current draw and limitation

[manicool271989]

Hey guys I'm new to this blog (well at least to the forums anyway) I just wanted to discuss something interesting a housemate of mine brought.

   See I use a plug in dc converter to power my breadboard projects at various (6, 12) voltages. One day my housemate asks me if there is a limitation on the current that can run through my circuit projects because of the ac to dc adapter? I told him no there shouldn't be a limit and my reasoning was that since the adapter gives out a particular voltage and with the simple relation that V = IR you can ideally have from 0 to infinite amount of current going through the circuit. That's what I said, but I began to have second thoughts because I kept hearing current being drawn from a power supply during lectures at school. I kinda feel ashamed to be asking such a basic material question, but i really would like to know if my initial reasoning is right or not and clarification on what current draw means? (My guess would be that a load, such as a resistor, when connected in a circuit has some current going through it and that current is the current being drawn?)

Sorry if this is posted in the wrong place, but again this is my first time posting on the forums.

-manicool271989

[Feanor]

You are on the right track. The load/resistance that you connect across the terminals of your voltage supply allows current to flow from one terminal to the other. This is the current drawn by the load.

So if you put a very low resistance between the terminals then you will get a very high current, from ohms law. I = V/R.

The current will never be infinate however, infact it will probably not be very high at all for two reasons.

First of all every power supply has what is called an internal impedance, this is a resistance built into the power supply, it comes from the physical way the supply is built, if there are copper wires in the supply then there is resistance in the copper, if it is a battery then there is resistance in the electrolyte, etc.

Secondly most power supplis have overload protection, this means that if too much current is drawn from them then the power supply will stop applying a voltge to the terminals. this can be done in many different ways, fuses, circuit breakers, altering PWM in switch mode power supplies.

[Mechatrommer]

in the dc converter there is transformer (coil)... bigger transf = bigger current limit, smaller coil/tranf=smaller max current (limit) but i cannot explain it technically or theoritically, just from experience. the biggest transformer/dc converter that i have built/own only capable of producing 18A max when shorted (DMM Ampere measures) at 12V. i have alot bigger transf, but not assembled yet. for discreet components dc conv. you short it? the chip will burst opened!

edited: i just checked my dear transf psu, its a 18V 8A max, not 18A as slashed above. well, doesnt make any significant difference isnt it

[manicool271989]

Thanks for your answers!

So I did a little more research (mainly wikipedia) to figure out how the internal resistance of the power supply would affect the the voltage at the output of the internal power supply circuit (where in my case I assume I get 12 or 6 volts depending on the setting). It seems that when the load draws current from the power supply the internal resistance (lets assume impedance is only because of resistors) multiplied by the current being drawn (IR) gives a voltage difference that is equal to the difference between the voltage at the output of the power supply when there is no load (which I assume is a short circuit) and the voltage at the output when there is a load connected. Is this a right understanding?

[Mechatrommer]

V is constant. only I is changing depending of load (R) you are connecting.

[manicool271989]

I see... so if the voltage is a constant 6 V, then regardless of the load, the output node of the power supply or adapter will always be at 6 V (because that is how the supply circuit is set up) and the amount of current drawn by the load would be 6/(load impedance). But this current can't get to infinity because there is an internal impedance to the supply circuit so that you have some max. current that is:

((voltage before power supply impedance) - (6 V)) / (supply impedance)

and I guess since I power my circuits with an adapter plugged into the wall socket it would be:

((120 V) - (6 V)) / (load impedance)

right?

[Mechatrommer]

pls google on working principle of a transformer, here is one http://en.wikipedia.org/wiki/Transformer. the switch you are talking about is the connection to center tap and full secondary coil winding if i'm not mistaken. and you can check your psu apart to see if its transformer based or ic based psu. i dont really understand with the word "impedance" since it will involve frequency affecting it, but "resistance" term is ok for me, so i cannot help you much with the technical. 120V in your equation should not be there, as i said, google a bit. and i suggest you learn how a full wave rectifier works http://en.wikipedia.org/wiki/Rectifier, its always work right next to the transformer, dont worry, its simple.

[manicool271989]

cool I'll check it out.

[Feanor]

You are getting a bit confused. For the purpose of understanding what internal impedance is and hence maximum current draw you do not need to try to resolve how the 120V AC is converted into 6V or 12V by your supply.

So just forget about the 120V for now.

Your power supply gives you either 6V or 12V (6V is shown in the picture). It also comes with a series connected internal resistance shown in the picture as 3 ohms(this will vary from power supply to power supply and could include inductance and capacitance). This is also known as the Thevenin equivalent circuit of the power supply.

This power supply is shown connected to three different loads.

1. No load - open circuit. This is the connection that will reveal the maximum voltage that the power supply can supply. The power supply terminals have nothing connected between them. The resistance is hence very large(infinity) like when you set your multimeter to measure resistance and then leave the probes hanging apart in the air.

The current that flows is equal to the voltage divided by the resistance. I = V/R = 6/(3+infinity) = 6/infinity = 0A. The current drawn is zero amps.
The IR voltage drop across the internal resistance is 0*3 = 0V. So the voltage on the terminals of the power supply is 6V.

2. Short circuit - The biggest possible load. This is the connection that will reveal the maximum current the load can supply. The two terminals are connected together directly. The current  that flows is equal to the voltage divided by the resistance. I = V/R = 6/3 = 2A. The IR voltage drop across the internal resistor is 2*3 = 6V.

So the entire 6V of the supply voltage is dropped across the internal resistor meaning that the voltage between the terminals is 0V. This makes sense because they are directly connected so how could they have a different voltage?

3. Light load - A load somewhere between infinity and 0 ohms is connected between the terminals(2997 ohms in the example).

The current in the circuit = I = V/R = 6/(3+2997) = 6/3000 = 2mA. The I*R voltage drop across the internal resistor is 2mA*3 ohm = 6mV. The rest of the 6V supply is dropped across the 2997 ohm load. So the terminal voltage = 6-2mV = 5.998V(a bit less than 6V).

I hope this helps to clarify what is going on. This is an extremely simplified model of your power supply. Notice that as we decrease the load resistance from infinity down to zero we get more current flowing which gives us more IR voltage drop across the internal impedance which leaves less IR drop across the load so reducing the terminal voltage.

[Mechatrommer]

ah! this one explained it! i was confusing this simple thing with some electromagnetic effect thing.