Jump to content
Electronics-Lab.com Community

newbie confused with voltage dividers


Recommended Posts

I'm learning electronics from a book and know im on the voltage dividers topic, i fully understand how it works and how perform calculations about it, but i'm confused using vd in real life, heres an example:
a have a 9v source and need to connect 2 devices, one needs 3 volts, and the other 5 volts, ok i can connect 2 resistors in series previous calculations and everything looks fine, if the total circuit current is ie. 500mA that means that 500mA and 3 volts exists in R1 and 500mA and 5 volts in the second resistor, ok if i connect a load(ie. a led) into R1 the current througth R1 will be the same? or i have a new circuit with R1 in parallel with the load?, if that is the case i must consider the device(load) resistance? how i know that device resistance? can i use a multimeter to measure every device resistance then design the circuit? with the new resistance found from the load, all devices have some kind of resistance?

how can i attack that problem? is there a formula or something to do this? is there a goodtutorial, guide, etc?

thanks in advance and please excuse my poor english

by the way the book only says how can i divide voltage if i need diferent values for diferent devices but dont take in account the current behivor with the load connected.

Link to comment
Share on other sites

The current is the same in a series circuit, but it equals the sum of all branches of a series parallel circuit. I think you already understand this from your explanation. However, an LED is a different calculation than the normal series parallel resistor calculations.
Here is an over simplified calculation:
If your supply voltage is 9V and your LED uses 2V and your current for the LED should be 20 mA (0.020A), then you would subtract the 2V from the 9V, leaving 7V. Then 7V/.020A = R or 350 ohms. You would then select a resistor at least this value to protecct the LED. Since the closest value is 330 ohms, you would go up to the next value.
I hope this helps to clear it up for you.


Link to comment
Share on other sites

I think the problem you are describing has to do with the fact that you cannot get a good voltage source from a divider. When the load current changes, the voltage will change. The amount of change is determined by the resistances you use to get the voltage. From the viewpoint of the load, the voltage will be the change in load current times the impedance. The impedance is the circuit that is seen by the load with the voltage sources replaced by ground.

Link to comment
Share on other sites

Yes, if you change the load, then the current requirement will change. If you have several devices on one line which have a changing load, they will interact. You will need to supply enough current for the worst scenario for all devices so that one device does not get too little current. V/R determines your current availability. If you have a specific schematic that you would care to share, everyone could make more specific recommendations to you. To this point, I can only make vague comments based on ohm's law for you.


Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...