• 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.

555 timer questions

Status
Not open for further replies.

lordv2500

New Member


Can someone please help me understand how these work.
In the first one the diode zener and 100k create a voltage of 3.3v. The voltage on the the 200ohm resistor is 3.3-0.7v (0.7v -voltage on the transistor)
Ie=Ic so that the elec current on the capasitor is 2.6v/200ohm.
The cap is charged and when it reaches 2/3Vcc it discharges (because of the transistor in leg 7 of the 555 timer). When it reaches 1/3 Vcc it is charged again and the same thing is repeated. So what do you get for the output of the 555 timer. Usually the trigger is a seperate pulse but this time it's connected to the 6th and 7th legs. Also how do I calculate the time of the output pulse t=r*c?

In the second one it looks like the same thing except without the diode zener. What is the function of the opamp at the bottom?
 

kinjalgp

Active Member
The circuit is that of ramp generator using constant current source.
The expected waveforms are shown below:
 

Attachments

lordv2500

New Member
Yeah, I understand that now. I was a bit confused because there was no sepperate trigger wave.
What about the second circuit?
Since there is no constant current source will the 555 work at all. I'm confused. The cap won't be charged properly it seems and the op amp at the bottom will amplify the voltage? so as a result it'll be like a low voltage of about 1mV at Vout? Please explain.
The curcuit will produce no result without the opamp right?
 

nettron1000

New Member
Like kinjalgp said , the transistor circuit is used as a constant current source for charging the capacitor .The opamp in the bottom schematic is
called a voltage follower which as a gain of 1 so there isnt any amplification, its purpose is to isolate the output from any connected load.

The constant current source is what gives the nice straight lines in the ramping voltage seen in the posted drawing. If the 555 was connected in its regular astable or monostable configuration, the ramping voltage would be more of a smooth exponential curve.
 

kinjalgp

Active Member
lordv2500 said:
Yeah, I understand that now. I was a bit confused because there was no sepperate trigger wave.
What about the second circuit?
Even the second circuit is a constant current source. The only duifference between 1st and 2nd is that for the 2nd one the output will depend upon stability of your power supply whereas 1st one is resistant against power supply variations due to the use of zener diode as constant voltage source.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top