# Help understanding Timer circuit

Discussion in 'General Electronics Discussion' started by Username99, Sep 12, 2014.

5
0
Apr 26, 2012
Hi,

As the title suggests I need help understanding the timer circuit in the schematic attached.

I built this project in college last semester, but did not design it.

It is a circuit designed to measure a users reaction time to an event.

The user presses the set button and after some delay the LEDs start displaying the count. The object is to stop the count as quick as possible so as to measure the users reaction time. The LEDs are then reset by pressing the reset button, and the LEDS will go to '000'

Just wondering how this works, Digital is not a strong point of mine, and I am trying to change that.

Thanks

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2. ### Gryd3

4,098
875
Jun 25, 2014
How much of it do you currently understand?
Have you looked at the data sheets for any of the ICs on your project?

5
0
Apr 26, 2012
I handed this project up last semester. I understand very little of it. We only had to solder the components in place. It was our first Digital project and now that I have a more little time I want to try learn a bit more about Digital.

I get that the Duty cycle is TimeOn/Total Time (correct me if I write something that is wrong) as this will give me a fraction between 0 and 1 and I multiply this by 100 to move the decimal 2 places to represent it as a percentile. So this tells me the amount of time during each cycle of the wave that the it is 'on' for.

I see the following in my circuit:

Resistors R1 and R2 are connected in series to capacitor C1 ( I am assuming we have the ability to charge and discharge C1 somehow) C1 can be fully charged therefore there is no voltage drop across R1 or R2 and 'discharge' (7), 'threshold' (6) and Trigger '2' are all held at +5V.

**I do not know the function of 7, 6 & 2* (please help here if you can)**

If C1 discharges; '7' will be at 5V( R2/ R1+R2 ) and both 2 and 6 will be 0V.

Alot of asumptions there, but it is a start.

Now '1' is always held at 0V.

The output of the 555 timer ('3') is I am assuming the oscillating waveform is appearing. It must have an On/Off ratio described by the equation above.

Reset (4) is constantly +5V and so is Vs (8).

I also do not know what 'astable' means.

That is pretty much where I am at right now, I am embarrassed to say,

4. ### Gryd3

4,098
875
Jun 25, 2014
Good start. The 555 in use will charge and discharge the capacitor through the two resistors in place. There is a formula available to determine the charge, and discharge rate.
The 555's output is determined based on the charge/discharge rate previously which is pretty close to half second on, half second off. This will provide a 1Hz output signal from the 555 which is the basis for 'timing' everything on the board.

I am currently unsure where the 1000Hz signal source is though. Can you take a picture of the bottom?

The decade counters merely change their output every time they receive a clock pulse, but they are cascaded together, so one must reach 10 before the next will change.
They change the output of half a byte of data (4, bits) to represent 0 - 9
0000
0001
0010
0011
0100
0101
...
1001

These 4 bits are input to the 7 segment decoders which will output the proper combination to represent the number on the 7 segment display.

You also have an 74ls74 which is used to control the set/rest portion of the circuit. This chip is a 'pair' of flip flops. It can almost be seen as two chips packed in one.

I can write up some more details later. In the mean time, I encourage you to google for the ICs that are on that board. The resulting data sheets will tell you what each pin does. It will also occasionally show you the internal workings of the IC to help get a better understanding of how it works.

5
0
Apr 26, 2012
O

Let me first give my reply a title for reference later:

Post Title: 555 Internal components.

This post if 555 internal components part (i)
555 internal components part (ii) will come soon after.

I have printed out the data sheets and had a glance through them already I will go off and look at them in more detail now. I saw the internal workings in the datasheet before and they looked like trigger is at the (-) terminal of what I assume is a comparator, an even bigger assumption I will make now is that a comparator is a box that compares two independent voltage levels and depending on what it sees then it gives a dependent output. This goes into a box 'F/F'.

***Just a short note: I have come back after finishing this post to put in a piece of information, that is more relevant at this stage) regarding the "Trigger" comparator; The voltage at the + terminal of the "Trigger" comparator is Vcc(1/3)

The 'F/F' box:
This has an input from the trigger terminal (T).
• There is also an input connected to the emmiter terminal of a BJT, I am assuming NPN) with the collector of this BJT connected to discharge (7).
• The is an input from another Comparators output. This comparator is comparing the voltage levels at Control voltage (-), which is dictated to by (2/3)Vcc (assuming R=R=R, i.e. Resistros are equal ohms) and Threshold (+).
• There is an 'Output Stage' box between F/F and the output terminal, I do not know what this does or if it does anything (Apart of course from our approx 50/50 1Hz duty cycle wave).
• There is another terminal to the F/F box and this is connected to Reset (4) through the collector of an PNP BJT, the emmiter is at some voltage reference, in my circuit I think is at some level whereby the 555 never resets because the base and collector are always in Reverse Bias because the Base (N region) of this BJT is connected to Reset (4) input which is constantly held at +5V.
The Flip Flop is starting to look like the beating heart of the 555 IC.
*** Assuming 'F/F' stands for FlipFlop, and if I remember back to my theory on JK flip flops that;
• If J and K are both 0, the output does not change.
• If K is 1 and J is 0 then the output goes to 0.
• If J is 1 and K is 0 then the output goes to 1.
• If both J and K are both 1 then the output toggles between 0 and 1.
[Again this is assuming for now that these are not D-Type or some other type of Flip Flop that I don't know about.]

So continuing on with all these assumptions (i just trying to figure it out as I go);

Zooming back out to the higher level of abstraction, the 555 timer as a box, if you will. We spoke about how we can charge and discharge the capacitor earlier. This is effectively putting;
• Threshold (6) and Trigger (2) at either 5V or 0V,
• Discharge (7) at either (5V)(1/10) = 0.5V when C1 is S/C i.e uncharged OR 0V when C1 is O/C i.e. charged.
• +Vs (8) is always connected to 5V.
• And 'Reset' is always connected to +5V.
• Terminal 1 is always connected to ground.
• Output (3) is where over output voltage levels come out, High or Low.

Last edited: Sep 13, 2014
6. ### Gryd3

4,098
875
Jun 25, 2014
Some additional notes since you are taking a look at the comparators.
There are two comparators connected to Threshold(6) and Trigger(2). Those pins are compared to an internal voltage divider. One comparator will be looking at 2/3 the supply voltage, and the other will be looking at 1/3 the supply voltage.
Here is a handy animation you may find interesting : http://www.williamson-labs.com/pu-aa-555-timer_fast.htm

Those comparators will be driven digitally with a very very high open-loop gain so that the output is either high or low depending on which input is higher. Later study into analogue devices will show you how to use opamps as buffers and amplifiers for analogue signals.