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Shunt Resistors

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alm:
I don't mind not knowing, but I get somewhat annoyed that I point to the clearly labeled solution somewhere within two ~20 page datasheets, and you seem unable to find it and wait for someone to point it out to you.

A shunt resistor is just a (usually) low-value resistor to sense the current. The voltage over the shunt is I*R (Ohm's law). Since the dissipation is I^2*R, the R is usually quite low for high currents, and it's often a power resistor.

The way an LM317 works is that the reference voltage between the output and adjust terminal is held at about 1.25V, this voltage is specified as 'reference voltage' in the electrical characteristics table (I'll use the OnSemi and National datasheets, the datasheet for the actual device you use might be different, but will be similar). This how the adjustable voltage regulator works: the adjust terminal is connected to a voltage divider between output and ground, the voltage over the top resistor is 1.25V, so the voltage between the output terminal and the ground is 1.25V/(R1/(R1+R2)) = 1.25 * (1+R1/R2). You'll find this formula (plus an extra term for the voltage caused by the current from the adjust terminal, but this is usually quite low) in the application hints/information section of the datasheet. This is where the 1.25V comes from. Not sure where you read that the input voltage should be 1.25V, the only thing I found is that the output voltage is a function of 1.25V and the resistive divider.

The solution that I had in mind is in figure 23 of the OnSemi LM317 datasheet (I wasn't sure initially if it was in the OnSemi or National one, so I mentioned both). Figure 26 is the very basic example: A current shunt R1, as soon as more than 1.25V is dropped over R1, the LM317 will decrease the output current. Figure 23 extends this idea. The JFET (2N5640) is configured as a current sink, it sinks a (fairly) constant amount of current (IDSS, stated in the datasheet, at least 5mA for the 2N5640). Any JFET with a similar IDSS would work, as would any other constant current sink (eg. with a bipolar transistor). This causes a constant current through R2 (ignoring Iadj), and develops another voltage (IDSS * R2) at the wiper of R2. The total voltage that the adjust terminal sees is R1 * (Iout+IDSS) + R2 * (IDSS). You set R1 so it can output the required max. current (eg. 1A), 1.25 ohm would allow 1A - IDSS (almost 1A) with the R2 wiper all the way to the top. The value of R2 is chosen so that even with the wiper all the way down, there will be no more than 1.25V over it (no point in less than 0A), but it should be fairly close if you want it to be able to adjust it all the way to 0A (the provided value will probably be variable between .5A and 1A or so). The diodes are there to limit the voltage across the pot to ~1V (shouldn't happen unless the current through the FET gets above ~10mA).

The reason that the FET is tied to a negative voltage is that because it should even be able to limit the current if the output voltage is close to ground. The reference pin must be 1.25V below the output voltage, so if the output voltage is 0V, the adjust pin should be at -1.25V. There will be some voltage drop across the FET, so a little margin can't hurt.

Figure 22 has this combined with an LM317 in voltage regulator configuration, with a similar setup to bring the adjust terminal below ground so the output voltage can be adjusted below 1.25V. It has a FET with a lower IDSS (2mA min, 10mA max), and a 1k pot instead of 100 ohm (more range). This is basically exactly what you need. If you have trouble understanding the circuit, it might be a good idea to build it on a breadboard and do some tests, or even simulate it in LTspice.

fsleeman:

--- Quote from: alm on May 17, 2010, 05:59:19 pm ---I don't mind not knowing, but I get somewhat annoyed that I point to the clearly labeled solution somewhere within two ~20 page datasheets, and you seem unable to find it and wait for someone to point it out to you.

--- End quote ---

I understand how you feel, but this is a very common problem. While it is trivial for seasoned vets to find this kind of information, it is often a problem for those new to engineering design. There is a ton of stuff on the internet that can help but a lot of it is either not labeled, mislabeled, misleading, wrong, or so poorly written that nobody can understand it. It does take some experience to sort out the crap. From my experience and understanding universities (in the US at least) do not mention how to find practical information about circuit designs. Most of what I have learned about data sheets and the like has been on my own, but learning how to learn on your own is very valuable on its own.