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The datasheet has a pretty thorough description of the use of the ADJ pin with $R_1$ and $R_2$: Since both $R_1$ and $R_2$ appear in the equation for the output voltage $$V_{\text{out}} = 1.25\text{ V} \times \left(1 + \frac{R_2}{R_1}\right) + I_{\text{ADJ}}R_2$$ you need both in order to realize an arbitrary output voltage. Depending on your ...

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Overview I'll avoid depending upon algebra as an explanation. (Because algebra, while providing quantitative answers, doesn't often help people understand something unless they are very fluent with mathematics.) Regardless, it's still helpful to have the datasheet available. So here is TI's LM317 datasheet just to make it convenient when needed. The best way ...

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See LM317 based power supply with current limiting on SE and alexan_e's answer. Read through all the answers and comments as there were some good suggestions. The circuit avoids high-current pots by using some trickery. Source: ON-Semi datasheet. Other circuits from the web: Bench power supply 0 - 25 V, 0 - 5 A. This one is medium complexity and looks ...

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Yes, that is clever, I think it would 'work', however the problem is that Iadj is 50-100uA so that you will not be able to get an accurate load current. First, it is large, so your 200uA current might actually be 300uA. Also, the temperature coefficient is fairly large: And it varies with input-output voltage. If you're looking to put a constant 200uA ...

11

Read the datasheet of the LM317, on page 9 it states: So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V. Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317. The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 ...

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This page lets you enter the desired voltage and then calculates optimal values for the resistors. You can choose the resistors' series. For instance, if you select the E12 series, it will find the best match with resistors from the E12 series. Obviously E96 series will give better matches. Some examples, with E12 (10%) resistors: 3.3V: 1k2 + 1k8 $\... 10 Yes, it is possible. Do not connect the outputs directly together but rather put a small resistor (0.5 Ohm or so) between each output and the summing node. Check: http://www.ti.com/lit/ds/symlink/lm317.pdf 8.3.11 Adjustable 4-A Regulator Circuit on page 15/16. You can also look at 9.3.13 which generates higher current with a single LM317, using instead ... 10 The 317's valid input voltage range is from a minimum of 3 V above its programmed output, to a maximum of 40 V (32 V for some types). Within that range, it will provide approximately the programmed voltage on its output, as long as ... its power dissipation is low enough for the heatsink you're using with it to keep its temperature within specification. the ... 9 Neither of your circuits is correct. The correct circuit looks like this: simulate this circuit – Schematic created using CircuitLab If you connect it incorrectly (as in the first circuit) it may act as a zener, resistor, zener + resistor, etc. and give you some voltage drop or whatever. I doubt you tested the I-V characteristics thoroughly over a ... 9 To drive 50 mA through 330 ohms, you need 16.5 V. For 50 mA through 1330 ohms you need 66.5 V. In the constant current mode, the maximum output voltage of the LM317 circuit is$V_{in} - 3\ {\rm V} - 1.25\ {\rm V}$. With 9 V input, that means you won't get more than about 4.75 V output. Your circuit should work pretty well for load resistance below ... 8 To get the LM317 down to zero volts you need to bring the control pin down to -1.25 V. Figure 1. This dual LM317 circuit consists of a current limiter based around LM317(1) and a voltage regulator based around LM317(2). The voltage regulator section is relevant to this post as it is adjustable down to zero volts. Source: ON-Semi datasheet. The control pin ... 7 When in doubt, look at the datasheet. Specifically, take a look at the Application Hints section. TI provides this general schematic: note: even though it's labeled as LM117, the LM317A is pin-compatible and functions similarly from an electrical analysis standpoint that we may as well treat the two as equivalent devices. There will be differences when it ... 7 Note A says the minimum load current for this circuit is 30 mA. If the circuit will not always have a load of 30 mA or more, you need a resistor there to draw 30 mA and satisfy the minimum load requirement. Without the 30 mA minimum load, the the circuit will not regulate correctly - the output voltage will probably rise. The LM317 by itself (without the ... 7 Although I've not specifically attempted a$15\:\text{A}$boosted LM317 before, this is along the lines of what I'd try out first. This is roughly taken from the Figure 23 you mentioned: simulate this circuit – Schematic created using CircuitLab In this case, I went for the D44/D45 series devices. (The PNP version has simply HORRIBLE Early Effect, ... 7 Since you say this is for a comnercial product, I'll be up front about this: Given the choice between buying your product with an LM317 regulator or a product from your competition using the LTC3652, I'd rather buy from your competition. You are saving a few cents, and delivering an inferior product. The LM317 is in all ways inferior to the LTC3652. The ... 6 There is a schematic in the ON-Semi datasheet It needs a negative voltage that feeds the two depletion mode JFETs, they operate as constant current sources and in conjunction with the two diodes connected to the output provide about -1.4V to the pot. 6 A somewhat nasty solution is to place two silicon diodes of suitable current rating in series with the LM317 output (after the voltage-setting resistor divider). This would reduce the output voltage by about 1.4 volts. This will degrade the voltage regulation slightly, as the diode voltage drop will vary with load current. A better solution is to provide ... 6 When you short the output, Q1 heavily conducts and basically connects pin 1 (ADJ) directly to 0V. Between Vout and ADJ internally is a 6V zener diode and a 50 ohm resistor: - It's very likely that the zener diode will fail short circuit (most of them do on over-current) rendering the device dead. If it can be tolerated a 1k resistor in series with the ADJ ... 6 You don't need U1. Both U1 and U2 have the same nominal current limit. Just readjust R1. You don't need D1, with its variable voltage drop, but in its place, you would need a diode from U2 out to U2 in, to power the input in the event that you connect the battery without input power. Once you have removed D1, set U2 to deliver 13.8v. This voltage will ... 6 According to MAX485 datasheet, for DIP/SOP packages the ground goes to Pin5, and +5Vcc goes to Pin8. You seem to have the chip powered upside-down. 6 Figure 1. As suggested by the datasheet. The LM317 works by adjusting its output to be 1.25 V above the voltage at the ADJ pin. With R1 = 240 Ω there is a current of$ \frac {1.25}{240} = 5.2 \ \text {mA} $running through it and R2. The constant current through R2 means that the voltage drop across it changes linearly with the resistance of R2. ... 5 It's a linear regulator. It can only decrease voltage, cap it, as you've said. It is working appropriately as you've described it. If you want to be able to go up or down from the input voltage you need a buck/boost topology. 5 A 24VAC transformer will give you as much as 40VDC with a light load. That's pretty marginal for an LM317 with zero output. Way too close for comfort, in my opinion. Anyway, if you were to use a third LM317 to regulate the input voltage down to +5VDC, you could use that 5V to power a 7660 charge pump to generate about -4.5V for your JFET current sink. ... 5 No, this won't work to give you +/- supplies, because of the common negative rail. You can, however, use the transformer you have to give you +/- adjustable supplies by using two half-wave rectifiers (2 diodes), double up on the capacitor values to keep the ripple at ~2Vp-p, and an LM317 for the positive and an LM337 for the negative supplies. With ... 5 Here is a quite short list of TI's Buck/Boost converters which have PDIP packaging. There are only 2 of 33 DC-DC converters which are available with though-hole package , and the only difference of the listed items is the Operating Temperature Range (C°). So as you can see and as @Dave Tweed said, through-hole design is not popular. Let's have a look on the ... 5 I would take a close look at the National Semiconductor (now TI) LM2575 family. Very, Very easy to use. LM2575: 1 Amp buck converter LM2576: 3 Amp buck converter I would suggest a couple of things, though. Have a bunch of simple PC boards made up that the students can assemble. Although this SMPS family is very forgiving, grounding DOES matter if you ... 5 True, but irrelevant in your case. An LED does not change forward voltage quickly- there is a slow (thermal) change, but nothing on the order of microseconds where the caps have some effect. The caps may effect better stability so I'd suggest using them. Edit: In simulation, with 100uH inductance (long wires with some loop area) in series with an LED load (... 5 If your LED lamp assembly is 10W at 900mA it needs about 11.1V to operate. The LM317 current source needs 1.25V in the resistor and a further 3V headroom to operate, put it all together and your circuit needs a supply of at-least 15.35V to behave as you intend. At that voltage and current the LM317 will be burning over 3.8W so you'll be wanting a heatsink ... 5 simulate this circuit – Schematic created using CircuitLab Figure 1. Transformer tap selector. This isn't a complete answer but will give you some ideas to work on. Whatever electronic solution you come up with is going to need a steady voltage to work with. This makes it awkward to derive from the transformer as you have a full-wave bridge rectifier ... 5 The correct nominal voltage according to datasheet calculations is 9.138V (min 8.775, maximum 9.50 resulting from the reference voltage tolerance, ignoring resistor tolerances and adjust current tolerance) The equation is:$V_O = V_{REF} (1+R_2/R_1) + (I_{ADJ} R_2)\\$ The 9.06V (nominal) calculator results from ignoring the effect of the 50uA nominal ...

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