# 555 in astable mode - Control pin role

#### ericgibbs

Joined Jan 29, 2010
11,154
hi Al,
This image shows even more detail.

Eric

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Assuming that by Ra you are referring to the resistor you have labeled R1 in your diagram above, you are correct but only to the extent that R4 is small relative to R1. If R4 is large, the R1 is not "discarded" and very much plays a role. In fact, R1 plays a role as long as R4 is non-zero, but the smaller R4 is, the less of a role it plays. Do the math.

Why and how are you concluding that "discarding" Ra results in less current charging the capacitor. How does the value of any voltage anywhere in the control voltage circuit have any impact on the current that is flowing in the capacitor (other than altering WHEN the circuit switches between charging and discharging modes)?

What happens if you set the control voltage to 75% of Vcc? What happens if you set it to 25% of Vcc? What observation can you make regarding your claim as a result?

Correct.

Look at the circuit! If you apply a voltage directly to the control pin, then what role does Ra play in determining the switching threshold?

Do the math! Whatever the voltage is at the control voltage (whether it is being overridden or not), the voltage at the trigger threshold is half of that. How long does it take a first-order circuit with a time constant tau to go from V1 to 0.5·V1?

Yes. And to see how, do the math!

I think I can finally answer to some of those questions...

I have told what I have understood from what someone explained me... Probably what was explained to me was more accurate than what I've understood.

Ok, so Ra is not discarded at all. But I was not understanding why that was told to me. Now I think I understand. By the fact that we have now a voltage applied to the Control pin, via V1 and R4, R1 is now dropping all (or almost all) voltage "coming" from V2. That's why I was told that R1 is kinda discarded!

Further, the capacitor charging time has, indeed, nothing to do with the fact of greater or less currents across the internal resistors that are part of the internal voltage divider. The fact is that, the voltage divider has now 2 resistors rather than 3 because of the reason stated above, and as these 2 resistors has the same value in Ω, then we have a pure voltage divider that will set the 2 new limits for the capacitor charge as v1 and v1/2. The dis/charging times are calculated with the 2 common capacitor voltage equations. I have done all the calcs and could prove all them with LTSpice!

If voltage at v1 (not V1) comes to 2/3 of Vcc, then the circuits behaviour is just the same as if the Control pin is left floating because we then have the same voltage, but coming from different sources!

Ok, sorry if I looked like I've abandoned the "fight", but no, I haven't.. In fact, I think I won the fight regarding the basic understanding of the Control pin role. It modulates the ON time of the output pulse. As a consequence, the frequency also changes because the OFF time is more or less constant because the discharge process has nothing to do with the Control pin. It discharges through one of the external resistors while the Control pin only has something to do with the internal resistors!

I hope I have not said any big non-sense anywhere!

#### WBahn

Joined Mar 31, 2012
26,305
You're getting pretty close, but it would be good for you to look at why the discharging time is not affected by putting applying a voltage to the Control pin. After all, the voltage that the discharge waveform starts out at and the voltage at which it stops ARE affected by the voltage on the Control pin, so why isn't the time that it takes to go between these two different pairs of voltages different as well?

#### WBahn

Joined Mar 31, 2012
26,305
hi Al,
This image shows even more detail.
View attachment 93221
Eric
That's too much detail for anything approaching this discussion -- a diagram needs to have a level of abstraction appropriate to what it is being used for.

Having said that, can anyone explain the orientation of Q1? Is that just an error in this schematic? I think it has to be since it seems like you would want it to mirror the orientation of Q4 in order for the threshold level to be set properly.

#### JoeJester

Joined Apr 26, 2005
4,390

#### MrAl

Joined Jun 17, 2014
7,849
Hi again,

Eric:
Thanks for posting that. Now we can make our own 555's all we need is about 25 transistors
With a little correction we can study it in more detail with that.

Richard:
Thanks for the little correction. I was wondering about that transistor too.

Psy:
I think I can finally answer to some of those questions...

I have told what I have understood from what someone explained me... Probably what was explained to me was more accurate than what I've understood.

Ok, so Ra is not discarded at all. But I was not understanding why that was told to me. Now I think I understand. By the fact that we have now a voltage applied to the Control pin, via V1 and R4, R1 is now dropping all (or almost all) voltage "coming" from V2. That's why I was told that R1 is kinda discarded!

Further, the capacitor charging time has, indeed, nothing to do with the fact of greater or less currents across the internal resistors that are part of the internal voltage divider. The fact is that, the voltage divider has now 2 resistors rather than 3 because of the reason stated above, and as these 2 resistors has the same value in Ω, then we have a pure voltage divider that will set the 2 new limits for the capacitor charge as v1 and v1/2. The dis/charging times are calculated with the 2 common capacitor voltage equations. I have done all the calcs and could prove all them with LTSpice!

If voltage at v1 (not V1) comes to 2/3 of Vcc, then the circuits behaviour is just the same as if the Control pin is left floating because we then have the same voltage, but coming from different sources!

Ok, sorry if I looked like I've abandoned the "fight", but no, I haven't.. In fact, I think I won the fight regarding the basic understanding of the Control pin role. It modulates the ON time of the output pulse. As a consequence, the frequency also changes because the OFF time is more or less constant because the discharge process has nothing to do with the Control pin. It discharges through one of the external resistors while the Control pin only has something to do with the internal resistors!

I hope I have not said any big non-sense anywhere!
What you could do is write the transfer function for the control pin. It takes a little thought, but you can then graph the various results such as discharge time, charge time, and total time for one cycle, and of course frequency, all controlled by the control pin voltage (x axis). The discharge time should be a constant horizontal line but everything else varies. You could of course also plot the duty cycle since that was what you were after i think, and that's very informative. You *might* find that the duty cycle has a lower and upper limit.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
You're getting pretty close, but it would be good for you to look at why the discharging time is not affected by putting applying a voltage to the Control pin. After all, the voltage that the discharge waveform starts out at and the voltage at which it stops ARE affected by the voltage on the Control pin, so why isn't the time that it takes to go between these two different pairs of voltages different as well?
Well, mathematically it's not dependant on the Control pin voltage because by the tOFF formula it only depends on Rb times C, in this case. And the Control pin only sets the new voltage references. But I might be missing here some detail!

Hi again,

Eric:
Thanks for posting that. Now we can make our own 555's all we need is about 25 transistors
With a little correction we can study it in more detail with that.

Richard:
Thanks for the little correction. I was wondering about that transistor too.

Psy:

What you could do is write the transfer function for the control pin. It takes a little thought, but you can then graph the various results such as discharge time, charge time, and total time for one cycle, and of course frequency, all controlled by the control pin voltage (x axis). The discharge time should be a constant horizontal line but everything else varies. You could of course also plot the duty cycle since that was what you were after i think, and that's very informative. You *might* find that the duty cycle has a lower and upper limit.
MrAl, I don't know if I would be able to do that. Anyway, that would be probably a very good idea, but I still have 2 more problems to solve and I don't have much time. So, that will have to wait. And to be honest, probably I won't be doing that because, I just don't have time.

I have two other assignments about signal processing or maybe I should say signal modulation and demodulation and as you can imagine, it has a lot of math and it takes a lot of time too! I'm really sorry!

#### MrAl

Joined Jun 17, 2014
7,849
Well, mathematically it's not dependant on the Control pin voltage because by the tOFF formula it only depends on Rb times C, in this case. And the Control pin only sets the new voltage references. But I might be missing here some detail!

MrAl, I don't know if I would be able to do that. Anyway, that would be probably a very good idea, but I still have 2 more problems to solve and I don't have much time. So, that will have to wait. And to be honest, probably I won't be doing that because, I just don't have time.

I have two other assignments about signal processing or maybe I should say signal modulation and demodulation and as you can imagine, it has a lot of math and it takes a lot of time too! I'm really sorry!

Hello again,

Ok, no problem

Then if i get a chance i'll work up the transfer function and post it. I think it would be good to have for anyone, including myself. I dont know why someone didnt do this yet anyway.
As you know, it's mostly about calculating the charge time as the discharge is always the same. Knowing the two voltage levels from the two-source resistor network allows us to calculate that too, from which we can easily get the total period and the duty cycle. There will be a slight non linear relationship for reasonable control voltage levels, but with an explicit formula we'll be able to calculated it quick.
If anyone has any objections to me posting it, speak now or i could just post a graph i guess.

#### WBahn

Joined Mar 31, 2012
26,305
Well, mathematically it's not dependant on the Control pin voltage because by the tOFF formula it only depends on Rb times C, in this case. And the Control pin only sets the new voltage references. But I might be missing here some detail!
You definitely are missing something. The tOFF formula very much depends on something other than Rb and C -- it depends on the voltage levels at which the falling waveform starts and at which it ends.

The voltage of a first-order waveform is given by:

v(t) = Vf + (Vi - Vf)e^(-t/tau)

For a waveform that is decaying toward zero we have Vf = 0 V, which leaves us with

v(t) = Vi · e^(-t/tau)

The time that the waveform spends in the falling mode before switching back to a rising waveform is the amount of time, Tf, it takes the waveform to fall from Vi, the initial voltage at the start of the falling period, to Vtrig, the trigger threshold voltage.

So what is the expression for Tf in terms of Vi, Vtrig, and tau?

When you change Vcntl, you change both Vi and Vtrig. So why does Tf not change? Show why mathematically.

#### WBahn

Joined Mar 31, 2012
26,305
MrAl, I don't know if I would be able to do that. Anyway, that would be probably a very good idea, but I still have 2 more problems to solve and I don't have much time. So, that will have to wait. And to be honest, probably I won't be doing that because, I just don't have time.

I have two other assignments about signal processing or maybe I should say signal modulation and demodulation and as you can imagine, it has a lot of math and it takes a lot of time too! I'm really sorry!
You are digging yourself a deeper and deeper hole. You are basically saying that, because you are struggling so much with the level of mathematics you are dealing with now that you feel you don't have time to learn how to do the mathematics well because you are going to need to spend even more time on ever harder math that you are increasingly less capable of doing. That will not end well.

The first thing to do when you are digging yourself into a hole is to stop digging!

What you need to do is take the time NOW to overcome your deficiencies in math at the level you should be at now. That will let you spend considerably LESS time on the math as you move forward.

#### MrAl

Joined Jun 17, 2014
7,849
You definitely are missing something. The tOFF formula very much depends on something other than Rb and C -- it depends on the voltage levels at which the falling waveform starts and at which it ends.

The voltage of a first-order waveform is given by:

v(t) = Vf + (Vi - Vf)e^(-t/tau)

For a waveform that is decaying toward zero we have Vf = 0 V, which leaves us with

v(t) = Vi · e^(-t/tau)

The time that the waveform spends in the falling mode before switching back to a rising waveform is the amount of time, Tf, it takes the waveform to fall from Vi, the initial voltage at the start of the falling period, to Vtrig, the trigger threshold voltage.

So what is the expression for Tf in terms of Vi, Vtrig, and tau?

When you change Vcntl, you change both Vi and Vtrig. So why does Tf not change? Show why mathematically.

Hi there,

Are we talking about the same circuit? I thought we already established that because the upper voltage before discharge is 2 times the lower limit value that the discharge time is:
t=log(2)*RC

and that is because for Vx the lower limit then the upper limit is 2*Vx so we have:
Vx=2*Vx*e^(-t/RC)

and of course dividing to simplify this we get:
1/2=e^(-t/RC)

and of course solving for t we get:
t=log(2)*RC

But maybe you were talking about discharging all the way to zero?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok guys, I have done all the math for the charge and discharge timings.

I can post here the file where I have all the math done but it is in Portuguese and I can't spend time translating it because I need that time to move on in the assignment! Actually the file that I can upload here is the assignment pre-lab work (theoretical work and LTSpice stuff), so it's kinda personal! Hope no one uses it with bad intent!

You can check the calcs and the simulations at point 2.3.

#### WBahn

Joined Mar 31, 2012
26,305
Hi there,

Are we talking about the same circuit? I thought we already established that because the upper voltage before discharge is 2 times the lower limit value that the discharge time is:
t=log(2)*RC

and that is because for Vx the lower limit then the upper limit is 2*Vx so we have:
Vx=2*Vx*e^(-t/RC)

and of course dividing to simplify this we get:
1/2=e^(-t/RC)

and of course solving for t we get:
t=log(2)*RC

But maybe you were talking about discharging all the way to zero?
No, we are talking about the same circuit, but it is pretty evident (or so it seems) that Psy does not recognize that the magic that results in the discharge time being independent of the control voltage is NOT because it is only dependent on the RC value, but, rather that it is because trigger threshold is always half of the control voltage whether the control pin is overdriven or not.

When analyzing the case of the control pin being left open, the analysis is almost always done with one threshold being 2·Vcc/3 and the other being Vcc/3. While it is often pointed out that this makes the circuit insensitive to the actual value of Vcc, it is seldom pointed out that when the control voltage is overridden that you end up with the same factor-of-two relationship between the thresholds that you had with the control voltage being internally generated and, further, that while this results in the discharge time being nominally unaffected, it changes the charge time precisely because the two thresholds, relative to Vcc, are locked to the same proportion.

#### WBahn

Joined Mar 31, 2012
26,305
Ok guys, I have done all the math for the charge and discharge timings.

I can post here the file where I have all the math done but it is in Portuguese and I can't spend time translating it because I need that time to move on in the assignment! Actually the file that I can upload here is the assignment pre-lab work (theoretical work and LTSpice stuff), so it's kinda personal! Hope no one uses it with bad intent!

You can check the calcs and the simulations at point 2.3.
As an aside (and I may have mentioned this before), your English is quite good.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
As an aside (and I may have mentioned this before), your English is quite good.
Thanks for the compliment! I'm not sure if you're talking about the wording or about the exclamation and question marks, periods and so on, but I always try to be as accurate as I can possibly be. I always try to do it also in my native/mother language. My mother was a teacher. 1st grade teacher, I mean, she used to teach young students since 1st to 4th grade... So she was also my teacher for those 4 years and I like to think that I always try to follow her steps in terms of speaking and writing the correct way!

Of course that, as English is not my mother language, probably there are a lot of mistakes and some misunderstandings when I read what others try to explain to me and also sometimes I find it hard to make my self clear because there are a lot of specific words with specific meanings that I'm not always aware! Anyway, I always try my best!

#### WBahn

Joined Mar 31, 2012
26,305
You do a very good job. In many cases it seems like your word choice or something else might be just a little bit odd, but usually when I try to look at it more closely I have a hard time identifying anything specific. It's close enough that I usually can't spot anything other than that it is just slightly different from what would seem "natural" to me.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok, nice, it is good to hear that!

Anyway, I'm on another problem right now! I need to control a servo-motor with the 555 timer. Maybe I should start another thread, no? I'm struggling a little bit to make it work on LTSpice!

#### WBahn

Joined Mar 31, 2012
26,305
I'd recommend starting a new thread for the new problem. Otherwise it becomes a confusing mess.