Resistors + Ohm's Law.

Thread Starter

Bod

Joined Sep 18, 2016
317
So just a quick question, what colour bands on a resistor what I need to drop the voltage for 9V to 5V. Now this is probably incorrect as I don't really know about resistors to well. Also I tried Ohm's law and I don't really understand so any help on that would be appreciated!

Thanks, Bod.
 

hrs

Joined Jun 13, 2014
394
Hi,

You should probably post a circuit diagram of what you want to achieve. The voltage that is dropped over a resistor depends on the resistor value and the current through the resistor per Ohm's law.
 

Thread Starter

Bod

Joined Sep 18, 2016
317
Hi,

You should probably post a circuit diagram of what you want to achieve. The voltage that is dropped over a resistor depends on the resistor value and the current through the resistor per Ohm's law.
Here (it's not very accurate as I could not find the exact components)
Screen Shot 2017-01-06 at 13.02.43.png
 

Thread Starter

Bod

Joined Sep 18, 2016
317
For segment driving, use a 330 to 560ohms.
I would do that but I am using components out of a kit and they come pre-connected. So I am deciding wether to cut the wires or not. A swell as mu circuit is not finished and is messy so it will be very hard to understand.
 

Dodgydave

Joined Jun 22, 2012
11,284
If you're using seven segments with resistors in them already, it wont need anymore additional resistors, if you want to put your own resistors in then any values from 330 to 1k ohms will do.
 

Thread Starter

Bod

Joined Sep 18, 2016
317
For segment driving, use a 330 to 560ohms.
Hold up Bod, I was going on a tangent and forgot what I was doing. I actually have a switch that shall go in the resistor (9 - 5v) to power a 5v dc relay.
 
Last edited:

MrChips

Joined Oct 2, 2009
30,704
You are going about the problem incorrectly.

If you are using a 9V source to drive a red LED, the voltage across the LED will be about 2V.
Hence the voltage across the series resistor will be 9V - 2V = 7V.

Now you need to know the current through the circuit.
If you want 1mA through the LED, then the series resistor = 7V/1mA = 7000Ω
For 10mA, R = 7V/10mA = 700Ω
For 20mA, R = 7V/20mA = 350Ω

The way you have drawn the circuit with the connections from COM to COM on a 7-segment display does nothing and has not effect except waste power.

Current limiting resistors go on each of the seven segment connections, not the common pin.

The trial and error approach is to start with high value resistors, 7kΩ and gradually reduce the value towards 330Ω. Choose the resistor value that gives you an acceptable brightness.
 

MrChips

Joined Oct 2, 2009
30,704
Hold up Bod, I was going on a tangent and forgot what I was doing. I actually have a switch that shall go in the resistor (9 - 5v) to power a 5v dc relay.
You are failing to tell us how much current the device takes.

Edit: why did you not say a 5V relay in your first post?
 

Thread Starter

Bod

Joined Sep 18, 2016
317
Sorry wrong diagram. Here is the right one. (I think I got the relay wrong.) Btw The long straight wire is to go to some other circuit.
Screen Shot 2017-01-06 at 13.35.05.png
 

MrChips

Joined Oct 2, 2009
30,704
No point in supplying you with anymore answers until you can provide an accurate schematic and the current required.
I'm out of here.
 

Thread Starter

Bod

Joined Sep 18, 2016
317
No point in supplying you with anymore answers until you can provide an accurate schematic and the current required.
I'm out of here.
Let me rephrase my answer. I will get a 9V battery supplying from 100mha to 1200mha. That will go through a finger activated switch. This will activate a relay starting up a motor. As I said I am using a 9v battery, but the relay is only 5v dc so I need a resistor that will lower the voltage to 5v. This is probably still unclear and if so you can just ignore the thread and I will waste my time (not yours) finding the right answer. Also as for a schematic I don't have an application that will do that so I had to use an online version which is less accurate (I think anyway).

Well.. I hope this helps, Bod :oops:
 

mcgyvr

Joined Oct 15, 2009
5,394
So you have a 9V battery, a switch, a 5V relay and a motor? correct?
Is this just an on/off toggle switch or is it momentary (meaning its only on when the switch is held down)

And you just want to use the switch to turn on the motor? Correct?
Does the motor run off the 9V battery too?
What are the specs on the motor (voltage/current requirements)?
What are the specs on the switch (voltage/max current)?
 

Thread Starter

Bod

Joined Sep 18, 2016
317
Wich relay are you using?
How much current does the relay need?
An Omron relay. It does not say what current, but I would of thought its not that high.
So you have a 9V battery, a switch, a 5V relay and a motor? correct?
Is this just an on/off toggle switch or is it momentary?
And you just want to use the switch to turn on the motor? Correct?
Does the motor run off the 9V battery too?
What are the specs on the motor (voltage/current requirements)?
What are the specs on the switch (voltage/max current)?
Yes correct.
Its momentary.
Yes just to turn it on.
No, its a 6 - 15V DC motor and using a 3pin switch I am supplying it 12v on one output. And unfortunately only 5v on the second.
Its a MFA Como Drills motor. Specs: http://www.mfacomodrills.com/pdfs/Miniature DC motors.pdf Mine is the RE - 360 (3 POLE) & RE - 360/1
There are no specs, just a generic switch but with two metal pads and when you place your fingers on both pads you become the connection.
 
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