my flashing led question 3

[circuitmangler]

Here's my (obligatory) flashing LED question. Consider the 2-transistor flashing circuit at [link]http://www.uoguelph.ca/~antoon/circ/2qflash.htm[/url]
without the diode. My understanding of the way the circuit
works is:

1. C1 charges through R1, R2 and R3, and as it does so the
potential at the base of Q1 rises from -VCC(?) up to .7V.

2. When the potential at the base of Q1 reaches .7V, Q1
turns on allowing C1 to discharge through the B-E junction,
R2 and R3. At the same time, Q2 is switched on which allows
current to flow through the LED.

3. When C1 is sufficiently discharged, Q1 switches off which
causes Q2 to switch off (turning off the LED) and C1 resumes charging.

Is this basically correct, or am I way off the mark? If this
is basically how the circuit works, the question I'm stumped
on is:

How is the circuit able to achive a negative potential
(wrt. ground) at the base of Q1?

I'd appreciate any help with this. Thanks!

 

[OperaHouse]

When C1 is sufficiently discharged, Q1 switches ON.

 

[nbucska]

you can buy flashing LED at Radio Shack

<nbucska@pcperipherals.com>

 

[ijl]

You are slightly off the mark. The operation of the circuit is roughly the following:

The circuit undergoes a cyclical operation, so we drop in when Q2 is not conducting and C1 is (roughly) discharged.

Current flows from +Vcc to the base of Q1 through R1.
Q1 starts to conduct and so does Q2. The potential over R3 rises. Because C1 cannot be charged/discharged rapidly, the potential of the -ve terminal of C1 rises roughly in step with the potential of R3. Current starts to flow in the path Led -> E-C of Q2 -> C1 -> R2 -> base of Q1. Q1 starts to conduct more and more driving Q2 into saturation. C1 is getting charged in this process until it is fully charged.
The +ve terminal of C1 is now at a potential of a couple of volts = Vcc minus voltage over Led minus saturation oltage of Q2. The -ve terminal of C1 is at about 0.7 V.

When C1 is fully charged, it cannot any more supply current to the base of Q1. The only current to the base of Q1 flows through R1. As a consequence, the C-E current of Q1 = the current to the base of Q2 decreases. The voltage at the collector of Q2 = voltage at the +ve terminal of C1 drops. Again, because C1 cannot be charged/discharged rapidly, the voltage at the -ve terminal of C1 drops similarly and pulls down the voltage at the base of Q1. The current to the base of Q1 drops further and Q2 stops conducting. The voltage of the +ve terminal of C1 is now at zero (Q2 is not conducting).

Because C1 is charged to several volts, and has not had
time to discharge, the voltage at the -ve terminal of C1
must be negative, which prevents Q1 from conducting. C1 discharges slowly through R3, R2 and D1 until it is (roughly) discharged, and the cycle starts again.

Hope this helps.

 

[circuitmangler]

Thanks for the detailed post, ijl! Apparently I was pretty
far off the mark This NPN-PNP combination is used quite
a bit in simple circuits, so it's really nice to finally
understand how it works.

One question though -- the circuit works without the inclusion
of D1 (just snip it out). How would C1 discharge then?

Thanks!

 

[IRstuff]

C1 can still discharge through the emitter-base diode junction of Q1.

TTFN

 

[IRstuff]

oops, sorry, wrong way. It must be discharging through R4.

TTFN

 

[circuitmangler]

Do you, perhaps, mean R3? or someother Ri for i = 1, 2 or 3?

 

[IRstuff]

drat, I hate that!!!

R3

TTFN

 

[ijl]

Leakage currrents will discharge C1 without D1, I guess:
- leakage of C1 itself, it is an electrolytic cap.
- leakage through B-E junction of Q1
- perhaps even leakage through the B-C junctions of Q1 and Q2. The latter one is forward-biased for the discharging current.
- Some leakage current may flow even through R2 - R1 - Led - Q2
The blinking rate should be lower without D1.