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Ohm's law continually confuses me in my attempts to use it. Right now I'm trying to supply a maximum of 800 mA by 1.5 V from a 5V source. Does that mean that I need a 1.875 Ohm resistor V=IR, 1.5=.8R, R=1.875, or am I misusing it like I think I am? How do you think conceptually of Ohm's law?

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  • \$\begingroup\$ The source is 5 volts so you need a series resistor to take up the voltage slack (5 volts - 1.5 volts = 3.5 volts) and at 800 mA that's a resistor of value 4.375 ohms. Thus total circuit is 6.25 ohms and this will take 800 mA from a 5 volt supply. \$\endgroup\$ – Andy aka Dec 11 '16 at 21:20
  • \$\begingroup\$ @Andyaka I appreciate your response, but can you explain why? And to clarify, I'd need a 1.875 ohm resistor and then a 4.375 ohm resistor? Why not just a 6.25 ohm resistor, or is it the same thing? \$\endgroup\$ – JavaProphet Dec 11 '16 at 21:24
  • \$\begingroup\$ Do you know Kirchoff's Laws? The Kirchoff's Laws are "universal" laws that apply to all lumped circuits. Ohm's Law is just a description of one particular type of circuit element: the linear resistor. So you should only apply Ohm's law to determine the I-V relationship of a linear resistor and not for any other purpose. \$\endgroup\$ – The Photon Dec 11 '16 at 21:56
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The 1.875 Ohm 'resistance' you calculated is based on the 800mA and 1.5V, so this wouldn't be a resistor you add, but is the equivalent resistance of whatever load if drawing that 800mA (which may not be a resistor). As @Tony Stewart mentioned, the load may not be linear, so e.g. it may draw more current at lower voltage, unlike a resistor which will continue following the V=IR formula as the voltage changes.

If you wanted to step down the 5V source to the 1.5V load voltage, you could use a series dropping resistor, calculated as @Andy aka showed. So you would have this:

schematic

simulate this circuit – Schematic created using CircuitLab

R2 would be your 'load', and R1 is used to create the required voltage drop.

(Please note that this is OK for educational purposes, but in a practical circuit you would not use a resistor to drop the voltage is it would be very inefficient.)

Many people use the water analogy to think of electrical concepts as it tends to be easier to visualize. In this particular case you would think of a water source at a high pressure (5V), and requiring a specific flow and pressure (800mA at 1.5V). If you were to connect the high pressure source directly to the load, instead of 800mA you would get 5 V / 1.875 Ohm = 2.67 A. But if you put a long narrow hose in between the source and output, this would restrict flow and cause a pressure (voltage) drop.

I hope that analogy helps you think about Ohm's law.

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You use Ohm's Law to calculate the voltage drop from 5V to 1.5V @0.8A is 4.375 ohm, but you can't say for sure the load is linear at 1.875, but at least you are given one operating point.

Thus the sensitivity with changes in R or 5V will not be linear with a non-linear such as a 1.5V motor or a 1.2V IR LED

Ohm's Law works for linear components and non-linear parts only with known assumptions.

For startup motor current it purely I=V/DCR of the coil CD resistance until EMF is generated to reduce the current, then the ZL(f)=2pifL impedance of the motor becomes reactive as RPM increases per volt and current rises only with losses and Load.

For the 1.2V IR LED (or any diode for that matter), after Vf is reached the diode is saturated beyond the dynamic curve and it's linear bulk resistance, Rs or ESR dominates the current rise with voltage.

You can apply Ohm's law to CMOS logic or any semi-conductor when conducting towards rated current and determine its equivalent DC driver impedance from the datasheet specs for Vol and (Vcc-Voh) vs Iout to get the incremental \$\Delta V/\Delta I=Rs = ESR = RdsOn \ (for\ MOSFET \ or\ CMOS) \$

You can also use Ohm's Law for Capacitors and Batteries and compute the losses from ESR or determine the ESR from Cap Dissipation Factor or directly in some datasheets or from the CCA cranking amp spec in the car battery for a rated drop from 12.5 to 7.5V so from Ohm's Law, \$ (12-7.5V)/CCA = ESR \$

You can also use Ohm's Law to estimate an LED's ESR from it's VI curve in the data sheet and then use that to limit current from any supply voltage. ESR in diodes is inversely related to its size and thus thermal power rating where size controls bulk resistance. For all single diode the ESR = 1/ Pd (+/-50% worst case) ( this tolerance can be <1% in the same batch and is generally closer to -50% in better quality power diodes.)

  • thus 68mW 5mm LED's typically are 15 ohms and 1Amp 3V LEDs which with suitable heatsinks can handle 3W are 330 m-ohms typical ESR.
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