There is 9v "on top" of R1/R2. However, there is a voltage "drop" across R1/R2 due to the current flowing through them.
E=IR, or Voltage = Current x Resistance.

The voltage at the junction of R2 and the timing capacitor is the capacitor's charge voltage at any given time. If the threshold pin weren't connected there or pin 7 were not connected, the capacitor would eventually charge to 9v.

 

There is 9v "on top" of R1/R2. However, there is a voltage "drop" across R1/R2 due to the current flowing through them.
E=IR, or Voltage = Current x Resistance.

The voltage at the junction of R2 and the timing capacitor is the capacitor's charge voltage at any given time. If the threshold pin weren't connected there or pin 7 were not connected, the capacitor would eventually charge to 9v.

Alright, This is what I was trying to get at the whole time

Just to be sure. You would consider it a RC circuit right? And as the capacitor fills with the charge, the voltage accross R1/2 decreases. If like SGTWookie said, the discharge pin were not there, then the Capacitor would fill to 9v and there would be no current flowing though that section at all. Is that correct?

 

Alright, This is what I was trying to get at the whole time

Just to be sure. You would consider it a RC circuit right? And as the capacitor fills with the charge, the voltage accross R1/2 decreases. If like SGTWookie said, the discharge pin were not there, then the Capacitor would fill to 9v and there would be no current flowing though that section at all. Is that correct?

Can anyone confirm this?

 

Yes, that's correct. It IS an RC circuit. The capacitor would eventually charge to nearly 9v - except that real-world capacitors aren't perfect; they always have some "leakage current" - but it would be very close to 9v, and the voltage across R1/R2 would be nearly zero.

 

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Why when the discharge pin is connected to ground, does the electricity not flow through Ra/1 to ground? At least that's what I assume that's what that part of the sim is showing.

 

Well, it does. Turning on the internal transistor makes a path for the charge on the timing capacitor, and lets current come through Ra. That does not matter, as the point between Ra and Rb is effectively ground. Pulling charge off that capacitor is the only thing that matters. Current through Ra makes no difference.

 

Well, Beenthere's basically correct, but the discharge pin does have a limit as to how much current it can sink, and Ra/R1 has a limit as to how much power it can dissipate.

If Ra/R1 is near 0 Ohms, as soon as pin 7 tries to go low, it'll blow the lid off the timer.

Best to keep the current from Ra/R1 below around 7mA. The exact current limit for pin 7 (discharge) is hard to find, but ST Microelectronics' datasheet for their bjt 555 equivalent mentions a 15mA limit. So, if Ra/R1 is sized to limit current flow to 7mA or less for 2/3 Vcc, you should be OK. By the same token, Rb/R2 should be sized appropriately.

 

I have a 6 volts power supply(reads 6.06 on the meter), and if I remember right I put a 150ohm resistor there. That would be 40mA running through Ra when the discharge pin is on. I've left the timer running for 24hours with no problem. Maybe I should consider changing out my resistors anyways.

2nd question. This might be a hard one to answer. I was doing some voltage measurements while the 555 was running. Just to really get a good handle on things. I'm using a 6volts wallwart to power it. R1 is 150ohms, R2 is 2.7Kohms. And C1 is 330uF. I'm running in astable mode.

So as I was poking around and I got most of the voltages I was expecting. Except for when I measured across R1. The voltage would swing from 0.2 volts to 5.4 volts. I understand where the .2 would come from(As the cap fills current across it is lowered as R2 takes up most of the voltage since it is 10x bigger). But where does the 5.4 come from? I assume I am getting that because the ground is connected to pin 7 and current is flowing from V+ through R1 to pin7/ground. But I don't understand why it would be 5.4 and not 6.

Is there internal resistance in the 555 or am I missing something....again. Can anyone shed some light on this.

 

Is there internal resistance in the 555 or am I missing something....again. Can anyone shed some light on this.

There has to be, since the monostable uses that transistor to short the capacitor.

I came in late on this thread, so I've missed a lot, and haven't read everything yet, but I've got another set of tutorials going, half finished at this point.

The 555 Projects

 

I have a 6 volts power supply(reads 6.06 on the meter), and if I remember right I put a 150ohm resistor there. That would be 40mA running through Ra when the discharge pin is on. I've left the timer running for 24hours with no problem. Maybe I should consider changing out my resistors anyways.

I just downloaded a fresh datasheet from ST Micro. Turns out the 15mA specification I remembered was not a limit; it was the current where they measured the saturation voltage (max 450mV with Vcc=15v) on the discharge pin. The only limit on current sinking via pin 7 is the package power dissipation - which wasn't specified, but it's an industry standard 8-pin DIP. You could probably safely sink 200mA when Vcc=15v, or around 80mA@ 6v. However, you'd really be better off using a smaller cap and larger resistors to achieve the same timing.

2nd question. This might be a hard one to answer. I was doing some voltage measurements while the 555 was running. Just to really get a good handle on things. I'm using a 6volts wallwart to power it. R1 is 150ohms, R2 is 2.7Kohms. And C1 is 330uF. I'm running in astable mode.

So as I was poking around and I got most of the voltages I was expecting. Except for when I measured across R1.

You mean from ground to the R1/R2/Pin 7 junction, right? Or are you measuring from Vcc to the R1/R2/Pin 7 junction?

The voltage would swing from 0.2 volts to 5.4 volts. I understand where the .2 would come from(As the cap fills current across it is lowered as R2 takes up most of the voltage since it is 10x bigger). But where does the 5.4 come from? I assume I am getting that because the ground is connected to pin 7 and current is flowing from V+ through R1 to pin7/ground. But I don't understand why it would be 5.4 and not 6.

Take a look at the internal schematic of the 555. There's a voltage divider network consisting of three 5k Ohm resistors, for a total series resistance of 15k. You can access the upper junction of this network at pin 5 (the CTRL pin). The voltage divider network establishes the upper and lower limits for the trigger levels, at 2/3 Vcc and 1/3 Vcc, in your case about 4v and 2v.

The discharge pin 7 will discharge the capacitor via R2 until the low threshold is reached. Then the discharge pin turns off, and C1 charges up again via R1-R2 until the high threshold is reached.

And I'm thinking you have a 1.5k resistor (black green red) instead of a 150 Ohm resistor (black green brown).

Measuring across C1, you're getting about 2v to 4v, right?

 

I just downloaded a fresh datasheet from ST Micro. Turns out the 15mA specification I remembered was not a limit; it was the current where they measured the saturation voltage (max 450mV with Vcc=15v) on the discharge pin. The only limit on current sinking via pin 7 is the package power dissipation - which wasn't specified, but it's an industry standard 8-pin DIP. You could probably safely sink 200mA when Vcc=15v, or around 80mA@ 6v. However, you'd really be better off using a smaller cap and larger resistors to achieve the same timing.


You mean from ground to the R1/R2/Pin 7 junction, right? Or are you measuring from Vcc to the R1/R2/Pin 7 junction?


Take a look at the internal schematic of the 555. There's a voltage divider network consisting of three 5k Ohm resistors, for a total series resistance of 15k. You can access the upper junction of this network at pin 5 (the CTRL pin). The voltage divider network establishes the upper and lower limits for the trigger levels, at 2/3 Vcc and 1/3 Vcc, in your case about 4v and 2v.

The discharge pin 7 will discharge the capacitor via R2 until the low threshold is reached. Then the discharge pin turns off, and C1 charges up again via R1-R2 until the high threshold is reached.

And I'm thinking you have a 1.5k resistor (black green red) instead of a 150 Ohm resistor (black green brown).

Measuring across C1, you're getting about 2v to 4v, right?

Ok I did some more test. R1 is 125ohms.

Across C1 is around the 2-4v range.

If I measure VCC to the Junction then I get the 0.2 to 5.4

If I measure from Ground the the junction I get about 0.65 to 5.95.

Does that help a little?

 

Ok I did some more test. R1 is 125ohms.

That's pretty low. While the discharge pin 7 is turned on (sinking current), there's around 45mA flowing through R1. That's over 1/4 W.

Across C1 is around the 2-4v range.

It's cycling back and fourth between the upper (4v) and lower (2v) thresholds.

If I measure VCC to the Junction then I get the 0.2 to 5.4

If I measure from Ground (to) the the junction (of R1/R2/Pin 7) I get about 0.65 to 5.95.

OK, so are you understanding why you're reading those values now?

You're sinking a lot of current via pin 7, which is increasing pin 7's saturation voltage to around 0.65v. If you increase R1 to perhaps 1.5k, the saturation voltage should decrease significantly.

 

That's pretty low. While the discharge pin 7 is turned on (sinking current), there's around 45mA flowing through R1. That's over 1/4 W.


It's cycling back and fourth between the upper (4v) and lower (2v) thresholds.



OK, so are you understanding why you're reading those values now?

You're sinking a lot of current via pin 7, which is increasing pin 7's saturation voltage to around 0.65v. If you increase R1 to perhaps 1.5k, the saturation voltage should decrease significantly.

So what are you saying? It should be 0V? What exactly do you mean by saturation voltage? It's sinking as much current as it can and that .65V is the "left overs"?

 

So what are you saying? It should be 0V? What exactly do you mean by saturation voltage? It's sinking as much current as it can and that .65V is the "left overs"?

Yes, it's measuring .65v because it's "saturated" with current flow. It won't go all the way to 0v (unless you disconnect R1/R2 from it), but the voltage will get a lot lower if you increase R1's value.