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Will LM317 circuit maintain steady output voltage in face of a change in input
Posted by ArtFromNY on 01 Feb, 2024 22:58 -
I realize that the standard LM317 holds the output voltage steady as the current drawn by the load changes.
My question is this: using a 15 volt input regulated down to 12 volts, driving a fairly constant load that draws ~50ma, how much change in that 15 volts input will the regulator tolerate before the 12 volt output drifts?
I realize that this is child's play to verify with hardware but my shop in my basement is too darn cold right now
I also realize that this is a simple noob question; I spent 50 years learning how to get defective TVs back to working as quickly and reliably as possible, and sadly neglected any learning of actual basic theory except as needed for that task.
Thanks all. Circuit I want to use is shown below.
Art
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it need something like ~2V from input to output to stay in regulation
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The specs you are looking for are "dropout voltage" (minimum input-output difference for reasonable operation) and "line regulation" (when input is safely above dropout).
The latter is generally fairly good and only matters in precision applications. The former is 2~3V for 78 and 317 type chips. -
Thank you both. I am using the regulator to feed a 50µa constant current source for calibrating Simpson 260 meter movements. Since that source seems to be happiest at 12V, I think based on the figures that were provided I will increase that source to 18V so that should assure the steady 12 volts on the regulator output.
Very much appreciate the help.
Thanks.
Art -
The load regulation could be as bad as 0.04% per volt, so a 12 volt output could change as much as 5 millivolts per volt of change in the input.
A good reference circuit would improve on that by at least a factor of 5 thousand, so less than 1 microvolt of output change per volt of input change.
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I'm confused what the 12V is for. You dial in 50uA with the large series resistor used
resource page Simpson260.com {slow to load} -
The load regulation could be as bad as 0.04% per volt, so a 12 volt output could change as much as 5 millivolts per volt of change in the input.
In the TI datasheet the line regulation is indeed .04%/V max at 25°C. But the dropout spec is quite vague, they just say "up to 3 volts."
OP's circuit is marginal in that respect even if in practice people say that it works down to ~2V.
Strictly speaking this circuit shouldn't work per the datasheet because as soon as the input voltage drops to 15V minus epsilon, the dropout is < 3V and the spec is violated.
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FYI - there is some confusion around the "dropout voltage", so I personally stopped call an LDO the LDO
https://www.eevblog.com/forum/metrology/lm399-based-10-v-reference/msg5311654/#msg5311654 -
I realize that the standard LM317 holds the output voltage steady as the current drawn by the load changes.
My question is this: using a 15 volt input regulated down to 12 volts, driving a fairly constant load that draws ~50ma, how much change in that 15 volts input will the regulator tolerate before the 12 volt output drifts?
I realize that this is child's play to verify with hardware but my shop in my basement is too darn cold right now I also realize that this is a simple noob question; I spent 50 years learning how to get defective TVs back to working as quickly and reliably as possible, and sadly neglected any learning of actual basic theory except as needed for that task.
Thanks all. Circuit I want to use is shown below.
Art
Hi,
That's a tried and proven circuit except that 240 Ohm resistor has been lowered for better performance. I think the recommendation now is 120 Ohms but you could check into that. That changes the bias current from 5ma to 10ma roughly.
As someone mentioned, a reference supply would allow even better performance. These things can not only be super accurate on the voltage set point, but also long term stable.
I had purchased a 4.096 volt reference and built a little circuit board. Then, I checked the output with two different very expensive 5 digit bench meters. They read within 0.001 volts of each other.
The output current is limited, but you can easily pump that up with an op amp in a simple circuit.
If you need an accurate voltage set point and long term stability, I would highly recommend one of these things. I can get you a part number if you are interested.
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FYI - there is some confusion around the "dropout voltage", so I personally stopped call an LDO the LDO
https://www.eevblog.com/forum/metrology/lm399-based-10-v-reference/msg5311654/#msg5311654
Hi,
The question is, who defined it that way as not all manufacturers are going to define it that way and this may even vary with device part number including the suffix.
For example, on one part Microchip defines the "drop out voltage" to be 100mv at the given data sheet output current level, down from a 1 volt differential voltage. That would mean for a 3.3 volt output you'd have to be sure the device always had at least 3.4 volts input.
However, that's not hard and fast either because there are temperature effects which could cause that to vary slightly, and also there is the specification of the actual output from part to part also. If there is a 1 percent difference in output from one part to another that could raise the DOV by maybe 33mv also bringing the total margin to 133mv for example.
However, if you designed for a minimum input of 150mv drop out voltage I would think it should work, but of course bench testing would be a good idea too. -
The load regulation could be as bad as 0.04% per volt, so a 12 volt output could change as much as 5 millivolts per volt of change in the input.
In the TI datasheet the line regulation is indeed .04%/V max at 25°C. But the dropout spec is quite vague, they just say "up to 3 volts."
OP's circuit is marginal in that respect even if in practice people say that it works down to ~2V.
Strictly speaking this circuit shouldn't work per the datasheet because as soon as the input voltage drops to 15V minus epsilon, the dropout is < 3V and the spec is violated.
With such a low load current, the dropout voltage will be more like 2 volts. A better regulator with lower dropout voltage could be used, like the LT1083, or even better, use a TL431 shunt regulator.
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..For example, on one part Microchip defines the "drop out voltage" to be 100mv at the given data sheet output current level, down from a 1 volt differential voltage. That would mean for a 3.3 volt output you'd have to be sure the device always had at least 3.4 volts input...
I would recommend to read the TI's slup239a.pdf about the dropout voltage and its meaning.
You will see it is not such an easy topic as you write.QuoteThis paper provides a basic understanding of the dropout performance of a low dropout linear regulator (LDO). It shows how both LDO and system parameters affect an LDO’s dropout performance, as well as how operating an LDO in, or near, dropout affects other device parameters. Most importantly, this paper explains how to interpret an LDO’s datasheet to determine the dropout voltage under operating conditions not specifically stated in the datasheet.
A perception that when an LDO have got "100mV dropout voltage" I will need >100mV higher input voltage (considering the Iout) to operate it, is absolutely wrong.. -
Not to confuse OP, but here sometimes people will go on a tangent about something. So keep asking questions about your project despite the side discussion.
The LM317 dropout voltage is 1.5V at low load and 2.3V at full load, so it is customary to give it a few extra volts input, at the expense of heat. The IC also has a minimum load spec of up to 10mA (which is usually provided by the divider load resistors).
I still don't understand your plan for making precise 50uA to calibrate the VOM or how the 12V or so is needed.
A bastard child of the LM317, the LM1117 started the "LDO" marketing title. The '1117 does not have the high drop Darlington pass transistor of the 317 and 78xx series. They use a Sziklai and thus one less VBE drop. All hail the technological innovation there. LM317 announced June 1976. -
I still don't understand your plan for making precise 50uA to calibrate the VOM or how the 12V or so is needed.
I bought one of those cheap little constant current PCBs from China. I hooked it up to 12VDC coming out of my Tek PS503 plugin. I put an exact 5000Ω load, which is the target resistance of the 260 meter movement, in series with a Keithley 195, across the output of the pcb.
I adjusted the little module until the Keithley read 50ua; repeated the test using a Fluke, readings were .03µa so close enough. Thus I can assume (please correct me if I am wrong) the constant current pcb will supply 50µa to a meter movement that measures exactly 5000 ohms.
I then used a Simpson 260-6 on the 50µa scale. The needle was 3 or 4 divisions off of full scale. I adjusted the shunt resistor in the meter circuit until I had full scale deflection, satisfying the stated parameter of 50µa full scale as per the calibration instructions.
Because I do not want to readjust the constant current board every time I go to use it (it will be built into a blank Tek TM504 plugin) I want to be very sure that the cc source 12 volts stays rock solid; thus, the desire to use the LM317 circuit. The 12VDC for the constant current source will not be provided by the PS503 once the fixture is in a actual use; I will rectify and filter the 17.5VAC supplied to the blank plugin by the TM504. -
..For example, on one part Microchip defines the "drop out voltage" to be 100mv at the given data sheet output current level, down from a 1 volt differential voltage. That would mean for a 3.3 volt output you'd have to be sure the device always had at least 3.4 volts input...
I would recommend to read the TI's slup239a.pdf about the dropout voltage and its meaning.
You will see it is not such an easy topic as you write.QuoteThis paper provides a basic understanding of the dropout performance of a low dropout linear regulator (LDO). It shows how both LDO and system parameters affect an LDO’s dropout performance, as well as how operating an LDO in, or near, dropout affects other device parameters. Most importantly, this paper explains how to interpret an LDO’s datasheet to determine the dropout voltage under operating conditions not specifically stated in the datasheet.
A perception that when an LDO have got "100mV dropout voltage" I will need >100mV higher input voltage (considering the Iout) to operate it, is absolutely wrong..
Hello again,
You've made this topic even more interesting now. I for one appreciate that.
I think I see what you are saying here. The way "dropout voltage" is specified is sort of back-asswards.
However, does it make a large difference.
What you are showing here is that if we had a 5 volt regulator with a dropout of 1.2 volts, we would probably assume we needed 6.2 volts input to get that 5v output, because the input to output differential is 1.2 volts and 5+1.2=6.2 volts.
In reality, if the dropout is specified as the differential input to output voltage WHEN the output reaches 100mv less than the nominal value, then the input voltage would really have to be a little bit higher to maintain a perfect 5.000 volts output (in theory, and output current at the specified level).
With that second statement in mind, that would mean we would measure the differential voltage when the output got down to 4.9 volts because that is 100mv less than the nominal value of 5 volts.
So let's go with this and see what happens.
If the differential voltage of 1.2v is specified at 800ma, and we know the output at that point is 4.900 volts, then the current would go up when we finally got that desired 5.000 volts output. It would go up to about 0.816 amps.
Assuming the minimum internal resistance holds over a small increase of the output, that would mean that to start with the internal resistance R would have been (6.1-4.9)/0.800=1.5 Ohms, and the new current of 0.816 amps means we would have a voltage drop of about 1.224 volts, meaning we would really need 5+1.224=6.224 volts input instead of 6.200 volts in order to get 5.000 volts at the output.
What this means is that we would have to add 44mv to that original estimate of 6.200 volts in order to be sure we got 5.000 volts at the output.
So now the question becomes, what would we get at the output with just 6.200 volts input?
If we assume the max current of 816ma at 5 volts even though we may have slightly less than 5 volts, that would be the worst case current. With the worst case current of 816ma the output resistance would be 6.127 Ohms, and using the voltage divider formula that would give us an output of close to 4.981 volts. That's about 0.38 percent (that is slightly greater than one-third of one percent). So we lose less than one-half percent if we assume the theoretically less accurate rule of thumb estimation.
The question now would become, what types of applications would suffer because of this inaccuracy.
Regular run of the mill applications would probably not be affected, but voltage reference type applications might suffer and so for that type the effect would have to be considered on a case-by-case basis.
Feel free to go over this and offer a correction if needed or whatever else you want to add.
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Neat discussion about the exact meaning of "dropout voltage"
But regardless of what the datasheet claim (or even what you measure) you probably should not operate your linear regulators so close to the dropout limit. Dropout voltage are usually only measured in one dimension, mostly input voltage where output start to change xx%. Other specs however start to degrade long before dropout voltage, sometimes to the point of uselessness. PSRR for example are getting worse the closer you get to the dropout voltage limit. -
The drop-out voltage is the difference between the input and output voltage, at which the regulator stops working properly. In the case of the LM317, it'll still output a voltage in the drop-out region, but it will no longer be regulated and changes in the supply voltage and load current will affect the output voltage.
There's a lot of talk about 2V or 3V, but in reality, it depends on the load current, temperature and the LM317 itself. Note when temperature is mentioned, it applies to the LM317 chip itself, which means the effects of self-heating need to be taken into account, at higher currents.
Here's a graph from a data sheet (ST I think) showing the drop-out voltage of the LM317 at different currents and temperatures. If it's always going to be operated in a high temperature environment and the output current is under 20mA, then you'll get away with 1.5V. but if it's operated at the full 1.5A and over the entire temperature range, then it's 2.5V.
To answer the original question: it depends. At normal room temperature, say around 20°C, the drop-out voltage will be around 1.6V at 20mA and 2.3V at 1.5A, so when set to 12V, it'll stop regulating when the input voltage falls below 13.6V when the load current is 20mA and 14.3V, when it's 1.5A.
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And as mentioned by ArdWar, you will find you need a little more for good ripple rejection because (internal) transistors are not that good at it when on the edge of saturation.
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Note that the criterion for regulator drop-out in the data sheet graph above is a change in output voltage of 100 mV to indicate regulation failure.
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Note that the criterion for regulator drop-out in the data sheet graph above is a change in output voltage of 100 mV to indicate regulation failure.
Good point, the way that dropout is measured and specified can be misleading. What I do when setting up a linear PSU with full-wave rectification (the standard sort of supply) is to put the expected maximum load on it and then lower the input with a variac until I just barely see the bottom of the ripple reflected in the output. If you look at that on a scope so you can see both the output and that ripple bottom, you'll come up with a dropout voltage number that is a bit different than shown in those charts--and in my view a lot more useful.
From that I can tell the OP that 3 volts is enough at his low output load and he'd actually be better off with a little more load. -
FWIW I gave the LM317 model included in one of those extended LTSpice libraries a try.
Regulation failure below 2V dropout
Same deal between -40°C and +80°C
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Note that the criterion for regulator drop-out in the data sheet graph above is a change in output voltage of 100 mV to indicate regulation failure.
And that the graphs are a 'typical' guide of trends, as opposed to the Max figures in the tables. -
With the output set to 12V, 100mV, is an error of 0.83%, which I don't think is that bad. It depends on the application, as to whether that's a problem or not.Note that the criterion for regulator drop-out in the data sheet graph above is a change in output voltage of 100 mV to indicate regulation failure.
Good point, the way that dropout is measured and specified can be misleading. What I do when setting up a linear PSU with full-wave rectification (the standard sort of supply) is to put the expected maximum load on it and then lower the input with a variac until I just barely see the bottom of the ripple reflected in the output. If you look at that on a scope so you can see both the output and that ripple bottom, you'll come up with a dropout voltage number that is a bit different than shown in those charts--and in my view a lot more useful.
From that I can tell the OP that 3 volts is enough at his low output load and he'd actually be better off with a little more load.
That's true. The graphs just give an indication of how it will behave in real life. Relying on those absolute worst case figures can result in over-engineering, so it's good to know you can aim lower, when a part isn't being pushed as hard. For example, if the load current is only 500mA, then it's a bit silly selecting filter capacitors which ensure a minium input-output differential of 3V, when 2V will probably suffice, unless it's extremely cold.Note that the criterion for regulator drop-out in the data sheet graph above is a change in output voltage of 100 mV to indicate regulation failure.
And that the graphs are a 'typical' guide of trends, as opposed to the Max figures in the tables. -
100mV error is always a problem because it means that the control loop is far from its normal operating point, loop gain is down the drain and the pass device is turned hard on so there is zero ripple rejection. Basically the regulator is a constant voltage between the input and the output.
Frankly, specifying dropout voltage under such conditions is plain cheating. -
100mV error is always a problem because it means that the control loop is far from its normal operating point, loop gain is down the drain and the pass device is turned hard on so there is zero ripple rejection. Basically the regulator is a constant voltage between the input and the output.
Firstly 2V of headroom with 500mA is not marginal, unless under extreme temperatures. It's a >100mV above the bare minimum, so the control loop will function fine.
Frankly, specifying dropout voltage under such conditions is plain cheating.
Secondly lots of circuits will work perfectly fine with 100mV p-p of ripple, so I wouldn't say operating it in near, or beyond drop-out is cheating, if the load isn't fussy.