5
\$\begingroup\$

I'm trying to make a high-current, adjustable-voltage regulator using LM317 data sheets. Specifically, I am using the schematic below.

LM317-based high-current adjustable-regulator circuit

So, what is the point of the resistor related to Note A? Is it supposed to represent the load? What does it mean that the minimum load current is 30 mA, and can or should that value change? Also, what determines that value? Basically, I'd be happy to learn as much as possible about that resistor.

Edit: The figure above is incorrect in that the TIP73 is depicted as a PNP transistor. The figure below is from a more recent LM317 data sheet, correctly showing the TIP73 as an NPN transistor.

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ The given answers are correct. I will add though that if you are building this circuit into a PCB with a known draw of more than 30mA, you do not need this resistor. \$\endgroup\$ – Trevor_G Mar 9 '17 at 19:15
  • \$\begingroup\$ It will be mounted on an Arduino prototype-board shield. Where could I learn if it has a known current draw? @Trevor \$\endgroup\$ – floydoom Mar 9 '17 at 19:42
  • \$\begingroup\$ Huh.. surely you know how much power what you are driving from that Vo pin requires? \$\endgroup\$ – Trevor_G Mar 9 '17 at 19:47
  • 1
    \$\begingroup\$ I misunderstood your original comment. Please disregard my question. :) @Trevor \$\endgroup\$ – floydoom Mar 9 '17 at 19:53
  • 1
    \$\begingroup\$ For linear regulation to start, you need at least 0.66v across the 22 ohm resistor, so that 2N2905 can turn on. So the "Note A" resistor provides the 30 mA, through the LM317, and through the 22 ohm resistor. \$\endgroup\$ – glen_geek Mar 9 '17 at 19:54
7
\$\begingroup\$

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 additional transistors shown here) has a minimum load requirement of 5 mA, which the recommended voltage adjustment resistors will draw, so no "extra" load resistor is required in that case.

\$\endgroup\$
  • \$\begingroup\$ The regulator will feed a servomotor, and the regulator will only be used while operating the servomotor. Am I right in saying that even an unloaded and stationary servomotor will pull enough current; therefore, I shouldn't need that resistor? \$\endgroup\$ – floydoom Mar 9 '17 at 19:39
  • 2
    \$\begingroup\$ If your load will always be connected, and will draw 30 mA or more (or perhaps if you don't care what happens to the voltage with very low loads), you don't need that resistor. \$\endgroup\$ – Peter Bennett Mar 9 '17 at 19:44
3
\$\begingroup\$

In case your load might draw less than 30 mA, you add this resistor to make sure the regulator is always outputting at least 30 mA.

If your load might be totally disconnected, you choose the resistor to draw 30 mA at your (minimum) output voltage.

If the load might at minimum draw only 27 mA, you choose the resistor to draw 3 mA at your (minimum) output voltage.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the helpful reply, Photon. I wish I could accept both your answer and Peter's answer. Although I feel your reply was more concise and practical, I think Peter provided a bit more information. \$\endgroup\$ – floydoom Mar 9 '17 at 19:35
3
\$\begingroup\$

The technical reason why 30mA must be drawn is to bias both transistors to the starting threshold of conduction so that they can begin to be active to bypass the regulator with a current sharing ratio defined by the other resistors.

This begins when the voltage drop across the 22 Ohm R *30mA = 660mV.

Plan B

Works better than original for variable output voltage.

  • Change TIP73 from PNP to NPN like 2N3055.
  • V drop & I determines power dissipation.

enter image description here

  • I have simulated the LM317 with an emitter follower.
  • I also simulate two input options a) fixed 5V out and b) adjustable 2.5 to 7.5 triangle sweep
  • I used an ideal transistor as a dummy active load.
  • Now the 330 Ohm pre-load can be removed completely (still shown)
  • as the bypass PNP is inactive with no load until the load current exceeds 20mA then both bypass transistor take all the extra current.
  • Does this closed loop have a name?
\$\endgroup\$
  • \$\begingroup\$ I'm trying to determine where heat sinks are required. How would you determine the "current-sharing ratio" between the transistors? I'm assuming most of the current would go through the TIP73 due to the 500-ohm resistor after the 2N2905; I mean, wouldn't the TIP73 almost act like a short circuit? Basically, I've already decided to put a heat sink on the TIP73, but would a heat sink be required for the 2N2905? Figuring out exact currents would help me with that. @Tony Stewart \$\endgroup\$ – floydoom Mar 10 '17 at 20:25
  • 1
    \$\begingroup\$ I overlooked a problem with the suggested design using adjusted voltage with need for a constant current. Now the load is just to offset bias current as a voltage divider to pull down as LM317 is basically an emitter follower with adjustable reference and feedback. Will update design shortly. \$\endgroup\$ – Sunnyskyguy EE75 Mar 10 '17 at 23:39
  • \$\begingroup\$ First, wow, this is above and beyond what I expected. Second, I'd like to note a few things: The TIP73 is an NPN (I'm using a TIP41C in my design), although the old diagram I originally posted incorrectly displays it as a PNP transistor. Furthermore, I think you're stating that the 330-Ω resistor is not needed, albeit removing this resistor causes the LM317 to dump to ground, dropping output voltage to 0 V. Also, why is there variable input to the Adjust input? Wouldn't we be more concerned with a variable input to the LM317's Input input? \$\endgroup\$ – floydoom Mar 11 '17 at 17:49
  • 1
    \$\begingroup\$ the ADJ bias R is the only R's needed. LM317 Output will regulate with no load. Did you try my simulation? \$\endgroup\$ – Sunnyskyguy EE75 Mar 11 '17 at 18:02
  • \$\begingroup\$ I did use your sim. I foolishly shorted the resistor instead of opening that part of the circuit. I now see that it's not required \$\endgroup\$ – floydoom Mar 11 '17 at 20:37
2
\$\begingroup\$

I am using the schematic below.

I'm pretty sure the schematic is wrong. as is, it is wired for a npn power transistor.

it can work with a pnp but the wiring is slightly different.

So, what is the point of the resistor related to Note A?

I guess they are trying to get a minimum amount of current so the pnp is conducting -> 22R * 30ma = ~0.7v.

Not necessary in my view.

\$\endgroup\$
  • 1
    \$\begingroup\$ I think it is wrong too. The power transistor there should be NPN to work. And TIP73 is NPN transistor, so it looks like it is mistakenly drawn in schematic as PNP. \$\endgroup\$ – Chupacabras Mar 11 '17 at 7:08
  • \$\begingroup\$ Oddly enough, I made a working prototype using a TIP41C instead of the TIP73, hooking it up exactly as it's displayed (before I burnt out the TIP41C from what I assume was a lack of sufficient heat dissipation). I was even powering various servomotors from the regulator. I didn't even realize it was depicted as a PNP in the schematic until you guys said something. I'm using other data sheets as well, and the transistor is correctly depicted in those; in fact, I'll update my original post with a diagram from a more recent data sheet @Chupacabras \$\endgroup\$ – floydoom Mar 11 '17 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.