[SOLVED] Can i use 5V 5A dc supply to connect to a board which require 5V 500mA supply??

[nkarnani]

I am new to circuit designing and don't much idea about power supplies..:|

Please
Reply ASAP..

 

[nitishn5]

A 5V, 5A supply should be capable of supplying up to 5A at 5V.
Therefore it is fine to connect it to a board requiring 5V, 500mA.

 

[crutschow]

Yes, you can always use a power supply of the correct voltage with a higher current rating (and that is what you want for design margin). The circuit will only take the current it needs.

 

[neddie]

Yes you can.
The power supply CAN deliver 5A if needed , but it will only deliver the current required by your circuit.
In your case your circuit requires 500mA to operate , so that is what the power supply deliver.
If your circuit however required 6A , then your power supply may not be able to deliver that much current.
Cheers
Ned

 

[nkarnani]

A 5V, 5A supply should be capable of supplying up to 5A at 5V.
Therefore it is fine to connect it to a board requiring 5V, 500mA.

Yes, you can always use a power supply of the correct voltage with a higher current rating (and that is what you want for design margin). The circuit will only take the current it needs.

Yes you can.
The power supply CAN deliver 5A if needed , but it will only deliver the current required by your circuit.
In your case your circuit requires 500mA to operate , so that is what the power supply deliver.
If your circuit however required 6A , then your power supply may not be able to deliver that much current.
Cheers
Ned

Thank you guys for your reply..:smile:

I still have one question..
I have heard that excess current to the circuit will burn the circuit could anyone tell in which case that scenario happens??

 

[nitishn5]

Remember that current is always 'drawn by' a circuit, not 'supplied to' a circuit.

A scenario where an excess current is made to flow through a circuit would be when say some part failed or when the wrong part is inserted. Lets say that to light an LED, you are supposed to put a resistor in series with it. But instead of putting a typical 4.7k you put 47Ohms. This would cause a large current to flow and will damage the LED and if the current is large enough, it can also damage the supply.

A circuit if designed and assembled well should not have such problems.

 

[dexter0904]

I have heard that excess current to the circuit will burn the circuit

Yah, thats true....

Theory = you can connect small load to to high current source with no problem as amount of current is 'drawn by' the circuit....
Practice= You need to Limit current drawn from high current (Very Low Impedance source) or you will burn your components.
Demo= Connect your circuit with a 12V-100AH Battery and see what happens use 7805 regulator if your circuit is at 5V....

 

[nkarnani]

Remember that current is always 'drawn by' a circuit, not 'supplied to' a circuit.

A scenario where an excess current is made to flow through a circuit would be when say some part failed or when the wrong part is inserted. Lets say that to light an LED, you are supposed to put a resistor in series with it. But instead of putting a typical 4.7k you put 47Ohms. This would cause a large current to flow and will damage the LED and if the current is large enough, it can also damage the supply.

A circuit if designed and assembled well should not have such problems.

Yah, thats true....

Theory = you can connect small load to to high current source with no problem as amount of current is 'drawn by' the circuit....
Practice= You need to Limit current drawn from high current (Very Low Impedance source) or you will burn your components.
Demo= Connect your circuit with a 12V-100AH Battery and see what happens use 7805 regulator if your circuit is at 5V....

Ok guys..

Now i m getting something.. that is current is always 'drawn by' the circuit..
That is if I connect a wire directly to + terminal to - terminal with some low resistance that will let more current to pass through it that will drain the battery faster and if the resistance is high so a small amount of current will pass through it that will make battery last longer.

now i have two questions.

1)But in which case more heat will be generated? low resistance or high resistance?

2)I have heard about the shunt resistance concept. why do we use it to pass excess current? if it is the circuit who decide how much to consume so, we can supply 5A or 500mA there should be no use of shunt resistance?

 

[goldberg]

Its possible, but you have to pay the 500mA limit with a simple fuse.
Thomas

 

[nitishn5]

1)But in which case more heat will be generated? low resistance or high resistance?

You can find that out for yourselves.
Power Dissipated by Resistor is given by
P = I²R or P = V²/R
In this case since your voltage source is fixed, use the second equation.
Higher Resistance means more power dissipation i.e. more heat.

2)I have heard about the shunt resistance concept. why do we use it to pass excess current? if it is the circuit who decide how much to consume so, we can supply 5A or 500mA there should be no use of shunt resistance?

Some Power supplies such as a switched converter using inductors require a minimum current to be passed continuously for it to work properly. In such cases a shunt resistor can be used.

 

[nkarnani]

You can find that out for yourselves.
Power Dissipated by Resistor is given by
P = I²R or P = V²/R
In this case since your voltage source is fixed, use the second equation.
Higher Resistance means more power dissipation i.e. more heat.

I have a confusion regarding this
as we know in case of low resistance the value of current will be high as compared to case of high resistance.

so,
I1 - High current with R1 low resistance.
I2 - low current with R2 high resistance.

using the expression,
P=I^2*R

P1=I1^2*R1

P2=I2^2*R2

So the net result will be P1>P2??

 

[crutschow]

I have a confusion regarding this
as we know in case of low resistance the value of current will be high as compared to case of high resistance.

so,
I1 - High current with R1 low resistance.
I2 - low current with R2 high resistance.

using the expression,
P=I^2*R

P1=I1^2*R1

P2=I2^2*R2

So the net result will be P1>P2??

With a fixed voltage supplying the current, that would be true. To better see that use the equation V²/R for power.

 

[prestonee]

The equations given nitishn5 are correct... but the conclusion he yielded baffles me..
maybe this is due to my lack of coffee today and lack of sleep last night but... R is in the denominator, meaning large R = less power, small R = more power. Hence why you get more heat being dissipated by a short then with a 1Meg ohm.
grab a large power resistor and connect it to the 5v 5a supply, set the resistance to max, tap off point being at the opposite end of the power resistor. then slide the tap point closer and closer, soon you will 'feel' the warmth of the resistor.

- - - Updated - - -

Now if you had a fixed I and a variable Voltage (current source), you would see that as you increase the resistance you would dissipate more power yielding more power. just remember you control 2 of the 3 variables, V I R. use the equation that does not use the variable you do not control. if controlling the voltage and load, use P=V^2/R. If controlling Current and load, use P=I^2*R. If you know your voltage and can measure your current, but unsure of the load (active loading, ex mos devices) use P=V*I

 

[nkarnani]

With a fixed voltage supplying the current, that would be true. To better see that use the equation V²/R for power.

The equations given nitishn5 are correct... but the conclusion he yielded baffles me..
maybe this is due to my lack of coffee today and lack of sleep last night but... R is in the denominator, meaning large R = less power, small R = more power. Hence why you get more heat being dissipated by a short then with a 1Meg ohm.
grab a large power resistor and connect it to the 5v 5a supply, set the resistance to max, tap off point being at the opposite end of the power resistor. then slide the tap point closer and closer, soon you will 'feel' the warmth of the resistor.

- - - Updated - - -

Now if you had a fixed I and a variable Voltage (current source), you would see that as you increase the resistance you would dissipate more power yielding more power. just remember you control 2 of the 3 variables, V I R. use the equation that does not use the variable you do not control. if controlling the voltage and load, use P=V^2/R. If controlling Current and load, use P=I^2*R. If you know your voltage and can measure your current, but unsure of the load (active loading, ex mos devices) use P=V*I

Okay guys thanx I get it how Power will be consumed..

Now coming back to the thread..
the thing is that
I have a supply from SMPS
DC 5V 5A

I have to connect it to GSM Module which have the following specifications..




I want to know that can i use that supply and it will not burn the module??

 

[dexter0904]

you can connect Constant current source(500mA) b/w GSM module and 5V-5A Supply...