I am using the following circuit to run an LED.

Page 9 Top left diagram
LINK

but the problem is the output voltage on which the LED runs is very low (≈0.4V when supply voltage is 12V. How can i make this LED run brighter i.e increase the output voltage?
Atm the values are -
Supply 12V
Capacitor 330nF
Resistors 1.2KΩ each

 

Can you post your circuit as we can see if you made something wrong.

 

The circuit is same as the one in the link, and LED is connected to output directly at the moment.

 

Page 9 Top left diagram

You of course mean upper right (Figure 2), which is sometimes referred to as the other left.

How are you measuring the voltage? Your duty cycle is less than 50%. A typical voltmeter will show a voltage considerably below the peak voltage in that case. Consider: 1) Use a NE555, not the CMOS 7555 for more current output; 2) Use one of the many circuits to get >50% duty cycle. Check out the Projects sub-forum and look at post #4 of the Simple PWM thread for one example.

Here is my favorite configuration that allows duty cycle control:
View attachment 4416


John

 

Here's one of mine...

 

Here's one of mine...

Thanks Bill.

Regarding your circuit how should i go about choosing values for resistors in series with LED.

 

You of course mean upper right (Figure 2), which is sometimes referred to as the other left.

How are you measuring the voltage? Your duty cycle is less than 50%. A typical voltmeter will show a voltage considerably below the peak voltage in that case. Consider: 1) Use a NE555, not the CMOS 7555 for more current output; 2) Use one of the many circuits to get >50% duty cycle. Check out the Projects sub-forum and look at post #4 of the Simple PWM thread for one example.

Here is my favorite configuration that allows duty cycle control:
View attachment 4416


John

Feel stupid as i cant distinguish between left and right.
Thanks for yours, i will try it.

 

Pick a current you're comfortable with, such as 20ma, and work from there. If you're dealing with 5V you won't have much to spare, as a LED drops around 2.5V (red) to 3.5V (green) or more (white or blue). Think in terms of the CMOS or 555 shorting across one bank, leaving the other to light up.

A CMOS 555 will be a lot closer to 50%, while there will be less perfection on a stock 555. A 7555 is generally a drop in replacement though.

 

Hi Bill,
The circuit dont seem to be working. No consistant frequency output.
I am trying to get a 1kHz square wave signal to the LED.
These are the values i chose-
C = 330nF
R between pin 2 and 3 = 2kΩ
R beween LED and ground = 400Ω
R beween Vdd and LED = 100Ω

i am only using one LED rather than 2. What am i doing wrong?
Thanks

 

You don't mention power supply voltage. Try upping the LED resistor by a factor of 10 to see if it works. Even with very low currents it will light up, and only use a resistor that connects between the LED, power supply, and IC. Eliminate the surplus resistor (the one gong between pin 3 and the one of the power supply rails), you don't need it.

Once you've established the circuit, then you can try increasing the current through the LED.

If you have a DVM you might try measureing AC voltage to see if it is something like a backwards LED.

 

Power supply is 12V.

 

Then you had WAY too low resistance there. You may have fried the chip. 20ma is around 1KΩ.

 

Thanks for your help so far mate.

Tried different things, and this is what i get-
1. If i measure the output across LED, it isnt square wave.
2. If i measure across the LED and the resistor, its a square wave.

Now this is really confusing me.
Is there any alternative cicuits you would suggest. For eg a BJT with 555 to drive LED etc.?

 

The CMOS timers can only source about 10mA and sink 100mA (Check the datasheet). Those are max ratings so you shouldn't operate at the max sink or source.

You should either use a transistor (FET,BJT) to drive your load off the signal generated from the timer.Or configure your load so that the IC is sinking current keeping within the data sheet max power disapation etc.

For sink , source see the pic below.

 

Good advice there. I need to experiment with 7555's myself. Bought some 2 weeks ago as a matter of fact. A classic 555 will sink source 200ma on either rail, but the + side has more voltage drop since it uses 2 transistors in a darlington configuration. A CMOS device, as long as the current requirements are met, can go rail to rail (really close to either power supply voltage).

BTW, I misspoke earlier, a 470Ω resistor would limit the current to 20ma.

 

The output of a Cmos goes rail-to-rail only when it has no load current.

 

A CMOS device, as long as the current requirements are met, can go rail to rail (really close to either power supply voltage).

The output of a Cmos goes rail-to-rail only when it has no load current.

I think I said that.