Electronics > Beginners
General question about linear regulators and transformer VA
rstofer:
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.
soldar:
--- Quote from: queennikki1972 on August 13, 2019, 02:03:58 pm ---I have a transformer of 24V 40VA for example, which I assume is 1.6 amps.
--- End quote ---
1.6 A assumes power factor = 1. In other words, a straight resistor.
Reactive loads and non linear loads less than 1.6 A.
Brumby:
... 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 ---Voltage is something you have. Current is what you take. Quick answer.
--- End quote ---
... but to spell things out more clearly, I offer the following:
* From the specification given, your transformer is capable of providing an output current of up to 1.6A 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, then that is 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
*1 Now, in the real world, the transformer has some physical limitations and while we like to consider the 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 the full rated current.
NOW - there is a question you may not have considered - but - it is this: Can a transformer supply more than it's "rated" current?
The simple answer is "Yes" - however there is a huge BUT associated 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.
mariush:
24v ac 40va = 40/24 = 1.66a
rectification 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
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