What is the output of this UJT relaxation oscillator circuit?

[Femme_physics]

Homework Statement

This is a relxation oscillator that's realized by UJT. The graph is included.

What is the output of the circuit?


http://img821.imageshack.us/img821/2274/ujtok.jpg

Homework Equations

http://www.allaboutcircuits.com/vol_3/chpt_7/8.html

The Attempt at a Solution

Well, to find the output I need VEB or VBB which are not given to me...therefor unless I make the assumption that VEB = 0.7 volts this cannot be solved. And I can just say that

Vout = VR2

 

[I like Serena]

Hey Fp!

Seems to me that without the specification of the UJT you are indeed limited in what you can do.

What you can do, is give the shape of the output signal.
You can also give the frequency of the output signal.
And you can calculate the value of the capacitor.

 

[Femme_physics]

Yes I know the value for the capacitor :) But thanks for confirming it's a dead end-- I thought as much!

 

[NascentOxygen]

Everything good again with your teacher, FP? Now that the dust has settled.

Well, to find the output I need VEB or VBB which are not given to me...

When designing with active devices, you'll never be "given" precise specifications, for the simple reason that the specs vary from one device to the next, from one production batch to the next, and from manufacturer to manufacturer. Generally, the best you can hope for is a range, or, next best, a typical value. With this, the designer does the best she can. As for VBB, you are as good as given it; if you know it to an accuracy of 10% then consider it is known.

unless I make the assumption that VEB = 0.7 volts

You feel that's an unreasonable assumption to make? :uhh:

It's not clear whether the question sheet referred to the allaboutcircuits site, and its figures for the 2N2647, or whether you added that on your own initiative, but I expect those figures should be fine for the calculations here. Note that the amplitude and width of pulses for triggering may not be of major importance, as the signal is either certain to be more than adequate for the need, or else it will be fed to a buffer or attenuator, so your not being able to specify an output level to an accuracy of 3 decimal places is not the calamity you may think.

just say that Vout = VR2

10/10 :rofl: :rofl:

 

[Femme_physics]

Thanks Nascent, I can write an explanation about the assumption I made at worst case scenario.

Everything good again with your teacher, FP? Now that the dust has settled.

Pretty much, I just think he needs more experience teaching. He's still a young guy. While we're on the issue of UJT and seeing it's such a short thread, I thought to throw in another question... http://img403.imageshack.us/img403/5214/question99.jpg

Given:
Ic = Ie
Vp = η x VBB + 0.63
Vv = 1.2 volts
VEB = 0.65
η = 0.4

Calculate the frequency of the circuit pulses. Neglect the discharge time of the capacitor.
ANSWER:

I wasn't sure how to answer it, so I picked at a notebook I acquired of someone who already took my electronics course and this is what he wrote:

http://img441.imageshack.us/img441/5263/answer99.jpg

Where is the formula for t0 and t1 taken from? And what does it mean?

 

[NascentOxygen]

Where is the formula for t0 and t1 taken from? And what does it mean?

The formulae will fall into place once you have figured out how it works.

First, you'd better identify the <oh, no, not this again> EMITTER of that BJT. :tongue:

Next question to ponder, what is that ZENER doing?

I just think he needs more experience teaching. He's still a young guy.

You'll soon fix that; you're aging him fast.

 

[Femme_physics]

First, you'd better identify the <oh, no, not this again> EMITTER of that BJT.

*sets hair on fire and jumps from Everest!*Well... The zener defines the voltage at the cross-section above the zener to the ground. *cleans snow from hair*... The emitters are the arrows. There is one emitter for the NPN transistor and one emitter for the UJT

 

[NascentOxygen]

You have a good understanding of the UJT oscillator, so move your focus to the BJT in this modified oscillator.

Providing the zener is adequately fed, it will maintain the base of the NPN at a constant voltage. What would follow on, as a consequence of this, to that NPN's voltages and currents?

 

[Femme_physics]

Before I answer this, let's clear the details...

This is not an NPN, this is a PNP!

And this is not a BJT, it's a UJT!

Or...am I blind?

 

[NascentOxygen]

Before I answer this, let's clear the details...

This is not an NPN, this is a PNP!

You're right.

PNP and NPN transistors are examples of Bipolar Junction Transistors.

And this is not a BJT, it's a UJT!

This circuit is a UJT relaxation oscillator, but it is augmented by a PNP BJT.

 

[Femme_physics]

This circuit is a UJT relaxation oscillator, but it is augmented by a PNP BJT.

I'm not sure what the last part means...if there's no BJT defined in the problem, then there is no BJT. How can this logic be flawed?
Bet let me try answering your original question

Providing the zener is adequately fed, it will maintain the base of the NPN at a constant voltage. What would follow on, as a consequence of this, to that NPN's voltages and currents?

Well, it will affect the amount of current passing through the circuit but this has to be calculated to figure out

 

[NascentOxygen]

There is sufficient information on the schematic to enable you to calculate the PNP's emitter current and voltage.

You can see that a resistor of the original UJT oscillator (that you examined in an earlier thread), has been replaced by a PNP transistor in the circuit you are examining here.

A PNP transistor is also known as a BJT. (The same goes for an NPN, it also is a BJT.)

 

[Femme_physics]

Can't they just call things one name? :( Fine.

I did find the current IE all by myself :)

http://img600.imageshack.us/img600/4098/ieic.jpg

And Ie = Ic according to the problem definition The problem is that they're asking for the " frequency of the circuit pulses"... I'm unsure how to relate it all, and that new formula is really confusing

 

[NascentOxygen]

I'll accept your word that IE is 3.2mA. Your approach looks right.

As you know, IE is approx equal to IC.

So, where does the path for IC lead to, after it emerges from the PNP transistor?

 

[NascentOxygen]

P.S. I shall have to depart shortly (in 10 mins). Back in 9 or 10 hrs.

 

[Femme_physics]

I'll accept your word that IE is 3.2mA. Your approach looks right.

As you know, IE is approx equal to IC.

So, where does the path for IC lead to, after it emerges from the PNP transistor?

well, to the UJT or the Capacitor, depends on the time.

P.S. I shall have to depart shortly (in 10 mins). Back in 9 or 10 hrs.

ahh...drat ;) Well, we'll rehash this tomorrow then if you will...thanks!

 

[NascentOxygen]

well, to the UJT or the Capacitor, depends on the time.

Just as with the basic UJT oscillator, for most of the time that source of current is doing nothing but charging up the capacitor. So, the only change we have essentially made here is to replace the resistor in the basic RC timing network with a constant current source. How is that going to affect the shape of the waveform at the EMITTER of the UJT?

The importance of encountering alternative technical terms was brought home to me as a student. We had an exam, and I was pleased with it as we had thoroughly covered the work just a few days before. But some of the other students were not happy with it. One of the questions asked us to "explain the need to bias a Bipolar Junction Transistor", and my classmates claimed we hadn't been taught that topic. I realized they were right, we had thoroughly explored biasing NPN transistors, but the lecturer may not have even once referred to them as BJTs. Though of course our textbooks did.

 

[I like Serena]

Hi Fp.

Do you have formulas for a capacitor being charged?

You should have something like:
Q = C V
where Q is the charge on the capacitor, C is the capacity, and V is the voltage across the capacitor.

And you should also have something like:
ΔQ = I Δt
where ΔQ is the change of charge on the capacitor, I is the constant current, and Δt is the time interval that the capacitor is charged.

In your case the capacitor is charged from the V_v voltage to the V_p voltage.

 

[I like Serena]

Btw, I liked how you found Ie. Good!

 

[Femme_physics]

:)

Then, I can just rewrite formula and voila!

http://img818.imageshack.us/img818/5133/qcvc.jpg

That's the formula for the circuited pulses, from Vv to Vp!

 

[I like Serena]

Yep. That's it! :)

(Assuming your V is actually ΔV for the difference between Vp and Vv.)

Btw, the formula ΔQ = I Δt only works if the current I is constant.

Did you have those formulas? Or other formulas like it?

 

[Femme_physics]

I had the Q = VC

I did not have the Q = IT anywhere in my notebook. Problem is we're way behind in catching up with all the material in industrial electronics before the test (poor planning of our courses), so I'm going over UJT's, PUT, LPF's and relays. Though we covered the stuff in electricity and digital electronics well. Final external electronics test is at 18th this month.

 

[I like Serena]

Perhaps you have something like this?
$$I = {dQ \over dt}$$
This is actually the definition of current.
(Or from an SI point of view, the definition of charge.)

 

[NascentOxygen]

That's the formula for the circuited pulses, from Vv to Vp!

Hence, you find frequency of oscillation using t₁.

You quoted also a formula for t₀. Was t₀ explained, either here or in the earlier circuit? It has a close resemblance to t₁ ...

 

[Femme_physics]

Yes, except you don't minus Vv in t0..since you start from zero! BTW,

I assume as in before Vout = Vp + Veb ?

 

[NascentOxygen]

Yes, except you don't minus Vv in t0..since you start from zero!

You probably have it, but the explanation is too brief, lacks detail.

I assume as in before Vout = Vp + Veb ?

Output levels should be much the same if UJT hasn't changed.

 

[Femme_physics]

You probably have it, but the explanation is too brief, lacks detail.

Well, in the exercise I'm asked to graph it not put it into words :)

Output levels should be much the same if UJT hasn't changed.

You mean what I said is correct, right?

 

[I like Serena]

Well, in the exercise I'm asked to graph it not put it into words :)

You mean what I said is correct, right?

I didn't see pictures?

How can we say if it's correct without pictures! :p

 

[Femme_physics]

It's the same exercise I posted in pg1 -- sorry, thought we're on the same pagehttp://img403.imageshack.us/img403/5214/question99.jpg

 

[I like Serena]

Not that picture!

I want to see other pictures. The graph of Vout for one.

 

[Femme_physics]

Oops, sorry! I guess I was so sure I'm right... ;)

http://img513.imageshack.us/img513/5575/vpvpvpvvvvvv.jpg

 

[I like Serena]

Hmmm... no flowers or anything!

190/100

 

[NascentOxygen]

I assume as in before Vout = Vp + Veb ?

Was this established somewhere before? If you mark "voltage arrows" on your schematic, using KVL, you'll be able to demonstrate whether this should be a + or a minus Veb.

You are deciding here how to sketch the Vout pulse, and you need Vout's maximum value in order to sketch it. We know Vp (from Vcc and the manufacturer's data), so what value are you going to use for V_EB in the above equation?

 

[Femme_physics]

Was this established somewhere before? If you mark "voltage arrows" on your schematic, using KVL, you'll be able to demonstrate whether this should be a + or a minus Veb.

You are deciding here how to sketch the Vout pulse, and you need Vout's maximum value in order to sketch it. We know Vp (from Vcc and the manufacturer's data), so what value are you going to use for V_EB in the above equation?

You're right, I confused the signs

http://img825.imageshack.us/img825/336/signsz.jpg

 

[NascentOxygen]

Can you check that last line ...