# Transistor question

Discussion in 'Electronic Design' started by Efthimios, Jan 1, 2009.

1. ### EfthimiosGuest

I found the following LED flashing electrical diagram in the net

http://www.discovercircuits.com/DJ-Circuits/3vledfs1.htm

I assume that the flashing bit is done by the transistors and the
friquency from the RC components.

1. Can some one explain me why these two transistors (2222 2907) will
make the LED to flash ????

2. Also it requires about 30uA. If i want to reduce this what should i
do?

E

2. ### EfthimiosGuest

Thanks Jan,

But what makes the 2222 transistor switch between on-off states?

E

3. ### MooseFETGuest

The transistors provide the amplification needed and the RC parts
provide the timing control.

Do you know basically what a transistor does?

Examine the 2N2222 and just the 4.7K on its base and the 10K on its
collector. If you wiggle the end of the 4.7K up and down, what
happens to the voltage on the collector?

Now look at the 0.68uF and the 680 Ohms. What happens to the AC part
of what is at the collector of the 2N2222?

What happens to the 2N2907 when the AC on the collector of the 2N2222,
goes down towards zero volts?

What does this do to the 2N2222?

Now try to do the above with DC instead of AC.

Now try to fit a "by how much" into each of the above answers.
Notice that the "by how much" leads to an interesting situation.

Can you stand a slower blink or less light?

You can gain a little by scaling the 4.7M up by some factor like lets
say 1.5 or 2 and the 0.68uF down by the same factor. You can't go
very far on this before the circuit will stop working.

4. ### EfthimiosGuest

Can work as a switch or an amplifier.
If you increase the resistor the voltage will drop so the amplified
signal output from the transistor will drop.
What? Voltage will oscillate?
N2907 works as a switch. It switch off and stops capacitors from
charging.
It terns 2N2222 off.
With DC when the capacitor is charged will switch 2222 to conducting
state?

5. ### Guest

An interesting circuit, seemingly it's not just you thats baffled. The
lithium cell has a very high internal resitance so the current for the
led is supplied by the capacitor across it. Once that has discharged
the circuit resets. If you were to run this from a power supply the
led would come on after a delay and stay on.

6. ### EfthimiosGuest

what is a blocking oscillator?

E

7. ### JamieGuest

on Initial power to the circuit, the capacitor is nearly discharge due to
the base and emitter on the PNP (2907) and through the 680 and 10KR's
and LED network
The initial voltage across the cap at best will be the forward break
down of the BE of the (2907).

You'll get an inrush of current flow through the capacitor from the
680 R and 10KR+LED that has the (+) rail. This will not be enough to
get the LED started.

Because the cap was discharged to start with on the initial cycle,
you'll also see (+) voltage at the base of the PNP (2907), this will
reverse bias the PNP and turn it off.

The NPN (2222) depends on the PNP(2907) to be on for forward
biasing.. So at this point, the NPN is also off.

When the cap finally gets charged enough and removes the (+) forward
inrush of voltage from the charge cap down to ~ 0.6 or less at the base
of the PNP, the 4.7 Meg R will then forward bias the PNP, this will
then turn on the PNP which will also turn on the NPN (2222).

Now, the initial on cycle has been complete, and this is what happens
next.
Because the cap has been fully charged and now the 2222 is coming
on, the 2222 will also shine the LED of course and also the cap will
start to discharge how ever, since there is a charge in the cap now, the
polarity on the other end of the cap going to the PNP base will generate
(-) voltage, not (+) as it did in the initial power on cycle.

This (-) voltage will force the PNP on even harder, thus biasing the
2222 more, until full saturation is accomplished on the 2222.

At some point, the cap will deplete and start losing output to the PNP
base, when this happens, the PNP will drive the 2222 less. when the
2222 gets out of saturation state, it'll start to turn off. The 10R and
LED network will start to raise on voltage and also start charging the
cap, this charge is like the first initial charge. because the cap was
drained, it'll now force (+) volts on the PNP base thus turning it off
and forcing the 2222 to go off even faster..

The trick is to not over bias the PNP with the R, that is why you see
4.7M there. it's more or less a tickler to get it started and handle
forward drop off voltage of the base on the PNP.

The 680R being like it is, I suspect the on cycle to be much shorter
than the off cycle.

http://webpages.charter.net/jamie_5"

8. ### EfthimiosGuest

Thanks Jamie for your detailed analysis.

E

9. ### MooseFETGuest

Right so the signal on the collector is bigger than the one on the
base and is also inverted (goes negative when the base goes positive).

Just imagine some AC is at the collector of the 2N2222 by magic and
then see what the 0.68uF and 680 Ohms will do with it. We'll get back
to how the AC voltage appears at the collector later.

No look again at the circuit. The 2N2907 has its emitter at the +3 and
is a PNP. The 2N2907 turns on when the collector of the 2N2222 moves
down towards zero.

Right for the backwards case above. 2N2907 on turns the 2N2222 on.
2N2222 on turns the 2N2907 on. As far as the AC case goes we have a
positive feedback loop.
No I meant for you to consider the capacitor as an open circuit to the
DC. This means that the positive feedback situation doesn't apply to
the DC case.

The interesting situation is that if the 2N2907 turns on a little, the
2N2222 turns one quite a bit which turns the 2N2907 on a lot which
turns the 2N2222 on hugely ... etc.

10. ### MooseFETGuest

The efficiency starts with a roughly 50% starting point because the
LED has a forward drop about half that of the battery. From that
point down, it isn't all that bad. The 4.7M passes about a microamp
and the transistors leak. For most of the time, that is the only
current in the thing.

Replacing the 47 Ohms with an inductor and a diode would greatly
improve the efficiency but the inductor size for the simple case is
way too big. Making the 2N2222 make a burst of oscillation would be a
better way to go than a 2ms pulse.

11. ### MooseFETGuest

No it doesn't. The circuit will oscillate with a zero impedance power
supply. The 2N2907 doesn't have enough bias current in the DC case to
keep the 2N2222 saturated so once the 0.68uF is discharged, the LED
goes off.

12. ### TolstoyGuest

the circuit has negative feedback and positive feedback.

negative feedback: when the led is off, the big cap charges and
eventually turns the transistors and led on;
conversely, when the led is on, it drains the cap and at some point
everything turns off. somebody pointed out that you need a high
impedance battery to make this work.

the positive feedback makes the circuit tend to snap either to full on
or full off condition. the cap and resistor from the npn collector to
the pnp base provide this feedback. in other words, when negative
feedback causes the npn transistor to start to turn off, this will
pull the base of the pnp down, turning the npn off even more -- what
you might call a "snowball effect." Conversely, if the npn transistor
starts to turn on, it will pull down the pnp base and make the circuit
turn on more and more, like a snap-action switch.

14. ### MooseFETGuest

This is not correct, the circuit works with a low impedance supply.

15. ### Jasen BettsGuest

feedback with no net phase shift will not cause oscillation

if you've got over unity gain in a phase-shift free loop you end up
'pegged' against one rail or the other.

16. ### JosephKKGuest

Care to try that analysis again? Charged through the 4.7 Meg,
discharge through Schmidt trigger 1st base.

17. ### JamieGuest

Is that what you think?

http://webpages.charter.net/jamie_5"

18. ### JosephKKGuest

John, it is a blocking oscillator, and about 90% efficient.

19. ### Sylvia ElseGuest

This would only be true if the circuit is phase-shift free all the way
down to DC. Though off hand I can't imagine a simple circuit that would
be phase shift free at higher frequencies, but not DC.

The circuit in question is not phase shift free at any frequency. It
contains a 360 degree phase shift, and unity gain, at its switching
frequency. The output waveform is not sinusoidal (by any stretch of
imagination) because of extreme non linearity.

Sylvia.

20. ### JosephKKGuest

No way in hell spud. The 4.7M is the only possible charge path. Or
did the reverse polarity designation fool you?