General question about linear regulators and transformer VA - Page 1
Forum section: electronics / beginners

author: queennikki1972
author_rank: Regular Contributor
timestamp: August 13, 2019, 02:03:58 pm
content: I understand that a linear regulator like the LM317T is a max output current of 1.5 according to the specs?I have a transformer of 24V 40VA for example, which I assume is 1.6 amps.Does this mean the transformer is to large and would overload the regulator? Or does it simply mean it will only put out 1.5 amps on the output of the regulator?Any links to explain something probably so simple are appreciated. I just cant seem to get a clear understanding..

author: Totalsolutions
author_rank: Contributor
timestamp: August 13, 2019, 02:25:02 pm
content: Voltage is something you have. Current is what you take. Quick answer.

author: bson
author_rank: Unknown
timestamp: August 13, 2019, 03:12:27 pm
content: A handy rule is that current in equals current out. So if you put a 0.5A load on the regulator its current out is 0.5A, its current in is 0.5A, the rectifier's current out is 0.5A, its input is 0.5A, and finally the transformer's secondary current is 0.5A both on input and output (it will have at least two terminals). Only the transformer primary will differ (it will have equal power to the secondary, minus some marginal efficiency loss). Current is determined by the load.Your transformer, however, is a bit too hot for the LM317T; it has a maximum input voltage of 30V. 24V RMS means the peak secondary voltage is just shy of 34V. Even with a little voltage drop in the rectifier this means the 30V maximum will be exceeded.

author: not1xor1
author_rank: Frequent Contributor
timestamp: August 13, 2019, 03:49:14 pm
content: Quote from: queennikki1972 on August 13, 2019, 02:03:58 pm1.6A is the maximum current you can draw. The amount you actually draw depends on the load.It is like a car. If the tachymeter range goes up to 200mph that doesn't mean you can only drive at that speed.In this very case, a transformer, the maximum specified current applies only to resistive laods.If you use a diode bridge and a levelling capacitor you must derate the maximum current value.With a 24V 40VA you cannot usually take more than about 1ADC (i.e. about 60% of maximum resistive load current).

author: Chriss
author_rank: Frequent Contributor
timestamp: August 13, 2019, 04:59:32 pm
content: Voltage is always what you can measure on a circuit under power.Current is measurable on a circuit when the circuit is under any load.The load determines how much current you can measure.So, in your case we can say:The transformer secondary side gives you a voltage of 24VAC but no current even if it is rated to give out 1.6A.If you connect anything to the transformer secondary winding lets say the rectifier diode bridge than you can measure a minimum current what is taken by the component.If you put a bulb 24VAC/24W then you would measure a current of 1A on the secondary side of the transformer and almos 0.6A will left unused and unmetered.The same story is for the linear regulator too, but dont try to connect you LMxx to the transformer cos you will kill the IC cos your input voltage is higher then the max Uin for the regulator.My best regards.Sent from my SM-J500F using Tapatalk

author: queennikki1972
author_rank: Regular Contributor
timestamp: August 13, 2019, 08:43:51 pm
content: Than k you everyone i think i understand now..

author: ledtester
author_rank: Super Contributor
timestamp: August 17, 2019, 03:35:29 pm
content: Quote from: bson on August 13, 2019, 03:12:27 pmAre you sure about this? All of the datasheets I can find say the LM317 has an output voltage range of 1.2 to 37V and a max voltage differential (in minus out) of 40V. This even applies to the TO-92 package LM317L.

author: rstofer
author_rank: Super Contributor
timestamp: August 17, 2019, 05:36:07 pm
content: Quote from: queennikki1972 on August 13, 2019, 08:43:51 pmAs long as you understand that all of the replies are talking about 'averages'. The peak rectifier current depends on the size of the filter capacitors and the load. Why?The rectifier only conducts when its output voltage is higher than the capacitor voltage. Given very large capacitors, this on time is very short. Yet the power delivered to the load must pass through the rectifier while it is forward biased. Larger capacitors, larger peak current, shorter on time. Conduction angle, Google for it. I didn't see anything very useful at first glance. The transformer sees the same peak current but is less likely to be bothered by it.So, smaller capacitors result in lower peak diode current but then the capacitor has to do more work by taking and giving significant amounts of charge. Small capacitors work harder and ripple voltage is higher.Someplace there is a comfort zone for both components.Maybe Google for 'selecting rectifier and capacitors for power supply', there are some videos. I haven't watched any of them.

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2019, 08:15:37 pm
content: Quote from: ledtester on August 17, 2019, 03:35:29 pmYes, it should be fine for normal operation but will not be 100% guaranteed to be short circuit proof, in all circumstances.Under light loads and high primary voltage, the secondary voltage of the transformer can be much higher than the specified value. For example if the transformer has a regulation factor of 20% and the upper tolerance of the mains voltage is +10% of the nominal, the secondary voltage will be 24*1.2*1.1 = 31.68V, which is a peak voltage of 44.8V. If there's a short circuit, causing the LM317 to overheat and shutdown, drawing very little current, there's a risk it could be damaged by over-voltage. If it's in an application where a short circuit is unlikely or there's no way the voltage on the rectifier will exceed 40V, then it'll be fine, but if it's a bench supply it's worth considering a higher voltage rating part or a transformer with a lower secondary voltage.

author: soldar
author_rank: Super Contributor
timestamp: August 17, 2019, 09:06:07 pm
content: Quote from: Totalsolutions on August 13, 2019, 02:25:02 pmVery good way of putting it. I should remember that.

author: rstofer
author_rank: Super Contributor
timestamp: August 17, 2019, 09:06:19 pm
content: The LM317 may be self-protecting, or not... I haven't really played with one.Here's the problem with linear supplies (almost all of them): You start with some maximum input voltage and you adjust the output to some minimum output voltage. Then you apply full load. You now have the maximum possible voltage drop across the regulator times the maximum current - and the plastic melts. Maximum power dissipation in the regulator occurs at low output voltages where there is a lot of voltage drop across the regulator.What needs to happen is for the transformer to have taps such that its output voltage can be selected to be much closer to the regulator output voltage. We're trying to reduce the voltage drop across the regulator. Commercial supplies, like the Rigol DP832 have such a feature.

author: soldar
author_rank: Super Contributor
timestamp: August 17, 2019, 09:11:28 pm
content: Quote from: queennikki1972 on August 13, 2019, 02:03:58 pm1.6 A assumes power factor = 1. In other words, a straight resistor.Reactive loads and non linear loads less than 1.6 A.

author: Brumby
author_rank: Unknown
timestamp: August 18, 2019, 11:54:30 pm
content: ... and you thought a simple power supply would be something simple to talk about.As you can tell from the above responses, there are quite a few things that can come into play - but let's step back to the question you ask about a fundamental topic.This is a good summary:Quote from: Totalsolutions on August 13, 2019, 02:25:02 pm... but to spell things out more clearly, I offer the following:* From the specification given, your transformer iscapableof providing an output current ofup to1.6A at an output voltage of 40V.* The output voltage will be 40V*1- a figure defined by the physical construction of the transformer and the input voltage on the primary.* The current drawn from this transformer will depend on the circuit connected to it. If you have a circuit that only needs 10mA, then that's all the current which will flow from the transformer. Likewise, if you have a circuit that needs 1.5A, thenthatis the current which will flow from the transformer.* If you have a circuit that sometimes needs 10mA and sometimes needs 1.5A, then sometimes the current flowing will be 10mA and sometimes it will be 1.5A. This just happens as a function of the physics - you do not have to do anything (nor can you do anything) to have this current flow at a rate other than what the physics determines. This is an extremely common situation where you may have a circuit which is monitoring some condition (and only needs 10mA to do that) and then switches on a heating circuit or a motor or some other load that needs 1.5A*1Now, in the real world, the transformer has some physical limitations and whilewe like to considerthe output voltage will be a constant 40V, it does vary slightly. Those limitations include the strength of the magnetic field within the transformer and the resistance of the wire. The general rule is: the voltage at no load will be higher than the voltage given for the transformer (sometimes notably higher, so watch out for that) and the output voltage will be the nominated value (or close to) when supplying thefullrated current.NOW - there is a question you may not have considered - but - it is this: Can a transformer supplymorethan it's "rated" current?The simple answer is "Yes" - however there is a hugeBUTassociated with that. Things included in that "BUT" are:* The voltage WILL sag (fall) to a value less than the nominal* The transformer WILL start to overheat. If this over current is only for a short time - for example a turn-on surge during the initial charging up of capacitors, where it's all over in less than a second - then this overheating will usually be insignificant. If you run over current for an extended period, you WILL stress the transformer physically. It can cook and fail. Lovely smell.

author: mariush
author_rank: Super Contributor
timestamp: August 19, 2019, 01:08:00 am
content: 24v ac 40va = 40/24 = 1.66arectification with full bridge rectifier results in Vdc peak = 1.414 x Vac - 2 x Vdiode rectifier = ~ 31v and current can be estimated to Id = 0.62 x Iac = 0.62 x 1.66 = ~1.04 A