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I need to drop 12V to 5V @80mA efficiently (i.e. least amount of energy converted to heat). Using any kind of switching DC-DC converter is not an option because it will introduce too much noise into the system. (This is for an analog synth) Should I just use a standard 5V regulator (i.e LM2805 in a TO-92 package?) Should I drop the voltage down first with some series resistors or diodes? | |
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You could just use a voltage divider with resistors, select a resistor with the appropriate wattage to handle your power needs, this would dissapate power as heat, but so does a linear regulator.
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Ahh, yes... back to my days as a power supply designer when a customer requests a design and one of the design conditions is that you can not use the only possible circuit that meets the requirement. A switcher of some type is the only way to drop a voltage down with decent efficiency. If you use diodes, resistors, transistors or any other passive/active linear device to lower the voltage, the total power lost is the same which is: (12 - 5) X IL which is 7V x 0.8A = 5.6W in your design. Any linear circuit will burn this amount of power. You can shift the power between diodes, resistors, transistors but the total power burned is constant. Last edited by bountyhunter; 2nd November 2009 at 11:31 PM. | ||
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except 0.8 = 800mA not 80mA :P So 7V*0.08A = 0.56W waaay smaller! Do I still need TO-220 package with a big heatsink? | |
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| | #6 |
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At TO-220 in free air (no heatsink) will be fine to dissipate 0.56W. A TO-92 package would get rather hot.
__________________ Carl Curmudgeon Elektroniker | |
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__________________ Carl Curmudgeon Elektroniker | ||
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| | #9 | |
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A TO-220 will work well with no heatsink, it's thermal is about 70C/W with no heatsink so t will only rise about 40C above ambient. Last edited by bountyhunter; 3rd November 2009 at 12:04 AM. | ||
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| | #10 | |
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http://www.linear.com/pc/productDeta...39,C1133,P1613 Last edited by bountyhunter; 3rd November 2009 at 12:16 AM. | ||
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| | #11 | |
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No, the devices I referred to are not the switched-cap converters, they are switching regulators with inductive filtering. Inductors have nothing to do with the noise in a switching regulator. The noise comes from the switched currents which are rapidly being switched between the transistor and the flyback diode. The inductors just serve to filter and smooth the switching currents. The Linear Tech devices get their low noise primarily by controlling the transistor current rise and fall times to minimze the harmonic noise.
__________________ Carl Curmudgeon Elektroniker | ||
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| | #12 | |
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Say for instance he was trying to get 6 volts from 12 and set up a voltage divider of two 1 ohm, high wattage resistors...high enough to handle the power they are subjected to. Obviously your circuit load will tap into the 6 volts created by the divider, probably from zero volts to the 6 volt point. Well, the load will be parallel to that resistor, thus the overall resistance at that point will decrease, thus dropping more of the voltage across the other resistor in the divider....by how much depends on the impedance of the load. For this to be workable you would have to have a stable, known load impedance and calculate the value of the divider resistors accordingly. I tried this a long time ago with a circuit I was playing with and that's when I truly learned about series/parallel circuits. I'm not saying it can't work, but generally it will only work in very specific applications and the voltage divider will either waste a lot of power, impede so much current as to not properly feed the circuit, or be too unpredictable a voltage source as I described. | ||
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Basic buck regulator operation is covered in the switcher section of a book I wrote for customer training some years back. Check page 34 of attached document. That's a bogus approach IMHO because it increases switching losses. We used it when we absolutely had to, but it wastes power. BTW: it's not "harmonic" noise that's the real problem, it's the very high frequency ringing at the switch turn off that generates the most obnoxious EMI. The fundamental frequency of that ring is not a harmonic of the switch frequency, it's dependent on parasitic R-L-C values in the assocuiated switch components. Last edited by bountyhunter; 3rd November 2009 at 06:24 AM. | |||||
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| | #14 |
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i all..i have problem in designing cmos temperature sensor.can anyone give some ideaa or link references..
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| | #15 |
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Bloodiemarie83: Appending your question to someone else's thread is a serious breach of good manners, and disruptive to the discission.
__________________ You don't need a quadraphonic Blaupunkt -- you need a curve ball. Last edited by BrownOut; 3rd November 2009 at 02:58 PM. | |
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| 12v, drop, efficiently |
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