How Do Linear Drivers Work?

I understand that you can boost voltage to meet the requiremnts, and buck the voltage so the diode isn't over-fed. But how does a linear driver work. How does the driver make sure that the diode isn't getting to much or too little voltage?

A linear driver only bucks ... it always has less Vout than Vin since it uses some power to run the internal comparator and FETs it uses to regulate.

It works by always adjusting the output voltage to keep the voltage drop between the Vout pin and Adj pin at 1.25V(on LM317 type linear regulators), where the set resistor(s) resides. If you tap your load before the set resistor(the Vout pin) you will get a constant voltage with as much current as your load wants up to what the device is rated for. If you tap your load after the set resistor (Adjust pin) you will get a set current with as much voltage needed by your load, up to your power source voltage minus any overhead used by the regulator.


Linear and Switching Voltage Regulator Fundamentals

jib77 said:

It works by always adjusting the output current to keep the voltage drop between the Vout pin and Adj pin at 1.25V(on most linear regulators), where the set resistor(s) resides. .....

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Sorry for the precisation ..... this is true only for the LM317 series, that have a Vref of 1,25V ..... and, basically, they adjust the voltage, not the current directly (the current is a derivation of a certain voltage on a certain resistence load)

More specifically, the linear driver adjust the output voltage in the way that the current through the load always produce the same voltage drop on the reference resistor ..... when the load resistance increase, it increase the voltage, and when the load resistance decrease, it decrease the voltage, so it keep the flowing current in the load / sense resistor assembly always the same .....

For the LM317 serie and similar regulators, it happens keeping the drop voltage at the sides of the sense resistor (that is electrically in serie with the load) stabilized on 1,25V, cause this is the internal reference voltage of this regulators serie ..... some others have different voltages, but the principle is always the same .....

For the "current sink" and "current source" regulators types, the principle is similar, the only difference is that you use a mosfet or power transistor for regulate the voltage, and an op-amp for drive it, driven from the voltage collected from the sense resistor ..... the advantage of this system is that the current regulation is more precise, and the dropout of the assembly is lower than the one you have with the LM serie (LM317 schematic needs at least 3,75V, better 4V, more than the voltage needed from the load, so if you use a 5V FV LD, you need at teast 9V for the assembly works correctly ..... with a good mosfet/op-amp sink circuit, you may end needing 6 or 6,5V for get the same results) ..... disadvantage is that the circuit is a bit more complex .....

Also, with LM serie circuits, the trimmer, if you want to make a variable current driver, need to hold the same current, or a discrete part of it, of the load, small trimmers just smokes away for high currents ..... with a sink or source current circuit, you can just use a 1/4W miniature trimmer without problems, is only the fixed sense resistor that hold all the current, and cause it can also be 0,1 ohm, it also need to dissipate less power (example for 1A, with a LM317, you need 1,25 ohm that need to dissipate at least 1,25W, with a sink circuit with a 0,1 ohm Rsense, the resistor dissipate 100mW) .....

^ Very good explanation for the linear driver. +1 Your schematics are always high quality as well.

Sorry for the precisation .....

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No problem ... corrected my brain fart. :beer:

is there such a device as a DC/DC down converter? I.E. think boost driver but backwards. What I'm thinking is that if you supplied it with 8.4V from a supply that is only capable of 650mA, would you be able to get about 5V @ 1A out of the converter?

yes, that is a buck converter (mentioned above)

Oh ok, so it bucks, that makes sense The technicalities are a little confusing (Haven't went through a physics course yet as the branch of the CC I go to doesn't offer it.) but I get the point, thanks!

There is a big difference between a switchmode buck converter and a linear driver.

Linear drivers dissipate the voltage difference between input and output, whereas buck converters attept to switch that difference using (typically) an inductor as a temporary storage.

Linear drivers offer advantages in speed for modulation, but are only attractive if the difference between supply voltage and laser diode voltage is not very large. If you are powering a red diode off two lithium cells in series, a buck converterting switchmode driver would be prefereable, but if you are driving a 445 diode from 2 liths a linear driver would probably be preferable.

^ Right, this must be remembered, too (I'm so used to considerate it so normal, that i forgot to say it before) ..... ALL the power that is not used, is dissipated in heat, from a linear driver, and ofcourse, this also depend from the current ..... like, if you have an IR diode working at 2,8V 2A, and you use a linear driver for power it with a 12V PSU, your driver ends dissipating in heat (12-2.8)*2=18.4W (ouch) ..... so, less is the in-out voltage difference, better it is (always considering the internal dropout, ofcourse) .....

HIMNL9 said:

^ Right, this must be remembered, too (I'm so used to considerate it so normal, that i forgot to say it before) ..... ALL the power that is not used, is dissipated in heat, from a linear driver, and ofcourse, this also depend from the current ..... like, if you have an IR diode working at 2,8V 2A, and you use a linear driver for power it with a 12V PSU, your driver ends dissipating in heat (12-2.8)*2=18.4W (ouch) ..... so, less is the in-out voltage difference, better it is (always considering the internal dropout, ofcourse) .....

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Based on this, the overall efficiency could be derived by: 5.6/(18.4 + 5.6) x100 = 23%. An good example of where a switch-mode buck regulator could vastly improve the efficiency. Some of these approach 95%.

I still don't understand what modulation is, TTL and such. (Those two are related, aren't they?)

Ha, but I go the idea good enough for the time being. Thanks Guys

TTL and modulation are essentially the same thing: a method to turn the laser on and off rapidly. Modulation can be analog too, so its possible to operate the laser anywhere between off and full power, like a dimmer on a lightbulb. Blanking is also the same thing, though usually it refers to TTL (on/off) modulation.

Kage said:

Based on this, the overall efficiency could be derived by: 5.6/(18.4 + 5.6) x100 = 23%. An good example of where a switch-mode buck regulator could vastly improve the efficiency. Some of these approach 95%.

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This depend from the supply voltage ..... for the same example as above, if you use 5V PSU, instead 12V, the same linear driver at the same current need to dissipate (5-2.8)*2=4.4W ..... is the UNUSED part that need to be dissipated, and is for this reason that, with linear current drivers, is always a good thing to use a supply voltage nearest possible to the minimum required one

Buck (step-down) switching drivers works in a totally different way ..... the power dissipated from the final transistor (or mosfet), is only the one due to its internal RdsON, and only for the "ON" time of the cycle (suppose you have a buck driver with a final mosfet with 100 milliohm of RdsON, working at 50% duty cycle for 2A output, the transistor dissipates only 200mW) ..... for this reason they are much more efficents, and preferred when possible ..... only, at hobby level, they are less easy to build and less cheap than linear drivers, so linears still are used in a lot of cases .....