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efficiently drop 12V to 5v?

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danjel

Member
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?
 

MOSFET KILLER

New Member
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.
 

bountyhunter

Well-Known Member
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.


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.
 
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bountyhunter

Well-Known Member
Should I just use a standard 5V regulator (i.e LM7805 in a TO-92 package?)
It will melt with 5.6W of power. You would need a 7805 in a TO-220 package attached to a heatsink whose thermal resistace is less than about 12C/W which is reasonably sized.
 

danjel

Member
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?
 

crutschow

Well-Known Member
Most Helpful Member
At TO-220 in free air (no heatsink) will be fine to dissipate 0.56W. A TO-92 package would get rather hot.
 

danjel

Member
At TO-220 in free air (no heatsink) will be fine to dissipate 0.56W. A TO-92 package would get rather hot.

yeah I was reading somewhere else that a TO-220 in free air should be fine for anything less than 1W (without a heatsink).
 

crutschow

Well-Known Member
Most Helpful Member
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)
If you really need the efficiency you could probably use a low noise switcher design such as from Linear Technology Linear Technology - Ultralow Noise Regulators. It requires a carefull layout and good filtering but you can get a switching regualtor with 10's of microvolts noise, which is good enough for most low noise applications.
 

bountyhunter

Well-Known Member
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?
Spank those naughty eyes. A TO-92 will still get too hot, the thermal on those is about 180C/W so it will rise about 100C above ambient temp.

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.
 
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bountyhunter

Well-Known Member
If you really need the efficiency you could probably use a low noise switcher design such as from Linear Technology Linear Technology - Ultralow Noise Regulators. It requires a carefull layout and good filtering but you can get a switching regualtor with 10's of microvolts noise, which is good enough for most low noise applications.
I can't get that link to open, but if those are their switched capacitor buck converters then they can have very low noise. They are very good for low power conversion with low noise because they have no inductors.

https://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1039,C1133,P1613
 
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crutschow

Well-Known Member
Most Helpful Member
I can't get that link to open, but if those are their switched capacitor buck converters then they can have very low noise. They are very good for low power conversion with low noise because they have no inductors.
Don't know why the link won't open for you. Try www linear.com/pc/viewCategory.jsp?navId=H0,C1,C1003,C1042,C1034

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.
 

ke5frf

New Member
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.

This isn't always a workable solution.

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.
 

bountyhunter

Well-Known Member
Don't know why the link won't open for you. Try www linear.com/pc/viewCategory.jsp?navId=H0,C1,C1003,C1042,C1034

No, the devices I referred to are not the switched-cap converters, they are switching regulators with inductive filtering. .
Then they are going to put out EMI

Inductors have nothing to do with the noise in a switching regulator.
Nope, you are wrong about that. Inductors ring when the switch gets turned off abruptly. Switching transformers also have leakage inductance and self resonance that tends to ring in the 50 - 100 MHz range. You wouldn't believe how many years of my life I spent fighting the noise that comes from those.


The noise comes from the switched currents which are rapidly being switched between the transistor and the flyback diode.
Actually, the problem is that the catch diode is not ideal which means it has a finite turn on and turn off time (off is worse) and the inductor rings while the diode gets it's act in gear.

The inductors just serve to filter and smooth the switching currents.
No, not true at all: in nearly every buck (step down) switcher, the inductor is the energy storage element in the converter: it releases stored energy during the transistor's off time. You may add L-C filters after that output, but somewhere in the heart of the converter is the inductor that stores and dumps energy.

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.

The Linear Tech devices get their low noise primarily by controlling the transistor current rise and fall times to minimze the harmonic noise.
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.
 

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BrownOut

Banned
Bloodiemarie83: Appending your question to someone else's thread is a serious breach of good manners, and disruptive to the discission.
 
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