# Reading more complex circuit schematics

#### fullNelson

Joined Nov 14, 2011
46
Hey guys.

For my question, consider, if you will, the simple schematic in the following link:

http://www.555-timer-circuits.com/flashing-led.html

If a 9VDC voltage is applied to the circuit, as per the diagram, is the voltage entering pins 4 and 8 also 9 volts or is the voltage reduced / affected by the two resistors in series?

#### praondevou

Joined Jul 9, 2011
2,942
Hey guys.

For my question, consider, if you will, the simple schematic in the following link:

http://www.555-timer-circuits.com/flashing-led.html

If a 9VDC voltage is applied to the circuit, as per the diagram, is the voltage entering pins 4 and 8 also 9 volts or is the voltage reduced / affected by the two resistors in series?
Since pin 4 and 8 are directly connected to the 9V power supply, yes there is 9V on them.
A 10nF capacitor is usually connected to pin 5.

• fullNelson

#### fullNelson

Joined Nov 14, 2011
46
Since pin 4 and 8 are directly connected to the 9V power supply, yes there is 9V on them.
A 10nF capacitor is usually connected to pin 5.
Thank you for the response. I ordered the parts to replicate the circuit for my own probing, but I am still new to reading schematics, and still need to build my skills at digesting/decomposing circuit schematics.

I see they omitted the 10nf cap on pin 5 as well. I wonder why.

I have another question in regard to this diagram. Is the voltage at pin 7 divided? Is that indeed a voltage dividing circuit via R1 and R2?

#### SgtWookie

Joined Jul 17, 2007
22,210
The 10nF capacitor from pin 5 (Control input) is not necessarily required, but it does help the timer to be more stable.

Pin 7 is the Discharge pin. It is an open-collector pin; which means that it either appears to be as electrically open (no connection), or as a short to ground.

Note that R1 should never be lower than about 100 Ohms per volt of Vcc, as otherwise you can wind up with too much current through R1 and pin 7.

As shown, the circuit will flash at about a 1.6Hz (cycles per second) rate; on time about 325mS, and off time about 307mS; just about a 51.4% duty cycle.

• fullNelson

#### fullNelson

Joined Nov 14, 2011
46
It does explain some parts of the circuit better. However, given my low knowledge of circuits, in general, some of the terms used don't help me understand the flow of current here.

For example, in "astable" mode, is the current that is discharged from C by pin 7 flow into BOTH ground and pin 5?

#### fullNelson

Joined Nov 14, 2011
46
Pin 7 is the Discharge pin. It is an open-collector pin; which means that it either appears to be as electrically open (no connection), or as a short to ground.
I thought IC pins were only I/O pins, are there different types of Input and Output pins? The wikipedia example didnt seem to indicate the pin type or the direction of current that well. Maybe I just don't know how to read it.

Thanks for your response as well, SgtWookie.

#### praondevou

Joined Jul 9, 2011
2,942
I suggest you have a look at this simulation:

http://www.falstad.com/circuit/e-555square.html A detailed explanation is included.

Use the internal diagram for reference. • fullNelson

#### SgtWookie

Joined Jul 17, 2007
22,210
I thought IC pins were only I/O pins, are there different types of Input and Output pins? The wikipedia example didnt seem to indicate the pin type or the direction of current that well. Maybe I just don't know how to read it.
Pin 7 is an output pin; it either sinks current, or it does not sink current. You can't change the functioning of the IC by changing how much current is flowing in to pin 7 - unless you exceed the power limits of the IC package.

Likewise, pin 3 is an output pin.

Some pins are input only, like the trigger and threshold (2 & 6).

Pin 4 (reset) is an input; if it's held low, the 555 timer will be held in the reset condition (pin 3 high, pin 7 sinking current).

All of them fall under the general category of I/O pins.

Pins 4 and 8 are power pins.

• fullNelson

#### SgtWookie

Joined Jul 17, 2007
22,210

#### fullNelson

Joined Nov 14, 2011
46
I have never ever ever seen current flow in a simulation before. I want to cry its so wonderful!!! It makes so much sense, but I have so many more questions based on the 555 timer simulation sgtWookie provided.

This shows external circuit to the 555 timer...sooo.....

It looks like there are two stages for current flow from the +9V: one is a complete near arrest of current through R1 and R2, the second is a blitz of current from the 9v past R1 and into the discharge pin....

My first question:

In the second "stage", why does it seem as if the discharge (pin 7) is "sucking up" current like a mad man past R1, despite the resistance? Its like it ignores the resistance all together.

Is this because of the op amp?

My second question:
When the current from the 9v is being sucked through pin 7, some of the current that was headed toward the ground suddenly looks as if it is being sucked/pulled in the opposite direction, as if to satisfy the current thirst pin 7 is having... is this right?

#### SgtWookie

Joined Jul 17, 2007
22,210
I have never ever ever seen current flow in a simulation before. I want to cry its so wonderful!!! It makes so much sense, but I have so many more questions based on the 555 timer simulation sgtWookie provided.
So does this mean you're OK with it? It looks like there are two stages for current flow from the +9V: one is a complete near arrest of current through R1 and R2...
That is when pin 7 is "open" (not sinking current) - the path of resistance from +9v to C1 is R1+R2= 1K+47k = 48k Ohms.
the second is a blitz of current from the 9v past R1 and into the discharge pin....
Because pin 7 is, at that point, ON - and is sinking current to ground via BOTH R1 and R2. The current flowing through R1 is wasted power at this point. The object is to discharge C1 via R2.

My first question:

In the second "stage", why does it seem as if the discharge (pin 7) is "sucking up" current like a mad man past R1, despite the resistance? Its like it ignores the resistance all together.
The current through R1 is 9v/1k = 9mA, or 0.009 Amperes. That is the highest current flow anywhere in the circuit. The current through R3 and the LED is lower because you have a ~1.8V drop across the LED, so there is ~7.2v to drop across R3, so 7.2v/1k=7.2mA.

Is this because of the op amp?
There are a couple of comparators internal to the 555 timers, one for the trigger, the other for the threshold. They are not opamps.

My second question:
When the current from the 9v is being sucked through pin 7, some of the current that was headed toward the ground suddenly looks as if it is being sucked/pulled in the opposite direction, as if to satisfy the current thirst pin 7 is having... is this right?
The current wasn't headed for ground, it was charging up the 1uF capacitor. When pin 7 is grounded internally, current flows through both R1 (1k) and through R2 (47k) towards ground.

The trigger input will cause pin 7 to turn on (sink current) when the voltage reaches roughly 2/3 of the supply voltage; in this case about 6v. The threshold input will cause pin 7 to turn off when the threshold voltage falls below 3v.

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• fullNelson

#### SgtWookie

Joined Jul 17, 2007
22,210
By the way - you can go ahead and make changes to that simulation I posted the link to. It doesn't matter how many changes you make; if you go back to my reply and click on that link again, it'll be exactly the same.

If you make changes to it that you'd like to save, then you can click on File, then Export - then right-click on the text in the box, then copy it out to a Notepad file.

Similar thing with Export Link - that's how I posted it to my reply.

• fullNelson

#### SgtWookie

Joined Jul 17, 2007
22,210
Here, I've made a couple of changes to the schematic:
http://www.falstad.com/circuit/#\$+1+5.0E-6+16.13108636308289+63+7.0+50 w+272+160+240+160+0 r+240+160+240+224+0+47000.0 w+240+224+272+224+0 w+240+224+240+256+0 w+240+256+272+256+0 c+240+256+240+320+0+1.0E-6+3.4984409129784177 g+240+320+240+336+0 r+240+160+240+96+0+47000.0 w+240+96+336+96+0 R+240+96+176+96+0+0+40.0+9.0+0.0+0.0+0.5 165+272+128+288+128+2+9.0 w+336+96+400+96+0 w+400+96+400+160+0 r+448+192+448+240+0+1000.0 w+400+192+448+192+0 g+448+304+448+320+0 162+448+240+448+304+1+2.1024259+1.0+0.0+0.0 O+448+192+480+192+0 d+192+160+192+224+1+0.805904783 w+192+160+240+160+0 w+192+224+240+224+0 o+5+32+0+35+10.0+7.8125E-4+0+-1

R1 and R2 are now both 47k Ohms - but the big deal here is that I added a diode across R2!

Before, the ON time was controlled by R1, R2 and C1, where the OFF time was controlled by just R2 and C1. This means that the ON time always had to be longer than the OFF time, because R1 could never be less than ~900 Ohms.

Now, ON time and OFF time can be adjusted completely independent of each other.
The larger R1 is, the longer the ON time.
The larger R2 is, the longer the OFF time.

Go ahead and try changing the values for R1 and R2.
Try R1 at 5k, then change it to 20k - and then change R2 to 5k, see what happens.

#### SgtWookie

Joined Jul 17, 2007
22,210
You might like this animation:
http://www.williamson-labs.com/555-circuits.htm
It's not really interactive, but you can choose from four animation speeds. Try starting with the 'slow' speed, and watch the timer go through it's cycles.

#### Ron H

Joined Apr 14, 2005
7,014
Pin 7 is an output pin; it either sinks current, or it does not sink current. You can't change the functioning of the IC by changing how much current is flowing in to pin 7 - unless you exceed the power limits of the IC package.

Likewise, pin 3 is an output pin.

Some pins are input only, like the trigger and threshold (2 & 6).

Pin 4 (reset) is an input; if it's held low, the 555 timer will be held in the reset condition (pin 3 high, pin 7 sinking current).
I think you meant to say "pin 3 low". All of them fall under the general category of I/O pins.

Pins 4 and 8 are power pins.