2
\$\begingroup\$

LM317 voltage regulator has internal short circuit protection alongside with its thermal overload auto-shutdown, but in the circuit below it will simply burnout when the output is shorted. And its adjust pin gets internally shorted with its output pin.

I know that a pass transistor would strip off the regulator's ability to stand a short circuit and limit current but whats wrong here ?

Li-Ion Charger Circuit

I built it more then once on different PCBs and they all end up the same way ( killed about 5 regulators till now as i though that something was wrong with my build)

one version of the PCBs i made :

my build

\$\endgroup\$
9
  • 1
    \$\begingroup\$ I'm seeing no output capacitor ... are you sure it's stable under those circumstances? (Do you have a scope?) \$\endgroup\$
    – user16324
    Aug 6, 2016 at 11:30
  • \$\begingroup\$ What is C1 doing? Could that create reverse voltage when the output is shorted? \$\endgroup\$
    – pjc50
    Aug 6, 2016 at 11:33
  • \$\begingroup\$ Where exactly was the short? \$\endgroup\$
    – Bruce Abbott
    Aug 6, 2016 at 11:34
  • 2
    \$\begingroup\$ The metal tab on an LM317 is connected to the output. It isn't apparent from the photo whether or not you have electrically isolated it from the heatsink, the necessity of which depends on what else the heatsink touches. \$\endgroup\$
    – Andrew Morton
    Aug 6, 2016 at 11:43
  • \$\begingroup\$ @BrianDrummond The TI LM317 datasheet shows no output capacitor in a suggested configuration for a battery charger. \$\endgroup\$
    – Andrew Morton
    Aug 6, 2016 at 11:44

3 Answers 3

Reset to default
6
\$\begingroup\$

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: -

enter image description here

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 pin will probably save it. The 50 uA ADJ pin current (normal operation) will cause a 50 mV error (times the standard R1/R2 feedback ratio) in the output voltage so there is a somewhat fluffy downside potentially.

\$\endgroup\$
6
  • \$\begingroup\$ thanks you for the answers, seems to be right but just want to ask wouldn't the exact thing happen if the potentiometer was turned all the way to the ground side in a standard lm317 variable PSU circuit ? this for example : i.stack.imgur.com/FP60P.png wouldnt that also short the ADJ to 0v and cause the same problem ? \$\endgroup\$
    – XEL
    Aug 6, 2016 at 12:36
  • \$\begingroup\$ Im no expert on the device but I'd say yes; this potentially can damage it too if the output we're shorted. Proof lies in the insertion of a resistor. \$\endgroup\$
    – Andy aka
    Aug 6, 2016 at 12:40
  • \$\begingroup\$ thanks for the answer but after searching the LM317 can normally handle 0v at its ADJ pin thats how you make it output 1.25v lm317 will always output 1.25v higher than whats on its ADJ pin the problem is that the current limiter cannot get the voltage lower than 1.25v in case of a short which means it cannot further decrease the voltage to decrease current \$\endgroup\$
    – XEL
    Aug 6, 2016 at 16:04
  • \$\begingroup\$ @XEL - you are absolutely right in standard configurations, that is how the LM317 would produce a regulated output of 1.25V but we're talking about the transient scenario (a few tens of nano seconds to the odd micro second) when the short is initially applied. If you have a reasonable capacitance built into the load that will hold-up the 8.4 volt output as the short starts to manifest itself (i.e. not fully shorted in the instantaneous) and this will turn on the transistor and apply 8.4 volts between output and ADJ. After all it's the LM317 that is becoming damaged according to your question. \$\endgroup\$
    – Andy aka
    Aug 7, 2016 at 8:32
  • \$\begingroup\$ @ Andy aka - but in this case the load is not a capacitor but a Li-ion battery. Imagine a short 'manifesting itself' on that! \$\endgroup\$
    – Bruce Abbott
    Aug 7, 2016 at 16:30
2
\$\begingroup\$

The problem with this circuit is that the current limiter (Q1,R1) doesn't work properly when the output is shorted. As the LM317 outputs 1.25V when its ADJ pin is at Ground potential, both Q1 Base-Emitter and R1 get 1.25V across them, the current limiter fails since you cant get the LM317 to output less than 1.25v ( atleast without negative voltage ) causing greater than 1.25A current flow. Without a resistor to limit Base current there is a good chance of Q1 burning out, possibly putting a continuous short on the LM317's output.

In normal use the output would never be shorted ( unless by mistake ) , so it isn't a problem. However for safety I would insert a resistor between Q1 Base and R1/R2, sized to keep Q1's Base current well below 0.5A under worst case conditions (100Ω should be high enough).

\$\endgroup\$
0
0
\$\begingroup\$

A few things here...

1 - Make sure you use an original LM317 and not a chinese spinoff... I experimented with chinese (wing shing) TL431’s and my conclusion is that the company is a fabless manufacturer that sources itself from many other chinese manufacturers. The result is a product that behaves very differently from batch to batch.

2 - The LM317 requires a 220 ohm (or less) resistor between output and adj. to compensate to its internal input-to-output leakage current. R3 at 470R is too high. Also reduce R2 to 1k and readjust your trimpot.

3 - Q1 requires base current limitation. Add a 1k resistor between Q1’s base and the 1R resistor for that purpose.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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