I have a AC to DC converter S8JX-G15024C which convert the input AC to get an DC output. The output would be 24V and 6.5A.

So, is it possible for me to connect directly the S8JX-G15024C which give an output current of 6.5A to the fuse F-10BS with rating of 3A? Would it blown the fuse.

Note: the fuse would then connected to a load which use 24V power input.

 

You sure that fuse is only 3A ?



Regards, Dana.

 

The question isn't how much current CAN the converter supply, but how much current WILL it supply.

If your load only draws 1 A of current at 24 V, then you'll only have 1 A of current flowing and your fuse will be fine. But it the load draws 5 A, then....

So what is the current draw for the load at 24 V?

 

Since the AC to DC converter S8JX-G15024C can give an output of 24V and 6.5A.

So, did it mean that, i can connect any load which draw current in the range of 0~6.5A to the converter directly. Then the converter would output an current to the load depend on the load current consumption?

 

Since the AC to DC converter S8JX-G15024C can give an output of 24V and 6.5A.

So, did it mean that, i can connect any load which draw current in the range of 0~6.5A to the converter directly. Then the converter would output an current to the load depend on the load current consumption?

Yes

 

Yes

OIC, thank you for your reply, it help me a lot

 

It’s a common misconception. The current specification of a power source is the maximum it can supply. Generally*, it is the load current consumption that one must consider. Your load must require no more than the source can supply.

Note that I said generally. In current driven loads, such as LEDs, your circuit must limit the load current, otherwise the load will attempt to use the maximum current of the source. As long as you’re aware of the load characteristics, the “general” rule will apply.

 

Since the AC to DC converter S8JX-G15024C can give an output of 24V and 6.5A.

So, did it mean that, i can connect any load which draw current in the range of 0~6.5A to the converter directly. Then the converter would output an current to the load depend on the load current consumption?

Yes. It's just like the battery in your car. It can deliver hundreds of amps of current -- and generally does when starting your engine. But you connect a dome light to that same battery when you open the door and it only pulls a small fraction of an amp from that same battery.

 

It’s a common misconception. The current specification of a power source is the maximum it can supply. Generally*, it is the load current consumption that one must consider. Your load must require no more than the source can supply.

Note that I said generally. In current driven loads, such as LEDs, your circuit must limit the load current, otherwise the load will attempt to use the maximum current of the source. As long as you’re aware of the load characteristics, the “general” rule will apply.

Since there is current driven load such as LED, so is there any voltage driven load which we need to limit the voltage pass through it?

And how we can distinguish electrical component whether it is belong to current driven or voltage driven load?

 

Since there is current driven load such as LED, so is there any voltage driven load which we need to limit the voltage pass through it?

And how we can distinguish electrical component whether it is belong to current driven or voltage driven load?

For an LED, you need to limit the CURRENT, not voltage. There are several ways t do this. Use a current source, a LED driver ... But for the simplest solution, use a resistor in series with the LED!

You need to know three values.

1. The supply voltage (Vs)
2. The LED forward voltage (Vf)
3. The LEF current required (Iled)

Supply voltage must be a few volts higher than Vf. The latter two parameters can be gotten from the LED datasheet. Note than Iled is often specified as a maximum value. Hence use a lower value in your calculations. I.e., if the value is 20mA, use 15mA. And, note that you must use the current expressed as amps. So 15mA is 0.015A.

The equation used to find the resistor needed is as follows;

R = ( Vs - Vf ) / Iled

And how do you know if your device is a current or voltage driven l9ad? You know your device! Read the datasheet.