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finding a buck regulator

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Berserk87

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new here. :D

Im trying to find a buck regulator and having trouble finding one that fits me.

i need one that can handle 50-70VDC input, and has a 25-35V @ 2 Amp output.

i found a couple that were close or dead on 60V, but i want to be able to have a margin of too high/too low input.

as a bonus it would be nice to be a part that is in multisims component lists so that i could try it in multisim.

so far ive just been using google, found a couple close matches, but figured it get better results in a community of electronics people :)
 

tcmtech

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Are you looking to by one new or build one?
We need information in order to help out!
What is the power source and what are you powering?
Any cost or design efficiency requirements?
 

Berserk87

New Member
it would be nice if i could buy a IC and a couple components and put it together myself, if its fairly simple in design.

my source voltage is 60VDC.

efficiency isnt a concern, ill work on that after i get a working project.

i would rather not have to buy a separate board to connect to mine, but its not a huge issue right now.
 

Berserk87

New Member
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Berserk87

New Member
wattage is voltage x current.

and voltage is current x resistance.

but my circuit has (theoretically) no resistance right now.

60V >> buck regulator >> LM7805s
 

dknguyen

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The LM7805 will have a load connected to it since it is meant to power something.
 
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Berserk87

New Member
The LM7805 will have a load connected to it since it is meant to power something.
yes, the LM7805 will be powering two usb devices, so 4.8V-5V @ 500mA each.

4 devices, which means 2 LM7805's.

20V and 2 Amps, which would mean that my buck regulator would have to have a 40W output wouldnt it?
 

dknguyen

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yes, the LM7805 will be powering two usb devices, so 4.8V-5V @ 500mA each.

4 devices, which means 2 LM7805's.

20V and 2 Amps, which would mean that my buck regulator would have to have a 40W output wouldnt it?
Correct...but...

All linear regulators work by burning off excess voltage as heat- the more excess voltage the more energy has to be burned off as heat for a given output current. Input current is equal to output current for a linear regulator, and the excess voltage is burned off as heat to produce the lower voltage output.

Switching regulators will take the power in the input voltage and current, and redistribute the power between voltage and current at it's output (ie. you sacrifice current for more voltage or sacrifice voltage for more current). This means a switching regulator is much more efficient and will not heat up nearly as much since it only takes the energy it needs. Use your buck regulator to get a voltage MUCH MUCH closer to the final output voltage so the linear regulator has to burn off as little excess voltage as possible. With your current setup, each LM7805 will rise to 300C above ambient because it needs to burn off 15V of extra voltage in 500mA of current (7.5W of heat). Or just get rid of the LM7805s altogether and make your buck regulator just output 5V to be used directly.

Do note that the outputs of LM7805s vary from device to device. Do NOT connect their outputs together. THe regulator's different output voltages will fight each other effectively causing a short-circuit where the power source is the voltage difference between their outputs. Power each device off one LM7805 and only one.

You could connect components at the output of each regulator so you can tie their outputs together...but this introduces problems and the consistency of the output. Diodes make it so only one regulator is ever working (The one with the higher voltage output), and resistors will dissipate energy but will cause the regulators to not fight each other as much (since they must fight each other through the resistor which allow each regulator to have a different output voltage), but then output voltage will vary with current draw which is no good for regulation.
 
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Berserk87

New Member
Correct. Do note that the outputs of LM7805s vary from device to device. Do NOT connect their outputs together. THe regulator's different output voltages will fight each other effectively causing a short-circuit where the power source is the voltage difference between their outputs.

Power each device off one LM7805 and only one.

im aware of that :)

anything beyond the buck regulator in my circuit im testing in multisim.

national semiconductors i noticed has "webench" which suggests/simulates certain parts/ICs for you, i put in my requirements and it still suggested the LM5576 despite digikey saying it only has a 2.5W output :S
although the temp simulations showed it getting ridiculously hot. (without heatsink)
 
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mneary

New Member
i put in my requirements and it still suggested the LM5576 despite digikey saying it only has a 2.5W output :S
That's why we have data sheets. The nominal power dissipation of the chip is 2.5W, the output is 2A at fairly high voltages.
 

Berserk87

New Member
That's why we have data sheets. The nominal power dissipation of the chip is 2.5W, the output is 2A at fairly high voltages.
im still confused :confused:

while i consider myself somewhere in the intermediate level of electronics, i still dont full understand the differences between some of the things on these datasheets, and the data sheets are 100+ pages sometimes, so im not sure where to look :S
 
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dknguyen

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The datasheet says the IC can handle 150C maximum and the power has a thermal resistance of 40C/W from silicon junction to ambient without a heatsink (it heats up by this much for every watt dissipated as heat). 40C/W is a typical unheatsinked value for a power package on a regular PCB without any special heat pads.

-That means we have an extra 150C-25C = 125C of headroom that we can heat up the IC by before exceeding the maximum temperature.
-if the IC heats up by 40C/W, that means that we can dissipate 125C/40C/w = 3.125W before exceeding the maximum junction temperature AT ROOM TEMPERATURE. If the room temperature is higher, then we can't dissipate as much heat since we have less temperature headroom.

You pick a worst case ambient temperature and the maximum junction temperature you are willing to tolerate (usually with at least some margin below the maximum rated on the datasheet) and calculate how much power you can dissipate like I Just did. Then you look at the losses in the IC and calculate how much current it can actually pass.

Continuing with my room temperature example, the datasheet says the worst case switch on-resistance is 340mohms. I'm going to assume 100% duty cycle operation as worse case (may or may not be true or even possible depending on your application or even the type of switching converter you are working with). THere are also losses from switching and since I don't know them I will just assume they are equal to the resistive losses. So I'm going to use a switch resistance of 680mohms. I neglect all other power dissipated because it's usually very small relative to the power switch.

3.125W = I^2*Rdson = (680mohms)*I^2
I = 4.6A maximum that can be allowed to pass through the switch.

For a buck converter, the highest switch current is equal to the highest output current (since buck converters make a lower voltage from a higher voltage. Conservation of energy means the input current is lower than the output current, thus making output current the maximum in the circuit). For a boost converter it is the highest input current.
 
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Berserk87

New Member
so how many watts would i be dissipating?

add:
since the classes ive gone through have mostly been low voltage circuits or very basic, we havnt gone over a bunch of stuff i guess =\
 
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dknguyen

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Most Helpful Member
The datasheet says the IC can handle 150C maximum and the power has a thermal resistance of 40C/W from silicon junction to ambient without a heatsink (it heats up by this much for every watt dissipated as heat). 40C/W is a typical unheatsinked value for a power package on a regular PCB without any special heat pads.

-That means we have an extra 150C-25C = 125C of headroom that we can heat up the IC by before exceeding the maximum temperature.
-if the IC heats up by 40C/W, that means that we can dissipate 125C/40C/w = 3.125W before exceeding the maximum junction temperature AT ROOM TEMPERATURE. If the room temperature is higher, then we can't dissipate as much heat since we have less temperature headroom.

You pick a worst case ambient temperature and the maximum junction temperature you are willing to tolerate (usually with at least some margin below the maximum rated on the datasheet) and calculate how much power you can dissipate like I Just did. Then you look at the losses in the IC and calculate how much current it can actually pass.

Continuing with my room temperature example, the datasheet says the worst case switch on-resistance is 340mohms. I'm going to assume 100% duty cycle operation as worse case (may or may not be true or even possible depending on your application or even the type of switching converter you are working with). THere are also losses from switching and since I don't know them I will just assume they are equal to the resistive losses. So I'm going to use a switch resistance of 680mohms. I neglect all other power dissipated because it's usually very small relative to the power switch.

3.125W = I^2*Rdson = (680mohms)*I^2
I = 4.6A maximum that can be allowed to pass through the switch.

For a buck converter, the highest switch current is equal to the highest output current (since buck converters make a lower voltage from a higher voltage. Conservation of energy means the input current is lower than the output current, thus making output current the maximum in the circuit). For a boost converter it is the highest input current.
You don't know how to calculate the power dissipated by a resistance? Or are you just having trouble applying it to a real situtation.
 
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