Hello All - Page 2 - Electronic Circuits Projects Diagrams Free

AtomSoft · 9th February 2008, 07:00 PM

ok i understand ohms law like 90%-100% now so i am going to learn
diodes
capacitors
transistors
op-amps.

Hero999 · 9th February 2008, 07:17 PM

Quote:

Originally Posted by quixotron

Well he is gonna need it if he wants to work in wireless. alot of the antenna and RFIC devices are in the s band region. a 216 mhz signal generator and standard oscilloscope won't cut it!!

Ever heard of wireless modules?

AtomSoft · 9th February 2008, 08:13 PM

Ok i get the concept of Diodes and Transistors. But im confused on how to use them. Ill ask about diodes first.

If i have a 3v input and 1 diode the output is assumed to be around 2.6v right?
Now to go down from 2.4v to 1.4V i need a resistor right. But how would i know what size resistor to use?

Do i have to use like a digital multimeter to get the A/mA(I) in the current circuit first?

How do i calculate what resistor would be needed? Or would i need to know the resistance of the diode and A/mA(I) of the circuit?

You see where i am confused?

Hero999 · 9th February 2008, 08:56 PM

Diodes very good voltage droppers because their voltage drop will vary with the forward current.

You can measure the current by connecting the multimeter in series with the circuit you're measuring.

Use Ohm's law to calaculate the resistor value required for the current needed to power the LED.

AtomSoft · 10th February 2008, 01:13 AM

So a diode isnt really 100% need right ? Like if your using D/C Current.

Brian Hoskins · 10th February 2008, 02:57 AM

The main application for a conventional diode is not it's forward voltage drop but the fact that current is only able to flow in one direction through them (apart from a small leakage current when reverse biased). As Hero said, the forward voltage drop is dependant on current, and to a smaller amount other factors such as temperature. As an example, if you read the datasheet for a general purpose diode like the 1n4001 you'll see a graph which plots forward voltage against current. The device is rated up to 1Amp, and from 0 - 1Amp the forward voltage varies by 0.4V - this is why they're not very good voltage droppers, unless the circuit current is known and remains fairly constant.

Brian

AtomSoft · 10th February 2008, 04:51 AM

ok im trying this and the LED is getting 1.75v . i am having trouble calculating the I = E / R. I get some wierd number ."0.0002459419" Can someone explain. or would this just be 245mA

Pommie · 10th February 2008, 05:01 AM

If you remove the 10k resistors and just keep the 330Ω you will then get a voltage of approximately 2V across your LED. The other 3V will be across the resistor and therefore the current will 3/330 = 9mA. To make your LED bright you want around 20mA through it and so the resistor should be 3/0.02 = 150Ω.

Mike.

AtomSoft · 10th February 2008, 05:21 AM

How can i calculate like you just did? Like how did you calc that when a 330Ω is placed there it would generate 2v on the led? I see how to get info from ohms law but like to create a outcome based on input is not yet understood lol.

AtomSoft · 10th February 2008, 05:26 AM

I found this:
http://www.doctronics.co.uk/resistor.htm

"A typical LED requires a current of 10 mA and has a voltage of 2 V across it when it is working. The power supply for the circuit is 9 V. What is the voltage across resistor R1? The answer is 9-2=7 V. (The voltages across components in series must add up to the power supply voltage.)

You now have two bits of information about R1: the current flowing is 10 mA, and the voltage across R1 is 7 V. To calculate the resistance value, use the formula:
R = V / I

Substitute values for V and I:"

So in mines it would be like

5V in and LED eats 2v so 3 left over
R = 3 / .010 = 300ohms

right ?

Now is it possible to use 3 100ohm in series to make a 300ohm resistor ?

Pommie · 10th February 2008, 05:32 AM

Correct, LEDs will conduct more and more current until the voltage across them is 2V. As more current flows a bigger voltage is dropped across the resistor. When the current is high enough the voltage across the resistor is 3V and the current stops increasing.

Mike.

AtomSoft · 10th February 2008, 05:36 AM

ok with this same concept i can manually control a 7 segment led right?

Also the forward voltage/current is the input the component actually needs to work right.

I tried this with a 7seg display works 100%. Give me a min to up a pic of it

AtomSoft · 10th February 2008, 06:37 AM

i like it lol

of course a microcontroller or pic w/e its called will replace the wiring in future. lol

Pommie · 10th February 2008, 06:43 AM

It's looking good. Just be aware that the segments are sharing the resistor and so the more segments lit the duller they will be. This can be rectified by having a separate resistor for each segment.

Mike.

AtomSoft · 10th February 2008, 07:01 AM

ok cool. Will note. Its just that its like 2am here and everyhere is closed so i didnt want to blow it lol. Um also i took a Dual 7 SEG LED off something i had around but its funny it only has 9 pins

I got it to light up tho just going to make a pinout now

9th February 2008, 08:56 PM	(permalink)
Hero999 Experienced Member	Diodes very good voltage droppers because their voltage drop will vary with the forward current. You can measure the current by connecting the multimeter in series with the circuit you're measuring. Use Ohm's law to calaculate the resistor value required for the current needed to power the LED. __________________ What's so bad about Microsoft? Get Opera it's simply a superb browser.

10th February 2008, 04:51 AM	(permalink)
AtomSoft Experienced Member	ok im trying this and the LED is getting 1.75v . i am having trouble calculating the I = E / R. I get some wierd number ."0.0002459419" Can someone explain. or would this just be 245mA Last edited by AtomSoft; 10th February 2008 at 04:54 AM.

10th February 2008, 05:26 AM	(permalink)
AtomSoft Experienced Member	I found this: http://www.doctronics.co.uk/resistor.htm "A typical LED requires a current of 10 mA and has a voltage of 2 V across it when it is working. The power supply for the circuit is 9 V. What is the voltage across resistor R1? The answer is 9-2=7 V. (The voltages across components in series must add up to the power supply voltage.) You now have two bits of information about R1: the current flowing is 10 mA, and the voltage across R1 is 7 V. To calculate the resistance value, use the formula: R = V / I Substitute values for V and I:" So in mines it would be like 5V in and LED eats 2v so 3 left over R = 3 / .010 = 300ohms right ? Now is it possible to use 3 100ohm in series to make a 300ohm resistor ? Last edited by AtomSoft; 10th February 2008 at 05:29 AM.

10th February 2008, 05:32 AM	(permalink)
Pommie Experienced Member	Correct, LEDs will conduct more and more current until the voltage across them is 2V. As more current flows a bigger voltage is dropped across the resistor. When the current is high enough the voltage across the resistor is 3V and the current stops increasing. Mike. Last edited by Pommie; 10th February 2008 at 05:46 AM.

10th February 2008, 05:36 AM	(permalink)
AtomSoft Experienced Member	ok with this same concept i can manually control a 7 segment led right? Also the forward voltage/current is the input the component actually needs to work right. I tried this with a 7seg display works 100%. Give me a min to up a pic of it Last edited by AtomSoft; 10th February 2008 at 06:18 AM.

9th February 2008, 07:00 PM	(permalink)
AtomSoft Experienced Member	ok i understand ohms law like 90%-100% now so i am going to learn diodes capacitors transistors op-amps.

9th February 2008, 08:13 PM	(permalink)
AtomSoft Experienced Member	Ok i get the concept of Diodes and Transistors. But im confused on how to use them. Ill ask about diodes first. If i have a 3v input and 1 diode the output is assumed to be around 2.6v right? Now to go down from 2.4v to 1.4V i need a resistor right. But how would i know what size resistor to use? Do i have to use like a digital multimeter to get the A/mA(I) in the current circuit first? How do i calculate what resistor would be needed? Or would i need to know the resistance of the diode and A/mA(I) of the circuit? You see where i am confused?

10th February 2008, 01:13 AM	(permalink)
AtomSoft Experienced Member	So a diode isnt really 100% need right ? Like if your using D/C Current.

10th February 2008, 02:57 AM	(permalink)
Brian Hoskins Experienced Member	The main application for a conventional diode is not it's forward voltage drop but the fact that current is only able to flow in one direction through them (apart from a small leakage current when reverse biased). As Hero said, the forward voltage drop is dependant on current, and to a smaller amount other factors such as temperature. As an example, if you read the datasheet for a general purpose diode like the 1n4001 you'll see a graph which plots forward voltage against current. The device is rated up to 1Amp, and from 0 - 1Amp the forward voltage varies by 0.4V - this is why they're not very good voltage droppers, unless the circuit current is known and remains fairly constant. Brian

10th February 2008, 05:01 AM	(permalink)
Pommie Experienced Member	If you remove the 10k resistors and just keep the 330Ω you will then get a voltage of approximately 2V across your LED. The other 3V will be across the resistor and therefore the current will 3/330 = 9mA. To make your LED bright you want around 20mA through it and so the resistor should be 3/0.02 = 150Ω. Mike.

10th February 2008, 05:21 AM	(permalink)
AtomSoft Experienced Member	How can i calculate like you just did? Like how did you calc that when a 330Ω is placed there it would generate 2v on the led? I see how to get info from ohms law but like to create a outcome based on input is not yet understood lol.

10th February 2008, 06:37 AM	(permalink)
AtomSoft Experienced Member	i like it lol of course a microcontroller or pic w/e its called will replace the wiring in future. lol

10th February 2008, 06:43 AM	(permalink)
Pommie Experienced Member	It's looking good. Just be aware that the segments are sharing the resistor and so the more segments lit the duller they will be. This can be rectified by having a separate resistor for each segment. Mike.

10th February 2008, 07:01 AM	(permalink)
AtomSoft Experienced Member	ok cool. Will note. Its just that its like 2am here and everyhere is closed so i didnt want to blow it lol. Um also i took a Dual 7 SEG LED off something i had around but its funny it only has 9 pins I got it to light up tho just going to make a pinout now Last edited by AtomSoft; 10th February 2008 at 07:12 AM.

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