Wondering if someone can help me by sugguesting the best way to drop a PCB design I am working on, to drop a 500ma input to 300ma.
Thanks for your help
What are you actually trying to do? What are the voltages involved? Normally if you have a switchmode voltage converter you’d specify it in terms of voltages…
Yes I have a 3.7 500ma voltage supply and would just like to drop it to 3.7 300ma. I don’t want to redesign my board just yet but would like to drop that current to 300ma output. I know if I put a Diode inline it will drop the voltage, will it also drop the current and how much?
Thanks
Why do you want to limit the current? It is possible to make a constant current supply, but they aren’t very common.
athnetix:
Yes I have a 3.7 500ma voltage supply and would just like to drop it to 3.7 300ma. I don’t want to redesign my board just yet but would like to drop that current to 300ma output. I know if I put a Diode inline it will drop the voltage, will it also drop the current and how much?Thanks
If the supply is a constant-voltage type the 500ma rating defines the “ampacity” of the supply: It is theoretically capable of supplying 500ma into a load whilst remaining in voltage regulation. It doesn’t mean that the supply will force 500ma into any load that’s attached. For these types of supplies, the load determines the current drawn.
I think you need to be more detailed and specific about the circuit and the load, your reasons for wanting to reducing the current-capacity of the supply etc. Post a schematic of your current setup if you can…
I suspect what you have is actually a 3.7 V supply that can deliver up to 500 mA. As Leon points out, constant current supplies are rare. It’s much more likely that you have a constant voltage supply.
If it is, in fact, a constant voltage supply and you’ve measured the current at 500 mA, that means the impedance of your load is 3.7 / 0.500 = 7.4 ohms. To hit 300 mA, you want the impedance to be 3.7 / 0.300 = 12.3 ohms. The difference is 12.3 - 7.4 = 5.1 ohms. Put a 5.1 ohm resistor in series with your load. The current will drop to 300 mA, but the voltage at the load will drop to (7.4 / 12.3) * 3.7 = 2.2 V.
Unfortunately, for a load of constant impedance, it’s impossible to change the current without also changing the voltage-- that’s what constant impedance means-- a constant ratio between current and voltage.
If you’re still puzzled, try explaining a little more about what you’re building and why you want what you want.
pingswept:
Unfortunately, for a load of constant impedance, it’s impossible to change the current without also changing the voltage-- that’s what constant impedance means-- a constant ratio between current and voltage.
Sure you can - add resistance in parallel.
As for the original question, if you really want to eat up 200mA from a 3.7V source, put an 18.5* ohm 1 watt resistor across it.
Orin.
*I know it’s not a standard value.