Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Simple question

Status
Not open for further replies.

alanzhao

New Member
Hi,

I have an output voltage of 1.5v(voltage drop 3v-1.5v=1.5v, right?)from a 555 timer powered by a 3v batter. However I want to have an output voltage of at lease 2v to flash LED, what can I do to increase the voltage?

What value of resistor do I need for the LED if I change the battery to 9v? How to calculate the resistor value?

Thankss!!
 

ljcox

Well-Known Member
The IC designers specified 4.5 Volt as the minimum supply voltage. What that means is that all ICs of that type will operate at 4.5 Volt.

But in order to achieve this, the IC must be able to work at less than 4.5 Volt.

So some may work at voltages down to say 3.5 Volt, others may work down to 2.7 Volt, etc. But they are not GUARANTEED to work below 4.5 Volt.

Len
 

alanzhao

New Member
now I change the battery to 9v, and the output voltage will be 7.5v. so I need a resistor for LED. how do I get the value of the resistor?
 

Russlk

New Member
The 555 output does not go all the way to 9 volts and the LED drops about 2 volts, so figure 8 volts driving current and 10 mA being sufficient to light the LED, then R = E/I = 8/10 = .8K 820 ohms is a standard value.
 

laroche73

New Member
low voltage 555 operation

Len's right, many parts will operate at lower voltages, there's just no guarantee that all 555's will. Try a CMOS version of the 555, like the 7555 or LMC555, if you want to run off a 3V supply. The high output voltage comes close to the positive supply (depending on how much current the load draws). CMOS 555s can't source much current, so you're better off sinking current when using normal LEDs (LED is on when the output is low - cathode connected to pin 3, anode connected to +VCC though a current limiting resistor).

https://www.learn-c.com/lmc555.pdf
 

alanzhao

New Member
Russlk said:
The 555 output does not go all the way to 9 volts and the LED drops about 2 volts, so figure 8 volts driving current and 10 mA being sufficient to light the LED, then R = E/I = 8/10 = .8K 820 ohms is a standard value.

What does E stands for?
Can you explain R=E/I?
 

Roff

Well-Known Member
Russlk said:
The 555 output does not go all the way to 9 volts and the LED drops about 2 volts, so figure 8 volts driving current and 10 mA being sufficient to light the LED, then R = E/I = 8/10 = .8K 820 ohms is a standard value.
Mmmm... Russ, your math is a little shaky. Not that it makes a big difference. The LED will still light up - just not quite as brightly. However, it may confuse the hell out of alanzhao.
 

alanzhao

New Member
Ron H said:
Russlk said:
The 555 output does not go all the way to 9 volts and the LED drops about 2 volts, so figure 8 volts driving current and 10 mA being sufficient to light the LED, then R = E/I = 8/10 = .8K 820 ohms is a standard value.
Mmmm... Russ, your math is a little shaky. Not that it makes a big difference. The LED will still light up - just not quite as brightly. However, it may confuse the hell out of alanzhao.

I think Russlk is right, 800ohm is the result, but 820ohm is the standard value I could find from the E12 assorted resistor set.

or do you mean the voltage drop actually is 1.5v?
 

ljcox

Well-Known Member
Russlk said:
The 555 output does not go all the way to 9 volts and the LED drops about 2 volts, so figure 8 volts driving current and 10 mA being sufficient to light the LED, then R = E/I = 8/10 = .8K 820 ohms is a standard value.

I think what Russlk meant is that the 555 output voltage is about 8 Volt for a 9 Volt supply and assume that 2 Volt is dropped across the LED. Thus, the resistor should be R = E/I = (8 - 2)/10 = 0.6k, so use 560 Ohm.

You need to learn about Ohm's Law.

R = E/I or R = V/I where V = Voltage which is also called EMF. This formulae and its variants (V = I * R and I = V/R) are basic Ohm's Law which is fundamental to electrical and electronic calculations.

Len
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top