Capacitor question and presentation

[shadow_2609]

Hi everyone! My name is Alex an this is my first post on the forum! I am into micro controller and digital electronics. I am a really low budget electronic hobbyist.
As i mentioned I am into digital stuff, so when I have to do something analog i am screwed! It could be a noob question, but here it is.
I was reading a tutorial for the transistor (my memory is like DRAM, it does need refreshing), and I reached the part of the oscillator, in particular the figure 42 of this page http://www.talkingelectronics.com/te_interactive_index.html ... here it says "When the power is applied, the 10u gradually charges and allows a voltage to develop on the base of the NPN transistor. When the voltage reaches 0.6v, the transistor turns ON and this turns on the PNP transistor. ", simple enough; but then he say "The voltage on the collector of the PNP transistor increases and this raises the right side of the 10u electrolytic and it firstly pushes its charge into the base of the NPN transistor. Then the 330k takes over then it continues to charge in the opposite direction via the base-emitter junction of the NPN transistor. " .
I already saw something similar on episode 708, the debunking of that overunity thing without understanding enough the cap ... My question is this, if i have 0.6 volts from X to Y, when Y is 4 volt, in X i have a raising from 0.6 to 4.6? i have 4.6 to 0.6?  if I charge plate X through the  22 ohm resistor, how can Y charge through X?
Thanks in advance
Alex

[BobsURuncle]

in particular the figure 42 of this page http://www.talkingelectronics.com/te_interactive_index.html ...

I see no figure 42, post the relevant text and images here.

[shadow_2609]

Sure, sorry wrong link!

This effect is called POSITIVE FEEDBACK and the circuit will get turned ON until it cannot turn on any more.
But we haven't joined points "X" and "Y" DIRECTLY (we have used a capacitor) so we have to start again and explain how the circuit works.
When the power is applied, the 10u gradually charges and allows a voltage to develop on the base of the NPN transistor. When the voltage reaches 0.6v, the transistor turns ON and this turns on the PNP transistor.
The voltage on the collector of the PNP transistor increases and this raises the right side of the 10u electrolytic and it firstly pushes its charge into the base of the NPN transistor. Then the 330k takes over then it continues to charge in the opposite direction via the base-emitter junction of the NPN transistor. This causes the two transistors to turn ON more. This keeps happening until both transistors cannot turn ON any more and the 10u keeps charging. But as it continues to charge, the charging current eventually drops slightly and this turns off the first transistor slightly. This gets passed to the PNP transistor and it also turns off slightly. This instantly lowers both leads of the 10u and both transistors turn OFF.
The 10u is partially charged and it gets discharged over a long period of time by the 330k resistor and when it starts to charge in the opposite direction, the base of the first transistor sees 0.6v and the cycle starts again.
The end result is a very brief flash and a very long pause (while the capacitor starts to charge again).
As you can see, there is very little difference between the high-gain DC amplifier we discussed above and the oscillator circuit just described.
That's why you have to be very careful when looking at a circuit, to make sure you are identifying it correctly.

[BobsURuncle]

My question is this, if i have 0.6 volts from X to Y, when Y is 4 volt, in X i have a raising from 0.6 to 4.6? i have 4.6 to 0.6?  if I charge plate X through the  22 ohm resistor, how can Y charge through X?
Thanks in advance
Alex

I think what happens is initially the Cap charges to -0.6V (using the cap polarity indicated in the schematic), that is to say the NPN Vbe is raised to +0.6V - turning on the NPN.  This turns on the PNP which charges the cap by, let's say, +3.4V (for the sake of illustration), so the Y terminal is now at +4V and X is still ~0.6V.   When the cap is fully charged it is no longer shunting current to the base of the NPN and the 330K can not feed sufficient base current to keep the NPN on so it shuts down and the PNP turns off, driving the Y terminal back down to 0V as it discharges through 22R and 330K loop.   Cycle repeats.

[shadow_2609]

But wait, doesn't the transistor start to conduct around +0,6 for an NPN transistor? So how can it be -0,6 volt on the base?

[orolo]

Hm, let me see if I got this right:

Initially the 10u cap is discharged. The 330k resistor charges it very slowly (3.3s time constant). When it charges enough to activate the NPN, the PNP also turns on. The circuit is designed to make that switching very fast, hence the 10n cap. So the voltage in Y will rise very fast. Since the voltage across a capacitor cannot change instantaneously, when Y rises, X will also rise. There is a positive feedback going on here: the more Y rises, the more X rises, and the harder the PNP switches on, rising Y even more. So this will go on until the PNP saturates. At this point, if Y is at 4 volts, X should be around 4.6 or a bit less. Now the 330k resistor will be unable to compensate the loss of charge through the base of the transistor, so X will start to go down in voltage. Even if X tries to pull Y down, it won't be able to do much, because that side of the cap is pinned to 4V by the saturated PNP, and the 22 ohm resistor diverts any transient trough the cap directly to ground (the time constant at the Y side is 220us). When the voltage in X goes down enough, the PNP will come out of saturation, and the voltage in Y will also go down. This will pull X down even further, closing the switch. At this point, the 330k will start recharging X, and the cycle will repeat.

[BobsURuncle]

But wait, doesn't the transistor start to conduct around +0,6 for an NPN transistor? So how can it be -0,6 volt on the base?

 
It's not -0.6V on the base  That's not what I said.

[shadow_2609]

It's not -0.6V on the base  That's not what I said.

Oh ok sorry I missunderstood! Now I know what u meant shame on me 

[shadow_2609]

Hm, let me see if I got this right:

Initially the 10u cap is discharged. The 330k resistor charges it very slowly (3.3s time constant). When it charges enough to activate the NPN, the PNP also turns on. The circuit is designed to make that switching very fast, hence the 10n cap. So the voltage in Y will rise very fast. Since the voltage across a capacitor cannot change instantaneously, when Y rises, X will also rise. There is a positive feedback going on here: the more Y rises, the more X rises, and the harder the PNP switches on, rising Y even more. So this will go on until the PNP saturates. At this point, if Y is at 4 volts, X should be around 4.6 or a bit less. Now the 330k resistor will be unable to compensate the loss of charge through the base of the transistor, so X will start to go down in voltage. Even if X tries to pull Y down, it won't be able to do much, because that side of the cap is pinned to 4V by the saturated PNP, and the 22 ohm resistor diverts any transient trough the cap directly to ground (the time constant at the Y side is 220us). When the voltage in X goes down enough, the PNP will come out of saturation, and the voltage in Y will also go down. This will pull X down even further, closing the switch. At this point, the 330k will start recharging X, and the cycle will repeat.

So, if I get it right, it does rely even on the fact that the 330k will be unable to provide sufficient base current to turn fully on the npn-pnp pair? So if I choose something lower the circuit will not oscillate or will oscillate in a higher rate (for the RC constant)?

[orolo]

The need for a high resistance happens after the pnp has saturated, making the LED shine. At that moment the base of the npn starts to discharge, because it has been pulled to its highest voltage by the rise of Y through the cap, and can only discharge through the base of the npn (since Y has maxed, it cannot pull X up anymore). If the resistor is too low, it will inject enough current to keep the npn active, and the pnp saturated, and the LED will never turn off. For a low enough resistance, this circuit is monostable: even if the LED starts off, it will turn on and never go off. Monostability happens a lot with ill-conceived relaxation oscillators; I went through hell with a PUT oscillator once for this reason. The current charging the cap depended on an homebrew electrochemical battery of very high but variable internal resistance. The LED flashed very irregularly, when it did.

To increase the frequency of oscillation it's a better idea to reduce the cap than to decrease the resistance. BTW, this circuit should not be used with more than the suggested 6V battery, since when the pnp cuts off, it pulls X to negative voltages, and a high enough reverse voltage in the BE diode can damage the npn transistor.

[stitch]

So, if I get it right, it does rely even on the fact that the 330k will be unable to provide sufficient base current to turn fully on the npn-pnp pair? So if I choose something lower the circuit will not oscillate or will oscillate in a higher rate (for the RC constant)?

I agree with orolo.  Said another way, if the circuit is working as intended, then the 330k is able to provide sufficient base current - but just sufficient and no more.  When the 10u cap charges it "draws off" enough current current so that the base's supply of current is insufficient and the NPN turns off.
If you choose a resistor with a value much lower than 330, then the base of the NPN will have "current to spare". So when the 10u cap charges, the "draw off" of current won't make any difference. If you choose a resistor with a value much higher than 330, then the base of the NPN will not have enough current to turn on in the first place.  Best to change the capacitor.

[shadow_2609]

I just find out that my beautiful cellphone didn't accepted my post for some reason... sorry!!!!
I just wrote thanks a lot for the help! now I understand a little more about bjt's feedback loop... or at least i think so haha! thanks a lot again bobsuruncle, orlo and stitch, you helped me a lot!
(again sorry for the very late reply but my old phone was a lil... )
Alex