Hi,

I just finished my first soldering project. I'm such a newbie. It's a JAMECO 3-D Christmas Tree. It all works fine but I don't understand how it all works. Here's the schematic for it.

https://www.jameco.com/Jameco/Products/ProdDS/211781.pdf

I can see it's an oscillator but can somebody explain the details of its operation?
I'm interested both in the operation and the numbers. I.e., what currents flow where. The LEDs are the tiny kind, I'm not sure what their voltage drop -- I should have measured yesterday, but was too tired.

So, R1 and R2 are current limiting resistors. R5 through and R8 are there to properly bias the bases of the transistors. R4, R6 control T2, T4 and R7, R8 control T1, T3.

The base current of T2, T4 is:

(9V - 0.7) / 82000 = 0.1mA

The base current of T1, T3:

(9V - 0.7) / 100000 = 0.83mA

It's hard to say according to the datasheet what the typical DC current gain is for this transistor because in the column where it should be listed, a "-" is put. But I assume the transistors will be fully open at 0.7 on the base.

The part I don't get is:

1) why would the transistors ever close (must have something to do with the caps but I don't get how it works)

2) why do R5, R6 and R7, R8 have different resistor values?

Many thanks in advance. And please remember, I'm new to electronics. So, if my calculations or analysis or questions evoke a fury of emotions, gently point out to me where I'm mistaken. That will be sufficient.

 

There's a pretty detailed explanation here:
http://en.wikipedia.org/wiki/Multivibrator#Astable_multivibrator

 

Oh cool! Thanks!!!

 

It's definitely easier to solder then to understand. I'm surprised there were no references in the manual to the theory behind the operation.

 

Only question: what's the voltage drops on those mini leds and what's the rated current?

 

Couple of volts usually. Depends on the color. There are charts online, and (thought I hate to say this) they are accessable through google.

 

Could be anywhere between 1.5 - 2.5V at 20mA which is about the maximum for 3mm LEDs. In this circuit they are being run at much lower current than that so the forward voltage will be lower than if they were run at full current.

 

Couple of volts usually. Depends on the color. There are charts online, and (thought I hate to say this) they are accessable through google.

Aren't the smaller LEDs different than the regular sized ones? And if so, I don't know what to look for exactly.

 

I've read a nice explanation here:

http://rayshobby.net/?p=1079

Questions:

1. if the capacitors change their charge polarity, why are polarized capacitors used? Isn't it bad to apply a plus where a minus should be? I don't get this quite.

2: The way the transistors are turned off is by the corresponding capacitor to be connected on the opposite side to ground. E.g, C1 turns off T2 when T1 is fully on and C2 turns off T1 when T2 is fully on. What I don't get here is how one plate of a capacitor going to ground can affect the other plate. Is it just a momentary impulse because of the voltage drop on the other side?

To rephrase, in one of the stages, the right plate of C1 is connected to the base of T2 and T1 is fully off. So, C1's left plate is +Vcc and its right plate is 0.6V because of the BE bias of T2. Now, when T1 is turned on by C2, the left plate of C1 goes to 0 and that turns off T2. But how come? Isn't the right side of the capacitor still at 0.6?

 

Doors and windows open and close. Transistors turn on and turn off.

The resistors have different values because if everything is exactly the same then some circuits will not oscillate (both transistors will turn on at the same time and stay turned on).