Electronics > Projects, Designs, and Technical Stuff
LM317 digital control
spec:
--- Quote from: imo on February 04, 2019, 10:53:59 pm ---
--- Quote from: spec on February 04, 2019, 10:22:50 pm ---
--- Quote from: SiliconWizard on February 04, 2019, 06:04:29 pm ---You can probably get away with a higher value for R2, so the opamp will have to sink a bit less current.
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The opposite is true. From the LM317 data sheet (link below), the minimum output current for an LM317 is 10mA, so R2 should be 125R maximum rather than 180R. A practical worst-case opamp dissipation would be 36V x 10mA =360mW, which is hardly a problem. Anyway, there are probably simple methods of reducing the opamp dissipation if required.
--- End quote ---
You may have 317' 10mA min current and 50uA opamp current (the ADJ current).
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Our posts crossed. :)
Yes you are quite right: 10mA is the minimum current for an LM317, not 7mA as I indicated on the schematic of reply #52. You can reduce the LM317 tail current by connecting a current mirror of 10mA to the output of the LM317, or you may get away with just a resistor, depending on the output voltage range in use.
not1xor1:
--- Quote from: spec on February 04, 2019, 10:54:40 pm ---
--- Quote from: SiliconWizard on February 04, 2019, 06:04:29 pm ---You can probably get away with a higher value for R2, so the opamp will have to sink a bit less current.
--- End quote ---
The opposite is true. From the LM317 data sheet (link below), the minimum output current for an LM317 is 10mA, so R2 should be 125R maximum rather than 180R. A practical worst-case opamp dissipation would then be 36V x 10mA = 360mW, which is hardly a problem, especially if you use a decent opamp. Anyway, there are probably simple methods of reducing the opamp dissipation if absolutely necessary.
Or are you suggesting that the LM317 output has current drawn off it by another means, so that R2 can be an open circuit, when the tail current would be only 50uA or so.
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Most LM317 work flawlessly with just 3-4mA of load. If one just needs to build 1-2 PSUs can test the ICs for stability on a breadboard.
A different but still valid approach is to provide a 10mA constant current sink for the LM317. Then the value of the resistor between output and adjust is no longer critical.
BTW a relative high power dissipation in opamps affects their performance.
Zero999:
not1xor1,
The LM317 definitely is specified for a minimum load of 10mA. Most devices will work with much lower minimum load currents, but relying on 4mA is marginal design.
--- Quote from: spec on February 04, 2019, 10:22:50 pm ---
--- Quote from: SiliconWizard on February 04, 2019, 06:04:29 pm ---Technically, you're still generating a negative supply.
(It's going to make a nice heater... ;D )
--- End quote ---
Very profound ^-^ The heat generated by the diodes, will be insignificant compared to the heat generated by the LM317 itself, not to mention the heat generated by the other four diodes in the bridge rectifier and the load itself.
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P = IV so irrespective of where the voltage is dropped, the same amount of heat, in terms of Watts, will be dissipated. I know what you mean though: it won't be as hot as dissipating all of the power in the LM317, as it's more spread out.
--- Quote ---Or are you suggesting that the LM317 output has current drawn off it by another means, so that R2 can be an open circuit, when the tail current would be only 50uA or so.
--- End quote ---
Yes, a current sink could be added, but in this case it's not needed: the resistor is fine since the op-amp effectively has a negative supply.
The main issue with this circuit isn't the power dissipation or LM317's minimum load current, but the voltage ratings of the components. What transformer voltage do you recommend? The original schematic showed a 22-0-22V primary, but no secondary voltage. Presumably it was inadvertently drawn mirrored and you meant 22V-0-22V secondary? That would give a DC bus voltage of nearly 62V, which would be far too high, so that doesn't seem right either.
To get 30V out, the LM317 needs at least 33V in, plus five diode drops, nearly 1V each at full load, giving a minimum DC bus voltage of around 38V. At first glance, a 30V transformer looks like a good fit, as it would give just over 42V out. The trouble is it's a marginal design. When unloaded, the transformer voltage will be around 10% higher than its fully loaded rating, the diodes will drop less voltage and the mains voltage could also be on the high end of the tolerance band, another 10% in the UK, causing the DC bus voltage to be much higher. VDC = 30*√2*1.1*1.1* - 1.2 = 50V. The LM317HV could be used, which is rated to 60V, but finding op-amps, which can do this is more challenging.
Another problem with the LM317 is its safe operating area protection kicks in when the voltage difference between its input and output exceeds around 15V. Using a centre tapped transformer and a relay or transistor to select between the centre tap, depending on the output voltage will alleviate this to some degree.
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