the large heatsinks is connected to it. The bottom NPN transistors are another current mirror but without a resistor. One would think you could just measure the current at the other end, so you get the opposite polarity from the start. But doing it this way makes it work properly at startup, when there is zero current. The two transistors on the right are a constant-current source where the current is defined by the two resistors. This current and the mirrored current from the LED are connected directly together so that they fight. The node they are fighting over is connected to the gate of the bottom-left MOSFET. The MOSFET is connected to the other heatsink. If the current through the LED is too low, the constant current source "wins", and pulls the gate voltage high, limited by the Zener diode to avoid overvolting the gate. If the current through the LED is too high, the mirrored current wins and pulls the gate low, shutting off current to the LED. When the current is shut off, the capacitor on the left continues feeding the LED until the current drops enough for the MOSFET to turn on again. This circuit also has a nasty tendency to get stuck in a steady state where the MOSFET is half on, causing an enormous amount of heat to be dissipated in the MOSFET and killing it. It might be possible to address this with an inductor to create oscillation and buffer current. But if I did it again, I would take another approach, and use an astable multivibrator to sample the LED current at discrete intervals instead of doing it continuously. This would allow time for the gate to settle either high or low so it doesn't get stuck in the middle.
    - aidenn0 12y ago
      Just in case you don't already know this, some MOSFETs are much better acting in linear mode than others; they are much more expensive than your typical power MOSFET though.
- nictinkers 12y ago
  I worry that this circuit has been presented without the relevant warnings. There's no isolation here, so you're exposing people who may go and build this circuit to live mains. Think about the pot, for example. If somebody built this, and stuck a normal potentiometer on the front of a box to adjust the voltage, they may not think to use a pot that has a high insulation rating between the handle and the wiper. You could get a nasty crack of line voltage DC through the pot handle via your squishy internals and down to ground. The pictures are also a bit concerning - the big metal heatsinks are almost certainly live as transistor tabs are usually connected to one of the pins and you don't seem to have isolating washers and grommets. I wouldn't want to thread a screwdriver down between live heatsinks to get to the adjustment pot. This sort of regulator is okay if it and everything it powers is sealed away so nobody can touch it, or if it's powered via an isolating transformer. It's an unusual configuration and it'd be unusual for people to know what must be done to be safe when building this circuit. That unusualness along with the nasty consequences of mistakes means you've got a moral obligation to take care to warn potential builders of some to these issues when presenting this circuit.
  - cnvogel 12y ago
    "presented without relevant warnings..." It's a completely un-annotated schematic. To build the prototype, someone has to select suitable electronics, put it together on perfboard, and last but not least strip a power-cable to have bare-ends to solder to the input of the bridge rectifier. You really think that this person needs an additional warning about the dangers of 120V mains?

There are circuits that can survive high voltages, but they are rare. You can buy power transistors (BJTs) which can handle 400V (VCE). The base can't take high voltages but it's simple to protect with a resistor. Here is a power supply I built that works directly with mains voltage, no transformer: https://frammish.org/realized.png https://frammish.org/realized.png https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-positive.jpg https://frammish.org/protoboard-positive.jpg

deutronium 12y ago
Very interesting, do you mind explaining roughly how it works? And what the output voltage is.
- oofabz 12y ago
  I'll try. This is a constant-current power supply, so the voltage depends on the resistance of the load. The LED on the left is the load. The circuit does not work well and I do not advise building it. It had extremely noisy output and tended to destroy the expensive LEDs I hooked up to it. The four diodes and large capacitor on the right are a full-bridge rectifier to provide 170V DC. The top left PNP transistors are a current mirror which reflects the current through the LED to the right. The 1k potentiometer on the base reduces the mirrored current in order for the logic to consume less power than the load. The top left transistor is mostly responsible for the total inefficiency of the power supply, because it adds 0.7V (VBE) on top of the voltage of the load. One of the large heatsinks is connected to it. The bottom NPN transistors are another current mirror but without a resistor. One would think you could just measure the current at the other end, so you get the opposite polarity from the start. But doing it this way makes it work properly at startup, when there is zero current. The two transistors on the right are a constant-current source where the current is defined by the two resistors. This current and the mirrored current from the LED are connected directly together so that they fight. The node they are fighting over is connected to the gate of the bottom-left MOSFET. The MOSFET is connected to the other heatsink. If the current through the LED is too low, the constant current source "wins", and pulls the gate voltage high, limited by the Zener diode to avoid overvolting the gate. If the current through the LED is too high, the mirrored current wins and pulls the gate low, shutting off current to the LED. When the current is shut off, the capacitor on the left continues feeding the LED until the current drops enough for the MOSFET to turn on again. This circuit also has a nasty tendency to get stuck in a steady state where the MOSFET is half on, causing an enormous amount of heat to be dissipated in the MOSFET and killing it. It might be possible to address this with an inductor to create oscillation and buffer current. But if I did it again, I would take another approach, and use an astable multivibrator to sample the LED current at discrete intervals instead of doing it continuously. This would allow time for the gate to settle either high or low so it doesn't get stuck in the middle.
  - aidenn0 12y ago
    Just in case you don't already know this, some MOSFETs are much better acting in linear mode than others; they are much more expensive than your typical power MOSFET though.
nictinkers 12y ago
I worry that this circuit has been presented without the relevant warnings. There's no isolation here, so you're exposing people who may go and build this circuit to live mains. Think about the pot, for example. If somebody built this, and stuck a normal potentiometer on the front of a box to adjust the voltage, they may not think to use a pot that has a high insulation rating between the handle and the wiper. You could get a nasty crack of line voltage DC through the pot handle via your squishy internals and down to ground. The pictures are also a bit concerning - the big metal heatsinks are almost certainly live as transistor tabs are usually connected to one of the pins and you don't seem to have isolating washers and grommets. I wouldn't want to thread a screwdriver down between live heatsinks to get to the adjustment pot. This sort of regulator is okay if it and everything it powers is sealed away so nobody can touch it, or if it's powered via an isolating transformer. It's an unusual configuration and it'd be unusual for people to know what must be done to be safe when building this circuit. That unusualness along with the nasty consequences of mistakes means you've got a moral obligation to take care to warn potential builders of some to these issues when presenting this circuit.
- cnvogel 12y ago
  "presented without relevant warnings..." It's a completely un-annotated schematic. To build the prototype, someone has to select suitable electronics, put it together on perfboard, and last but not least strip a power-cable to have bare-ends to solder to the input of the bridge rectifier. You really think that this person needs an additional warning about the dangers of 120V mains?