IC 555

[Rleo6965]

Wow...It costs thousand!! if converted to RM..I will use my lecturer's Oscilloscope first

How about Logic Probe. Its cheaper. I'm sure you can afford this.

 

[vick5821]

How about Logic Probe. Its cheaper. I'm sure you can afford this.

But how that will help in my understanding ? How much ?

 

[Rleo6965]

http://www.techdose.com/electronics/Using-a-Logic-Probe/395/page1.html

Buy Logic Probe that can be use in 5V to 12V supply. Can catch and store one time pulse. Have small switch to select TTL and CMOS level signal. If possible have two tone signal to identify Hi and Low Signal.
You need to buy tools to help you learn electronics easier.

You can also build one. But, Its better to buy one.

 

[vick5821]

http://www.techdose.com/electronics/Using-a-Logic-Probe/395/page1.html

I have check the price..around RM 160+ from RS component ..Need to think for that..So, thanks for the link yea ^^

 

[vick5821]

http://clarkson-uk.com/555-timer/operation/frames3.html

Delay-On monostable circuit..I have built it..My question is why they connect a diode to it ?

 

[Rleo6965]

http://clarkson-uk.com/555-timer/operation/frames3.html

Delay-On monostable circuit..I have built it..My question is why they connect a diode to it ?

Hmmm. It does'nt make sense to me. Diode was reversed biased and this will not do anything w/ + 1/3V VCC.. That is, capacitor was already charged and Timer Output was High voltage Level to control something.

Probably a discharge path for the capacitor if Timer was power off.

 

[vick5821]

Hmmm. It does'nt make sense to me. Diode was reversed biased and this will not do anything w/ + 1/3V VCC.. That is, capacitor was already charged and Timer Output was High voltage Level to control something.

Probably a discharge path for the capacitor if Timer was power off.

This is a delay on circuit..when power is supply, it will delay for some times before ON

 

[(*steve*)]

This circuit pretty much uses the 555 as a Schmitt trigger.

Here is an almost equivalent circuit (the major difference is the amount of current that can be sourced or sunk from the output).

A schmitt trigger can do almost everything that a 555 can do, and is probably easier to understand. Sometimes you may need more than one, but since they come six to a package, it's not a problem.

I *think* they're easier to understand too.

 

[(*steve*)]

http://clarkson-uk.com/555-timer/operation/frames3.htmlMy question is why they connect a diode to it ?

My guess is that it will allow the capacitor to discharge via the 555's voltage divider when power is removed.

 

[jackorocko]

Probably a discharge path for the capacitor if Timer was power off.

My guess is that it will allow the capacitor to discharge via the 555's voltage divider when power is removed.

Now I feel dumb for not posting it, I knew it after all

 

[Rleo6965]

This is my first time see such diode placement in NE555 circuit.

But steve gave the accurate answer.

 

[vick5821]

My guess is that it will allow the capacitor to discharge via the 555's voltage divider when power is removed.

how ? charge flow from where to where?

 

[(*steve*)]

Now I feel dumb for not posting it, I knew it after all

Not really. It's not overly clear to me that the diode is necessary (or for that matter sufficient) to safely discharge the cap when the power is removed.

I wonder if the comparator inputs might have diode junctions which would perform a similar function.

Looking at the schematic of the 555, it appears that there is a Collector-Base junction between gnd and pin 2 that would effectively perform the same task as that diode.

There are a couple of issues though:

1) Different 555's (e.g. a CMOS 555) might not have an equivalent diode
2) The junction may be damaged by carrying the discharge current
3) The internal structure may be different to the schematic shown.

Even if the internal diode is present, it may have a lower Vf than the diode you insert in the circuit and it may carry a majority of the current.

The placement of this capacitor is opposite to normal, and you may wonder if that makes a difference (and maybe it does). There is no equivalent path to Vcc, leaving the discharge via the charging resistor(s) and the internal voltage divider -- which would be relatively slow.

 

[Rleo6965]

Discharge current path

 

[(*steve*)]

how ? charge flow from where to where?

In reverse order:

From one side of the capacitor to the other.

Via the diode and any resistance across the supply rails. The 555 provides 15k of resistors here, so that is the maximum resistance.

This is a surprisingly non-trivial issue, so if you don't follow, don't worry too much. Just believe us and figure it out for yourself when you understand more.

It's somewhat akin to wanting to know how to spin a car into a parking bay when you're just learning to drive. It's not something you do often, and it's pretty much irrelevant to typical technique.

 

[(*steve*)]

Discharge current path

Your arrows are around the wrong way

 

[Rleo6965]

Your arrows are around the wrong way

Ha ha ha

 

[vick5821]

Ha ha ha

lol...so the arrow diagram wrong ?

 

[Rleo6965]

Don't mind the arrow.

 

[vick5821]

Discharge current path

So, this is wrong path ?