Question about a 555 timer power pin setup

[uk.casmith]

I'm following an electronics book and I was wondering if someone could help answer some questions for me:

I've got 9 volts connected to a 2.2K resistor going into PIN 2 of the 555 timer. In the book it states:

"The 555 timer receives positive power through the 2.2K resistor, which is 2.2K. Because the input resistance of the timer is very high, the voltage on pin2 is almost a full 9 volts. "

1. Can someone explain to me why the voltage is a full 9 volts because the pin resistance is so high? Wouldn't the 2.2K drop the voltage as well? Why does it seem to not have any effect?

Also, connected to PIN 2 is an open switch connected to a resistor (I don't have the value with me right now) then to ground. When the switch is pressed this creates a voltage divider.

2. I know on paper what a voltage divider is, but if the resistance on PIN 2 is very high will this not interfere with the voltage divider?

I've only seen voltage dividers not connected to anything and on paper.

Thanks for your time,

Cameron.

[odessa]

Pretty sure pin 2 goes to one of the non inverting inputs of the internal comparators, which has a very high impedance ( not the same as resistance). Google 555 internal circuit and you'll see what i mean.

[dazz]

1. Can someone explain to me why the voltage is a full 9 volts because the pin resistance is so high? Wouldn't the 2.2K drop the voltage as well? Why does it seem to not have any effect?

Look at it this way: that's like having that 2K2 resistor in series with pin 2's input resistance, so the equivalent is a very high resistance. Very high resistance means very little current flow through the 2K2 resistor and into pin 2, and by Ohm's law, very little voltage drops across the 2K2 resistor.

Also, connected to PIN 2 is an open switch connected to a resistor (I don't have the value with me right now) then to ground. When the switch is pressed this creates a voltage divider.

2. I know on paper what a voltage divider is, but if the resistance on PIN 2 is very high will this not interfere with the voltage divider?

I've only seen voltage dividers not connected to anything and on paper.

Precisely because pin 2's resistance is so high, it's almost like there's nothing connected to the voltage divider. Very little current flows into pin 2.

Check this out: https://electronics.stackexchange.com/questions/28897/how-to-choose-value-of-resistor-in-voltage-divider

[mikerj]

"The 555 timer receives positive power through the 2.2K resistor, which is 2.2K. Because the input resistance of the timer is very high, the voltage on pin2 is almost a full 9 volts. "

1. Can someone explain to me why the voltage is a full 9 volts because the pin resistance is so high? Wouldn't the 2.2K drop the voltage as well? Why does it seem to not have any effect?

"High impedance" means very little current will flow into/out of Pin 2.  If you check the 555 datasheets it shows trigger current is only 0.5 microamps for the bipolar version, and a tiny 10 picoamps for the CMOS version.

Ohms law says V=IR (voltage = current * resistance). 

For the bipolar part V = 0.5E-6 * 2200 = 1.1E-3 or approximately 1 millivolt drop across the resistor.
For the CMOS part V = 10E-12 * 2200= 0.22E-9 or approximately 2 nanovolts drop across the resistor.

Also, connected to PIN 2 is an open switch connected to a resistor (I don't have the value with me right now) then to ground. When the switch is pressed this creates a voltage divider.

2. I know on paper what a voltage divider is, but if the resistance on PIN 2 is very high will this not interfere with the voltage divider?

I've only seen voltage dividers not connected to anything and on paper.

Since almost no current flows into or out of pin 2 then the operation of the divider is virtually unaffected.

[uk.casmith]

Thank you that was an excellent answer and links!

[alsetalokin4017]

Now do you understand what the switch and resistor to ground does?

[6PTsocket]

Lets look at it this way. If 2.2K is in series with an input that is high resistance, you have a voltage divider. The 9 volts will
will be proportionally divided between the external and "internal resistor". If the "internal resistor" inside the 555 is much higher than the
2.2k resistor almost all of the voltage will be at the input pin.

Sent from my SM-G900V using Tapatalk