Electronics / Engineering Gurus QUESTION

[Davetronz]

Hey boys, who wants to help me out. I need solutions to this fucking problem.

Relaxation Oscillator

In the above circuit the neon bulb with fire (conduct) if the voltage reaches 68V. Prior to firing the bulb looks like an OPEN, once the bulb has fired it looks like a SHORT. The bulb will stay conducting as long as the current through the bulb remains over 1 mA. The switch is closed at t=0. Plot a graph of the VOLTAGE ACROSS THE CAPACITOR, Vc, for two complete cycles. Show, (calculate) all MAXIMUM and MINIMUM voltages across the CAPACITOR and the TIMES that these voltages occur. Show calculations



Anything that might help me will definately help... Take a look at the diagram I have attached.

R1 = 2 Mega Ohms
C1 = 1 Micro Farad
R2 = 10 Kilo Ohms
E = 100V
Neon Bulb

[rsxrsx]

Is this for Kelly Mahn's Class? Where are you taking electronics at SAIT?

[Dirty_SOHC]

haha if you are I sell you my book "Cheap". I took that 1st year course too

[rsxrsx]

Alright, well first off you can limit out R1 as this is the internal resistance of the switch (should affect your calculations). Next, work out the time contants and plot the capacitor charging and dis-charging graphs, as you know already know when the cap. is fully charged it will act like a power supply (but not constant) that's why you have to calculate it out, the cap will dischagre through R2 the 10k Ohm resistor thus illuminated the neon bulb as soon as the cap discharges all it's voltage your neon will turn off.

The key to solve this problem is work with tau (time constant) then my friend you'll figure everything out, good luck bro

[Davetronz]

Originally posted by Dirty_SOHC
haha if you are I sell you my book "Cheap". I took that 1st year course too

That person isnt me, I think you are thinking of mrmattyk, I know he was selling his books this year.

Thanks for the help boys, I am still working on her...
In response to the question, no I am in Colin Chamberlain for lecture and Glen Stephens for lab... So in other words, I havent learnt SHIT this year... Oh and I had Rick Koinberg for one day, and all I got was second hand smoke...

I have my final in 3 days and I am sooo fucking lost, along with all of my classmates...

[sml]

Ok, this is my take on this problem:
R1 and C1 makes up the time constant to charge C1 when the switch is closed since neon bulb is open circuit when it's not conducting. t=RC=(2M)(1u)=2seconds to charge the cap to 63% of the initial voltage (if we assume we start at 0V, that'd be 63V). Now, since the neon bulb works at 68V and 1mA and since current stays the same inside a loop, there should be 1mA running through R2. Vr2=IR=(1m)(R2)=1V. So, there's 68V going through the neon bulb when it's conducting and 1V going through R2, that'd mean 69V is required to turn this puppy on at C1. Now, since 63V < 69V, this would mean that it'd take just a bit longer than 2seconds to charge C1 to 69V. When C1 reaches 69V, the neon bulb will conduct meaning that it's a short to ground. That means you are technically discharging C1 through R2. t=RC=(10k)(1u)=10mseconds. So, from t=0s, you have 2s where it's chraging C1 to 69V, then for 10ms it will discharge from 69V back to 43V (37% of 69V discharge) and the cycle repeats... I'm sure if you REALLY want to get into the nity grity of it all, you can calculate how the cap is still being charged while it's being discharged and that'd make the discharge time longer, etc, etc, etc... but you get the big picture, how this circuit's suppose to work. Hope you can take it from here and plot your graph. I dunno how much detail you're being asked to do, but if you really want, you can plot the exponential charge and discharge curve using the equation and calculate the exact time to charge to 69V (probably 2.2s), but in reality, it's damn close enough and nobody actually will care! Hope this helps.

[sml]

Originally posted by rsxrsx
Alright, well first off you can limit out R1 as this is the internal resistance of the switch (should affect your calculations). Next, work out the time contants and plot the capacitor charging and dis-charging graphs, as you know already know when the cap. is fully charged it will act like a power supply (but not constant) that's why you have to calculate it out, the cap will dischagre through R2 the 10k Ohm resistor thus illuminated the neon bulb as soon as the cap discharges all it's voltage your neon will turn off.

The key to solve this problem is work with tau (time constant) then my friend you'll figure everything out, good luck bro

If R1 is the internal resistance of the switch, then that's a pretty useless switch since the whole purpose of it is to create a connection (short circuit) across 2 points and a short circuit is 0 ohms, nowhere near 2M ohms.

[rsxrsx]

Originally posted by sml


If R1 is the internal resistance of the switch, then that's a pretty useless switch since the whole purpose of it is to create a connection (short circuit) across 2 points and a short circuit is 0 ohms, nowhere near 2M ohms.

You don't want the switch to act like a load, that's why the resistance is so high, that's what i gather from that circuit anyway, look how the cap will charge and dis-charge, it will discharge through the 10k resistance, R1 won't even be effected...so why are you confused?

[91Accord]

Glen's a joke... he doesn't seem to know much, I had that guy years ago... and colin, dear god he was a bad coordinator

[gpomp]

I have no idea what you guys are talking about but I bet SML is right!

[Davetronz]

Originally posted by 91Accord
Glen's a joke... he doesn't seem to know much, I had that guy years ago... and colin, dear god he was a bad coordinator

Colin is horrible.... If you thought he was a bad coordinator... Wait till you see the guy teach... We have 3 chapters in AC to cover with 2 hours left... We are gonna die in the final... Which Glen wrote by the way...

[sml]

Originally posted by rsxrsx


You don't want the switch to act like a load, that's why the resistance is so high, that's what i gather from that circuit anyway, look how the cap will charge and dis-charge, it will discharge through the 10k resistance, R1 won't even be effected...so why are you confused?

I agree that the cap will discharge through R2, but how do you think it charges up? According to the question, it says that the neon bulb acts like an "open circuit" when it's not conducting (initiate state) as shown in fig1.

As you can see, if R2 is open circuit (ie not connected), then R2 is floating in mid air... no current is gonna flow through it if the other end is not connected to anything. What you then have is E in series with R1 and parallel to C1. R1 is what we usually call the current limiting resistance. It current limits since I=V/R and V and R = constant, then I is also a constant of known value.... so why are you still confused?

[GTS Jeff]

hmm hah yes....

nice vid!

[Davetronz]

Thanks SML, what you said makes PERFECT sense and it works out quite well on paper! Thanks for the pointers bro!
Much appreciated...
And I will see if I can allocate the funds to buy your bike off you, but I wont need the gear cus it wont fit a bigger white boy like me...

Dave

[rsxrsx]

Originally posted by sml


I agree that the cap will discharge through R2, but how do you think it charges up? According to the question, it says that the neon bulb acts like an &quot;open circuit&quot; when it's not conducting (initiate state) as shown in fig1.

As you can see, if R2 is open circuit (ie not connected), then R2 is floating in mid air... no current is gonna flow through it if the other end is not connected to anything. What you then have is E in series with R1 and parallel to C1. R1 is what we usually call the current limiting resistance. It current limits since I=V/R and V and R = constant, then I is also a constant of known value.... so why are you still confused?

Ok, Ok, Ok, This is what I'm trying to say, I look at the cap. charging almost instantly due to the fact that R1 has such a high resistance, of course R2 and the neon will not be affected during charge up, but how is a switch made up i.e contacts and the internal resistance of the actual swtich, now for schooling purposes yes, you would take into consideration the internal resistance of the switch, but in the field for example you would simply assume that the cap charges instaneously, our biggest concern is the neon in this case, I'm not saying by any means that you are wrong, i agree with you 100%, all I'm trying to say is there any harm in assuming the cap has instant charge on it. Hey, that's just my prespective on this problem, if you teacher wants you to due it the was that you explained, then do it, don't listen to me, trubo was just asking for some help, so i tryed to the best of my albilites to help him out, that's all, sorry.

[sml]

Originally posted by rsxrsx


Ok, Ok, Ok, This is what I'm trying to say, I look at the cap. charging almost instantly due to the fact that R1 has such a high resistance, of course R2 and the neon will not be affected during charge up, but how is a switch made up i.e contacts and the internal resistance of the actual swtich, now for schooling purposes yes, you would take into consideration the internal resistance of the switch, but in the field for example you would simply assume that the cap charges instaneously, our biggest concern is the neon in this case, I'm not saying by any means that you are wrong, i agree with you 100%, all I'm trying to say is there any harm in assuming the cap has instant charge on it.

Well, don't take this the wrong way, I'm not being aggressive or anything, but after working in the field for a couple of years you get to learn some times. I'm not saying I'm 100% right, but quite frankly I think I'm at least 95% right. Just for argument sake, I'll try to explain why I think you cannot assume that the cap charges up "instantaneously"... You should not ever connect a cap directly to the power supply, doing this will QUICKLY bring the 2 voltage sources to the same potential ie.) 100V. Check this equation for I=Cdv/dt. C is a constant, dv/dt is the change in potential for that cap, in this case if it's "instantaneous", then that means a very big number, right? So, if C is constant and dv/dt is a big number, then that means I is also a very big number. Then what you'll get is a little puff of smoke and deaf ears. Most caps have a max current rating and if you charge it up too fast, which means that it'll exceed it's current rating and it'll blow up on you. You probably don't have to worry much about little tantalum caps, but when you start working with electrolytics and lead acid (car batteries), then then little bang sound you hear will be followed by a nice shower of sulphric acid. Yummy. That's why R1 is there to limit the current so that the cap does not charge up instantaneously. It's also what you'd use in your calculations to find the time constant (tau, which you mentioned before)... t=RC=2seconds in this case. If you read the subject title to the question, it even says that it's a relaxation oscillator. Which to me implies that it's an oscillator (goes in a cycle) and it relaxes (charges/discharges)... Here's a diagram of what I think the capacitor output will be close to (not to scale cause 2 seconds is not the same size as 10 mseconds):

[Davetronz]

Ok boys, the question has been graded, sml's concept of it was 100% correct!
Thanks for all your help guys! Everyone that helped!

Dave