I may be dense but I don't understand how to calculate the voltage and current drop when an adjustable 3 terminal regulator is used as a current regulator, e.g. single resistor on an LM317 or 338.
If it's there in the datasheet or easily calculated from other datasheet parameters please just say so.
If you want application background, it's not a difficult service, just a 1A or 2A current regulator on a car electrical system, so as low a voltage drop as possible would be preferred.
If it's there in the datasheet or easily calculated from other datasheet parameters please just say so.
If you want application background, it's not a difficult service, just a 1A or 2A current regulator on a car electrical system, so as low a voltage drop as possible would be preferred.
Here is a data sheet and to calculate the power dissipated in the device, use Kirchhoff's law.
http://www.ti.com/lit/ds/symlink/lm317.pdf
http://www.ti.com/lit/ds/symlink/lm317.pdf
For linear regulators, calculation of the voltage drop is easy. The "current drop" (current going straight to ground before the regulator output) is normally little compared to the output current, except for really low-power applications.
The series voltage drop is the difference between the input- and output-voltage. The series power loss is this voltage drop times the output current. If you then compare that to the output voltage and current (output power) you can calculate efficiency.
For efficiency and heating reasons, you want the voltage drop across the series regulator to be as little as possible. But, the voltage drop must not (at any moment) go below the minimum voltage needed to keep the regulator active (non-saturated), even temporarily when the input voltage sags due to loading or voltage ripple. If the voltage regulator saturates, you no longer have a regulated output.
On top of that, you have a maximum voltage the regulator can stand.
The series voltage drop is the difference between the input- and output-voltage. The series power loss is this voltage drop times the output current. If you then compare that to the output voltage and current (output power) you can calculate efficiency.
For efficiency and heating reasons, you want the voltage drop across the series regulator to be as little as possible. But, the voltage drop must not (at any moment) go below the minimum voltage needed to keep the regulator active (non-saturated), even temporarily when the input voltage sags due to loading or voltage ripple. If the voltage regulator saturates, you no longer have a regulated output.
On top of that, you have a maximum voltage the regulator can stand.
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If you want low dropout, I don't think the LM317 is the right device. The regulator itself needs 3V input to output, and the reference voltage = voltage across the sensing resistor is 1.25V. Total dropout = 4.25V. The LT3080 http://www.analog.com/media/en/technical-documentation/data-sheets/3080fc.pdf would allow you to halve that, or make one yourself and you could get easily below 1V
Brian
Brian
Absolutely right, as long as you ensure that the input voltage at no moment in time sags to a level where the (lower) series-drop of the low-dropout regulator is not respected. The voltage ripple on the buffer capacitors (just before the regulator) may easily be several volts (with high loading) such that the difference in using an LM317 or a low-drop regulator becomes less pertinent.
A current regulator will drop whatever voltage is needed to maintain the required current in the load, if it is able to do so. Hence a current regulator does not 'have' an efficiency; the whole circuit has an efficiency.
Low voltage drop in a current regulator probably means that the current regulator is not doing very much and so is probably unnecessary.
Do you actually want a current regulator, or a current limiter? These are not the same thing.
Low voltage drop in a current regulator probably means that the current regulator is not doing very much and so is probably unnecessary.
Do you actually want a current regulator, or a current limiter? These are not the same thing.
Indeed, what is really needed I guess is a low dropout current limiter. Got an example?
I have at the back of my mind the commercial battery isolators used in dual battery auto systems, which range in capacity from 100mA to 50A, and AFAIK are made of single current limiting diodes, but I do not know if the components are readily available, nor are the best choice necessarily for DIY.
I have at the back of my mind the commercial battery isolators used in dual battery auto systems, which range in capacity from 100mA to 50A, and AFAIK are made of single current limiting diodes, but I do not know if the components are readily available, nor are the best choice necessarily for DIY.
No. They are two plain one-way diodes. Goes from the one alternator to two batteries, engine and 'house'. You can drain the house battery flat and the engine battery is not depleted at all. The numbers are the maximum expected current. (Any current-limiting, if any, is in the alternator system.)
Well, the only way the current "can" get limited is to reduce the voltage to the load enough to maintain the maximum set current. The difference between input and output voltage will (and can) only be whatever it needs to be, to cap the current load.
'Fraid you can't really get away from Ohm's law
... Unless there might be some really niche / specific current-limiting buck converters that can have (near) zero dropout or something...
'Fraid you can't really get away from Ohm's law
... Unless there might be some really niche / specific current-limiting buck converters that can have (near) zero dropout or something...
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TI and others make e-fuse / load switch parts with settable current limits. This is what you want I think.
Yeah I forgot about that, but there are a few parts that let you control the action taken when the current limit is reached.
http://www.ti.com/lit/ds/symlink/tps2660.pdf
This part may work, I think it can limit the current without shutdown until the thermal limit is reached.
http://www.ti.com/lit/ds/symlink/tps2660.pdf
This part may work, I think it can limit the current without shutdown until the thermal limit is reached.
Wow, you guys are really invested in this one. Is the crossword all done? Are the other threads boring?
It's really an unimportant application, nobody's losing sleep over it.
I want to be able to plug in 12VDC devices on my utility trailer behind my car. Things like USB chargers, trouble lights, travel coolers, air compressors. Some are over 2A, but there's always trial and error to see what I can make work. I just have the 4 wire trailer wiring, so I have to use the running light feed. Tapping into that circuit is only good for a few amps, so that's what I want this circuit for.
Please don't start with the oh no's. No I won't be upgrading to a 7 wire connection with aux power. Yes, it is safe, it is already fused now, I only use waterproof components, and I want something that will protect over and above the fuse.
I want something common and low parts count.
It's really an unimportant application, nobody's losing sleep over it.
I want to be able to plug in 12VDC devices on my utility trailer behind my car. Things like USB chargers, trouble lights, travel coolers, air compressors. Some are over 2A, but there's always trial and error to see what I can make work. I just have the 4 wire trailer wiring, so I have to use the running light feed. Tapping into that circuit is only good for a few amps, so that's what I want this circuit for.
Please don't start with the oh no's. No I won't be upgrading to a 7 wire connection with aux power. Yes, it is safe, it is already fused now, I only use waterproof components, and I want something that will protect over and above the fuse.
I want something common and low parts count.
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