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I try to construct a shut down pin or a feature which can set the output of LM317 or LM78XX with negative voltage that is created through a transformerless buck converter.

A buck converter having a positive input voltage can be convert its output to a negative output voltage without a transformer. But will it differ from a negative voltage created through a transformer center tap.

Can I hook the negative voltage from the buck converter to the resistor divider bottom leg . Some designs use current buffer to induce current in milliamp range to the resistor divider network. Why or for which conditions is this necessary?

Below is an implementation without a current buffer but with a diode, section 8.3 page 11. Why is a diode needed from the divider network to the ground?. enter image description here

I am trying to construct a shut down switch for lm317 so that its output is pulled to 0V. In shut down state, the load on the lm317 output can be as low as 0.1 ohm. So i try to find an optimal solution so that the output stays lower than 10mV when shut down pin is enabled. (if my load decreases down to 0.1 ohm when lm317 is in shut down state and the lm317 output stays below 10mV my leakage current will be less than 100mA which is acceptable in my design) I send a schematic and simulation.

I face a problem that my optocouplers output transistor has a VCESAT of 0.1-0.2V (according to datasheet). So I tried to compensate it by increasing the negative voltage from -1.25V to -1.5 and adding a compensation resistor R4 (120ohm). When shut down pin is disabled. lm317 will face a load of 0.1ohm and drive itself to its thermal equilibrium. By attached heatsink it delivers a maximum current of 1.5A. I use the lm317 as a current limited regulator. The current limitation is through thermal limit of lm317.

The lm317 works as current source. It will be operated at its thermal equilibrium. So that it can not go beyond 1.5A. With precision heatsink, it is stabilized at 1.56A at 0.2ohm. It functions as a current source between loads 0.2ohms and 0.3 ohms. at the final stage, I plan to omit 0.2 ohm and decrease this value to 0.05 ohm, so that precision heatsink defines the maximum current lm317 can deliver. It is 1.56A with a large heatsink. (I tested it , it does not go under 1.56A after 24 hours at 45 degree ambient. This is a scenario that will not happen in the operating enviroment.)

enter image description here

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    \$\begingroup\$ I have to admit being a bit confused by your writing (not the schematic, though... just you.) Can you provide a block diagram using the schematic editor to help clarify your words? (Or just re-write them or expand on them more?) I feel like you have inserted several different ideas, at one time, and it makes a tangle in my mind about the totality of your goals. What is your final goal, ignoring the details about how you see achieving it? \$\endgroup\$ – jonk May 21 '17 at 20:58
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    \$\begingroup\$ You go on in your text about transformers and transformerless but in the end you have no transformer. So why mention it ? But will it differ from a negative voltage created through a transformer center tap. No it doesn't, a negative (DC) voltage is a negative (DC) voltage, how it is created does not matter. If you disagree, show the schematic and explain why it is different. Some designs use... show an example instead of just mentioning something is done. There are many ways to do "something" show an example schematic. \$\endgroup\$ – Bimpelrekkie May 21 '17 at 20:59
  • \$\begingroup\$ I added a schematic. My goal is to implement a shut down switch for lm317 without touching the power path. I try to keep the output below 10mV when shut down. \$\endgroup\$ – rxpu May 21 '17 at 22:16
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    \$\begingroup\$ -1 for massive stream of conciousness text with no breaking up of ideas. Also you still haven't explained why you think two kinds of negative voltage are different, or shown the designs you're comparing. \$\endgroup\$ – The Photon May 21 '17 at 22:42
  • \$\begingroup\$ if you look at the root locus curve of both systems (transformerless negative voltage v.s transformer center tap driven center tap voltage, it may be predicted that the poles may tend to interact more with the poles of lm317.(stablity issue). But i wanted to hear your opinion abaout it, not mine. Anyway if you think they are exactly the same, no need to discuss further abaout negative supply source, I would rather then hear more opinions abaout the problems with the negative bias on adjustent pin. \$\endgroup\$ – rxpu May 21 '17 at 22:58
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A buck converter having a positive input voltage can be convert its output to a negative output voltage without a transformer. But will it differ from a negative voltage created through a transformer center tap.

Probably. Most buck converters are regulated, so the voltage should be reasonably stable. However voltage created from a center tapped transformer will not be stable unless it is passed through a voltage regulator. Any change in this voltage will cause a corresponding change in the LM317's output.

Why is a diode needed from the divider network to the ground?

If the circuit is working properly it shouldn't do anything. However if the -10V supply fails for any reason then the diode will keep the bias voltage down to ~0.6V, limiting voltage rise on the LM317's output.

At switch-off the negative supply may start to drop before the main supply has fallen below the LM317's dropout voltage, causing the output voltage to spike as the power supply shuts down. This could be bad news if it is powering a sensitive device. To prevent this you want the negative supply to have a longer switch-off time constant than the main supply.

Here's part of the circuit for a bench power supply that I made in the 1980's (still working perfectly today!):-

schematic

simulate this circuit – Schematic created using CircuitLab

D3 and C2 produce ~-16VDC, which LED1 drops down to a relatively stable -1.6V. R3 drops this by 0.3V to make the bottom of VR1 -1.3V, thus making the LM317 produce an output voltage of almost exactly 0V (actually 0.05V) when VR1 is turned all the way down.

Q1 shorts out VR1 and R3 to shut off the output when the current limit (not shown) activates. The extra 0.3V that R3 normally removes also compensates for Q1 C-E saturation, ensuring that the ADJ pin will be pulled down below -1.25V. Short circuit current is adjustable down to less than 1mA.

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  • \$\begingroup\$ Dear Bruce, thank you for your detailed description. Just one point. The bc547 has a VCESAT of 200mV to 600mV (datasheet). How have you achieved to find a working point near to 0V. It may swing to +200mV or in worst case maybe -200mV (because of -12V). You probably reduce -12V through R1 (2.7kOhm). Can this cause more fluctuation regarding the uncertainity of the VCEsat of BC547. Can it be better to use a lower negative voltage such as -1.5V and a smaller resistor to decrease the fluctuation on adj pin. \$\endgroup\$ – rxpu May 22 '17 at 11:05
  • \$\begingroup\$ The BC547 only has to pass about 3mA, so VCEsat is less than 100mV. The RED LED acts like a Zener diode, stabilizing the negative voltage at about -1.6V. If I was designing this circuit today I would probably use a negative voltage regulator (eg. LM337L) in place of the LED. IC's are much cheaper and easier to obtain now than they were back then (IIRC the LM317 cost me about $25 - equivalent to over $50 in today's money!). \$\endgroup\$ – Bruce Abbott May 22 '17 at 18:35
  • \$\begingroup\$ the fet output of microcontroller can pull the gate of BC547 to gnd which does not satisfy to shut down the BC547 completely which has its emitter at -1.6V. But anyhow this is another issue i should solve. Optos are also very cheap nowadays and suitable for level shifting. I built as you described and works really without problems. The only problem is that at very light load, when the adj of the lm317 is pulled negative to shut down, there exists a bigger voltage approx 100mV or more compared to heavy load. I believe this can only be corrected using an opamp as @TodorSimeonov suggested. \$\endgroup\$ – rxpu May 22 '17 at 20:32
  • \$\begingroup\$ "at very light load, when the adj of the lm317 is pulled negative to shut down, there exists a bigger voltage approx 100mV or more compared to heavy load." - LM317 needs a small output current to maintain regulation. Measure voltage between ADJ and OUT. If >1.25V the value of R2 might be too high. \$\endgroup\$ – Bruce Abbott May 22 '17 at 23:29
  • \$\begingroup\$ Thank you for the valueble information. I will hang a fix dummy load of 2.2kohm attached to the output of lm317 so that lm317 draws at worst case 3-4mA when not shut down but at light load. And when shut down it will help that the output will be below 30mV.Good tip. \$\endgroup\$ – rxpu May 23 '17 at 6:27
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I would not use this circuit even published on TI datasheet.

Let's suppose the potentiometer is turned down to zero and the negative voltage moves from -10V to -12V. There are plenty real scenarios for this to happen. Then the output of LM317 will become negative and its resistor divider will sink current from your circuit. Same situation may occur when you use it as a "shutdown". There is also a situation when your -10V supply is turned off for some reason - fault or current protection and its voltage is close to 0V. I think this is the reason to place a diode there. It will somehow limit the voltage overshoot, but even with this diode a 5V output circuit can jump up to 16V-18V when negative supply fails. This is not a shutdown.

Better find a LDO with a shutdown pin. The other way is to place a P-MOSFET in the positive rail after LM317 to turn ON and OFF. You can control it's gate with a small signal NPN and this way you can make your "shutdown" input TTL-compatible.

Update: There is way to do what you want using a -5V supply and a OPAMP/Comparator like LM393 (LM2903). Its main benefit here is its open collector output. In the circuit below when the control optocoupler (sorry couldn't find an optocoupler model in this editor) is turned on the negative input of the opamp will be close to -5V, less than GND and comparator's output will have no effect on LM317's resistor divider. As soon as the optocoupler is released, comparator's (-) will jump higher than its (+), the comparator will pull down its output and this way Lm317's output will be forced to go as low as current through shunt resistor is about 0. This schematic pulling ADJ pin down will simulate that maximum voltage set by diveder 240/2.4k is reached and will force LM317 to stay at the same voltage as the load is.

The feedback network should be redefined in practice for stability when turned off (pulled to 0V) and for minimal negative overshoot when turning off. I am not sure you will get 10mV precision but something around +/-30mV is achievable.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed May 22 '17 at 18:58

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