LM78XX to adjust CURRENT - Page 1
Forum section: electronics / beginners

author: LinkZ
author_rank: Contributor
timestamp: April 16, 2015, 02:04:14 pm
content: Hi, I have a question about LM3812. I want to make current output adjustable (from 0 to 1000mA) and i've found this schematic:I want to replace R1 with a POT, but what about the watt dissipated from R1/pot? Does it matter? Will a standard 1/2 watt pot do the job?

author: Simon
author_rank: Unknown
timestamp: April 16, 2015, 02:05:50 pm
content: The voltage on the resistor will be 1.25V usually so for 1A your talking 1.25W at least.

author: TimFox
author_rank: Super Contributor
timestamp: April 16, 2015, 02:34:19 pm
content: For a7812, the output voltage (pin 2 to pin 3) is 12 V, so the power in R1 at 1 A is 12 W.You can reduce this with a lower-voltage unit, such as the 7805.You could use a high-power fixed resistor with a higher-value pot across it, moving pin 3 from the output terminal to the pot wiper.The uncertainty in that circuit includes the relatively high quiescent current of the 7805/7812, which flows through pin 3. Search for more modern devices with lower quiescent current.Also: very important. The 78xx series requires a local capacitor (> 0.33 uF) from pin 1 to pin 3 for stability. I use 0.47 uF/X7R, as close as possible to the device.

author: bookaboo
author_rank: Frequent Contributor
timestamp: April 16, 2015, 02:38:59 pm
content: Why not use an LM317 ?Its made for the jobhttp://www.ti.com/lit/ds/symlink/lm117.pdf

author: Zero999
author_rank: Super Contributor
timestamp: April 16, 2015, 02:39:06 pm
content: No, the voltage across the potentiometer will be 12V so that's 12W at 1A.The input voltage needs to be above 14V to 15V (depending on the temperature and current setting) for it to regulate the current properly.Use the LM317 which needs 4V of headroom to regulate the current properly.

author: mariush
author_rank: Super Contributor
timestamp: April 16, 2015, 03:06:49 pm
content: So obviously this is not the smart way to limit the current.Right from the start, you won't be able to output 12v with 12v-37v, depending on the output current the minimum input voltage would have to be at least about 13v.The difference between input and output voltage would be dissipated as heat, so if your devices would consume a lot of power like 1A for example, then you really can't give the voltage regulator input voltages way higher than a few volts above the minimum needed. For example, even with 18v at the input, with 1A of current the regulator would dissipate (18v-12v)x1a = 6 watts.Even with a heatsink on the regulator, you still want to keep the power dissipation below around 5 watts.Then, obviously you won't be able to use potentiometers because of the power dissipated in them. Even if you'd use a resistor instead of a potentiometer, 12w at 1A is a huge amount of power dissipating in a resistor, you'd need a resistor rated for 20-25w just to stop it from being hot enough to burn you.In addition, potentiometers usually have very poor characteristics, as in the value will change with the temperature of the potentiometer... and with so much power dissipated, the pot will be hot and the current limit won't work quite right.If your input voltage can vary that much, it makes more sense to use a switching regulator circuit to output 13-14v all the time no matter the input voltage, or lower the voltage if the output current goes above your limit... basically limit the current with the switching regulator. Then, use the dumb 7812 to output a stable 12v with minimal energy wasted in the linear regulator if your circuit needs to have smooth (low noise/ripple) power.Do some research into using an opamp coupled with a small value resistor (like 0.1 ohm) to measure the current.. have a look at application notes and various linear regulator datasheets, they'll have example circuits.

author: macboy
author_rank: Super Contributor
timestamp: April 16, 2015, 03:10:42 pm
content: Replace 7812 with LM317, then you have only 1.25 V across the resistor, instead of 12 V, and that will drop the power in the resistor by a factor of nearly 100.For 1000 mA you want the resistor to be 1.250 Ohms. For minimum current, you want the resistor to be "high". For example, with 1k, the output current will be about 1.25 mA. You can't get down to zero current with this design. I'd recommend using a 100 ohm pot since going higher will really limit your ability to dial in the higher currents. Then your minimum current will be about 12 mA. With this simple circuit you can't really stretch it much more than that.Consider the power dissipation at higher currents; it will be significant. Also consider that the high current will be at a low resistance, so only a small part of the pot will be passing current, meaning that the power dissipation will be concentrated into a small section of the pot. With that in mind, I would use no less than a 2 watt wirewound pot. Don't even try to use a basic carbon pot. It will burn up. Put the pot in series with a 2 watt 1.0 ohm to 1.2 ohm resistor. That resistor will provide the current limit and will dissipate most of the heat when delivering higher currents. A logarithmic (audio taper) pot may be nice if you can find one at 100 ohm. You will still want to consider lowering the maximum current to say 500 mA, which will be much easier to deal with. Then only change necessary is to change that 2 W 1.2 ohm series resistor to a 2.5 ohm one. With that change, the maximum power dissipated in the pot is cut in half - you might even get away with a basic carbon pot then (though maybe not).

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: April 16, 2015, 03:18:46 pm
content: This is not a LM3812. And not with that pinout. Pin 3 is out, pin 2 is tab/gnd. Iq flows out pin 2, not 3. Where is your load?

author: Rick Law
author_rank: Super Contributor
timestamp: April 16, 2015, 09:08:24 pm
content: LM7805, LM7812, LM7809... are fixed voltage output device. Typically, with these fixed voltage output device, you can limit what comes out by limiting what current goes in. So, if you feed 1A into an LM7812, you will get at most just under 1A out.LM317 is adjustable. With LM317, or a buck like LM2596, you can limit current by lowering the voltage output. Both LM317 and LM2596 controls output via a feedback or an adj pin.Bare in mind LM317 is just a linear regulator, so excess voltage (energy) are burned off as heat and it gets to be a problem when you want to use a high Vin - way too much heat to deal with. Buck converter is better in this regard since it converts that energy and most of it goes to the output instead of dump as waste heat - dissipation of waste heat is much much less an issue.The link below is a current limit implementation for LM2596 like buck. You can adapt this for LM317 using the same strategy. I started that circuit for LM317 and LM350, and decided to modify it for LM2596 so I know that circuit (with modification) will work with LM317/LM350 and similar regulator.https://www.eevblog.com/forum/beginners/adding-cc-to-a-cv-buck-to-make-it-cccv/If I remember the modification correctly, with V(fb) verse V(adj),they are in reverse. Forcing Vfb up on an LM2596 will force Vout down so it will in turn lower current. Whereas, forcing LM317 Vadj up will force Vout UP and thus more current. So with LM317 you do the opposite - you need to force Vadj down to lower Vout. So, one of the modification is to swap the OpAmp inverted and non-inverted input. Also, the strategy cannot work below 1.25V since. You can at max force Vadj to 0.00V, but Vout at the point would be 1.25V.Good luck.Rick