I found the following LED flashing electrical diagram in the net
I assume that the flashing bit is done by the transistors and the friquency from the RC components.
On a sunny day (Thu, 1 Jan 2009 11:54:35 -0800 (PST)) it happened Efthimios wrote in :
It is in phase feedback with a gain > 1, so it will oscillate. When the 2222 is conducting, then the C is charged via the base emitter of the 2907 and the 680 Ohm resistor, with a short time constant. When the 2222 is off, then the cap is discharged via the 4M7 resistor, long time constant.
os
Thanks Jan,
But what makes the 2222 transistor switch between on-off states?
E
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.
On a sunny day (Thu, 1 Jan 2009 12:28:50 -0800 (PST)) it happened Efthimios wrote in :
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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?
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.
This circuit probably has a power efficiency in the 30% range. Something like a blocking oscillator could easily double that.
John
what is a blocking oscillator?
E
What is a google?
John
A kill-filter object ;-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Postings via gmail, yahoo, hotmail, aioe, uar or googlegroups, and wild-cross-posts are now automatically kill-filed using Agent v5.0 To be white-listed, send request via the E-mail icon on my website
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.
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Thanks Jamie for your detailed analysis.
E
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.
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.
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.
Hey, we could start another holy war around the validity of the concept of "averaged power efficiency."
John
I thought your New Year's Resolutions forbade that ?:-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Postings via gmail, yahoo, hotmail, aioe, uar or googlegroups, and wild-cross-posts are now automatically kill-filed using Agent v5.0 To be white-listed, send request via the E-mail icon on my website
mios
r,
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.
On a sunny day (Fri, 2 Jan 2009 09:49:11 -0800 (PST)) it happened Tolstoy wrote in :
No that is not correct. The meaning of 'negative' and 'positive' is the phase. The phase shift is 0 degrees, and stays that way for this circuit. With a phase shift of zero, feeding back the output to the input, and a gain of >=1, the circuit will oscillate.
Somebody was wrong.
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