Tweet
OK, guys. I read through this quickly, so if I missed some key fact, correct me.
As Leo said, the 125V / 2A is the switch rating. The pot is rated for 0.25W across the whole thing, as noted by JM. What isn't clear to many people is that (also as noted) that's for the whole pot length. It is quite possible to burn out part of a pot. If the pot is 1/2W rated for just the pot element, it can support a voltage across it of V = SQRT(P*R) = SQRT(0.25*1E6) = 500V. At that voltage (RMS or DC) across the whole pot, it's letting through a current of I = 500/1E6 = 500uA. The way to read that is that **any part of the pot**, from one end through the wiper cannot carry more than 500uA without overheating that section of the element.
That's not a serious problem with this one, but just remember the next time you try to make a pot carry lots of current. The wiper contact often has its own current rating too, and that may be harder to find or compute.
On to the circuit. That's not a VVR. It's a source follower fed a constant(ish) input voltage by the pot wiper. The MOSFET has a large transconductance and a near-infinite current gain, so whatever voltage is on the gate is fed out the source, minus a tiny sag of the source, just enough to make that MOSFET let the current through. Power MOSFETS have transconductances on the order of units of amps per volt, so if you're feeding a half amp of current through the thing, it sags the threshold voltage ( maybe 4V) to get any current to flow, then about 0.5V more to get the current to flow. So the source follows the gate voltage, but a few volts lower.
The MOSFET can get hot. As noted I think, the power it must dissipate is the voltage across it times the current through it. To figure your heatsink, estimate your maximum load current at the most voltage across the MOSFET and multiply. That's the watts to get rid of. The inside of your amp gets hot, and all the heat from the parts must go out through the air in there. Make a guess at maybe 40C air temps in there. MOSFET junctions start to die if they are hotter than 150C, so you can only let the MOSFET die get 110C hotter than the air inside the amp.
The number of 110C/P is the total thermal resistance you need to keep. There is a thermal resistance from the semiconductor die to the heat sink tab. That's often about 2C/W for TO-220s. So if you could somehow keep the heatsink tab down to 40C, the most power you could dissipate is P = 100C/2C/W = 55W. This is physically impossible since there is an additional resistance to heat flow from the tab to the air, and that's what your heat sink does. Once you know your load current and the voltage range across the MOSFET, figuring out what heat sink to use is straightforward, if tedious.
Most TO-220s have a real-world dissipation limit of about 20-30W on a biggish heat sink.
Also as noted, the heat sink tab is connected to the drain. It's running at the full input voltage. You must insulate it from the chassis. You can insulate it from the heat sink and tie the heat sink to the chassis, or let the drain short to the heat sink and insulate the heat sink from the chassis. It's more common to ground the sink and insulate the power devices from the sink with one of the several insulators and other parts that are available.
You MUST use either a thermally conductive rubber pad or heat sink goo to let the heat get from the device to the sink and not be obstructed by the openings filled with air between the device and the sink surface.
As Leo said, the 125V / 2A is the switch rating. The pot is rated for 0.25W across the whole thing, as noted by JM. What isn't clear to many people is that (also as noted) that's for the whole pot length. It is quite possible to burn out part of a pot. If the pot is 1/2W rated for just the pot element, it can support a voltage across it of V = SQRT(P*R) = SQRT(0.25*1E6) = 500V. At that voltage (RMS or DC) across the whole pot, it's letting through a current of I = 500/1E6 = 500uA. The way to read that is that **any part of the pot**, from one end through the wiper cannot carry more than 500uA without overheating that section of the element.
That's not a serious problem with this one, but just remember the next time you try to make a pot carry lots of current. The wiper contact often has its own current rating too, and that may be harder to find or compute.
On to the circuit. That's not a VVR. It's a source follower fed a constant(ish) input voltage by the pot wiper. The MOSFET has a large transconductance and a near-infinite current gain, so whatever voltage is on the gate is fed out the source, minus a tiny sag of the source, just enough to make that MOSFET let the current through. Power MOSFETS have transconductances on the order of units of amps per volt, so if you're feeding a half amp of current through the thing, it sags the threshold voltage ( maybe 4V) to get any current to flow, then about 0.5V more to get the current to flow. So the source follows the gate voltage, but a few volts lower.
The MOSFET can get hot. As noted I think, the power it must dissipate is the voltage across it times the current through it. To figure your heatsink, estimate your maximum load current at the most voltage across the MOSFET and multiply. That's the watts to get rid of. The inside of your amp gets hot, and all the heat from the parts must go out through the air in there. Make a guess at maybe 40C air temps in there. MOSFET junctions start to die if they are hotter than 150C, so you can only let the MOSFET die get 110C hotter than the air inside the amp.
The number of 110C/P is the total thermal resistance you need to keep. There is a thermal resistance from the semiconductor die to the heat sink tab. That's often about 2C/W for TO-220s. So if you could somehow keep the heatsink tab down to 40C, the most power you could dissipate is P = 100C/2C/W = 55W. This is physically impossible since there is an additional resistance to heat flow from the tab to the air, and that's what your heat sink does. Once you know your load current and the voltage range across the MOSFET, figuring out what heat sink to use is straightforward, if tedious.
Most TO-220s have a real-world dissipation limit of about 20-30W on a biggish heat sink.
Also as noted, the heat sink tab is connected to the drain. It's running at the full input voltage. You must insulate it from the chassis. You can insulate it from the heat sink and tie the heat sink to the chassis, or let the drain short to the heat sink and insulate the heat sink from the chassis. It's more common to ground the sink and insulate the power devices from the sink with one of the several insulators and other parts that are available.
You MUST use either a thermally conductive rubber pad or heat sink goo to let the heat get from the device to the sink and not be obstructed by the openings filled with air between the device and the sink surface.
Comment