Here's my take on it -- this is just from looking at it; running it through SPICE would probably be a better idea :-) :
It would probably be a little clearer if the author hadn't stuck "+" and "-" on C1 -- ignore those. I believe you really want an unpolarized capacitor there anyway...
OK then... prior to the transistors starting to turn out, the current flow through C1 and R3 is limited by the 10M resistor -- hence the current will be limited to something in the ballpark of 12V/10M=about 1uA. 1uA through R3 is about half a millivolt, so the right-hand side of C1 is effectively at ground, and the left hand side charges up slowly to around 0.6V.
As you correctly summarized, at that point Q1 and Q2 switch on. As this happens, current flows through the LED and Q2 and R3, so the voltage on the right side of C1 begins to increase, which causes the voltage on the left side to increase as well. This causes Q1 (and hence Q2) to turn on *even harder*. In other words, it's a form of positive feedback -- very quickly, the voltage across R3 will move from ~1/2mV to something around 9V (12V minus the drop across the LED and Vce of Q2); as this happens, the left side of C1 will move from ~0.6V to ~9.6V, which will Q1 fully on. Now, C1 begins discharging through R2 and Q1's base, all the way until the left side of C1 is back down to ~0.6V.
Note that the right side of C1 is still at ~9V, so there's ~8.4V across C1 (from the right side to the left).
At this point, Q1 begins to switch off, so Q2 does as well. Suddenly, there's very little current through R3, so the right side of C1 rapidly heads back towards ground... which means that the left side of C1 is rapiding heading towards -8.4V. Now, D1 begins to conduct, and current from ground through D1, R2, C1, and R3 begin to discharge the capacitor back towards -0.7V on the left (the drop of D1) and something close to ground on the right (depends on the value of R2 -- if R2 is 100K, the right side of C1 is still effectively ground, but if R2 is only 1K, now you have a voltage divider...).
Once the left side of C1 is around -0.7V, D1 stops conducting, and it's up to R1 to "finish" the job of getting the left side back up to 0.6V to begin all over again.
Note that it's R2 and C1 that control the rate at which the capacitor goes from having 8.4V across back to (around) 0V, whereas R1 only controls the rate at which the capacitor goes from (around) 0V to 0.6V -- this is why the text claims that R2 and C1 control the flash rate: While R1 has an influence, it's perhaps only 10% of R2 and C1 do.
Without D1, you end up exposing the base of the transistor to higher negative voltages (hence the advice that it's OK to remove it for lower power supply voltages, since you have a better chance then of not killing Q1), and discharging C1 also then takes place just via R1. This forces a long "off" time that's dependent on the value of R1, which manifests itself as the "lower duty cycle" mentioned in the text.
I trust plenty of people will let me know if the analysis here is incorrect. :-)
Joel, thank you very much! - your explanation is very clear. Much kudos to you for helping me understand this.
Can you withstand a follow up question?
There is a similar circuit here:
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sP1.html
This has no diode, and extra 10nF capacitor (which I assume is not doing much), but has the electrolytic capacitor the other way round. As you mentioned the capacitor's polarity, I am wondering how much of a concern this is? What happens if you use the capacitor the 'wrong way round'?
I wanted to make a simple flasher that would 'last forever' (or as long as possible) - so I am quite interested in not wearing out the electrolytic.
Joel, Your explanation is fine; but it's a crap design... with the right betas, the top of R3 will simply stick high... FOREVER ;-)
Test this by simply omitting R2 and C1, then examine the quiescent operating point.
...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 |
I love to cook with wine Sometimes I even put it in the food
This has no diode, and extra 10nF capacitor (which I assume is not doing much), but has the electrolytic capacitor the other way round. As you mentioned the capacitor's polarity, I am wondering how much of a concern this is? What happens if you use the capacitor the 'wrong way round'?"
I *think* the rationale is that since you're just charging and discharging the capacitor, although we're talking relatively low frequency (single-digit Hz), there's no actual DC bias on the capacitor, and thefore it doesn't actually matter which away the capacitor's polarity is oriented.
I suspect the 10nF capacitor does just give the NPN transistor a little extra "kick" to get going at first and that the circuit would work without it, but I'll again wait for someone else to comment on that. :-)
"I wanted to make a simple flasher that would 'last forever' (or as long as possible) - so I am quite interested in not wearing out the electrolytic."
There was a pretty long thread on SED here late last year, I think it was, with some of the talented designers like Winfield Hill and John Larkin trying to build the best (lowest power consumption) LED flashers -- schematics were posted and there was plenty of debate.
Colin Mitchell, the guy behind Talking Electronics, often has somewhat unique and interesting viewpoints on circuit behavior -- I think this stems from his truly understanding the electronics (usually!), but his formal training was as a technician, so he's forced to explain how things like oscillators really work rather than just waving around some phasor airthmetic and citing the Barkhausen criteria and calling it good. :-)
Might I suggest Colin's nomination as ignoramus of the year? Check out his explanation of soft starting a regulator at the top of this page...
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...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 |
Stormy on the East Coast today... due to Bush\'s failed policies.
"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...
Not a case of his "usually" understanding the electronics, then. :-)
It looks like he snagged that circuit from here:
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Could you explain the differnce in the two circuits? I'm finding that I don't have a particularly *good* explanation for why the top circuit is better than the lower one -- just some hand waving/guessing. Or are they both pretty bad?
...and it looks like the original (?) circuit form zen.co.uk then ended up here:
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... where there's plenty of additional useless hand-waving: "...the electrolytic capacitor charges... this type of capacitor is significant in low frequency and high current electrical circuit... an electrolytic capacitor contains filters that allow very low cutoff frequencies due to its very high capacitance..." -- sure, that's great, but not relevant to the circuit's operation...
Perhaps if I'm feeling ambitious I'll put up a "spot the mistakes in spot the mistakes." :-) Should be a good learning exercise...
I first bumped into Colin's work while visiting a Dick Smith Electronics store in New Zealand -- his projects are available in various small notebooks -- somewhat similar to the Forrest Mims mini-notebooks that Radio Shack once sold -- and I picked up a few. (I also remember when Dick Smith -- briefly -- had a U.S. presence... at the time Radio Shack had already gone seriously downhill and Dick Smith was looking attractive.)
Dick Smith Electronics in Oz/NZ has taken a course similar to Radio Shack in the U.S. in that they don't have nearly the hard-core electronics they once did, having been (understandably, I suppose) replaced by cell phones, computers, TV, etc... but they're still "useful." Meanwhile Dick Smith himself
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seems to be in semi-retirement, after flying around the world and otherwise engaging in a "well-heeled adventurer" lifestyle.
:-) I'm confident John could get his to work, even if it wasn't quite so few parts as he was initially thinking!
It's called a "speedup" capacitor - it helps turn on the PNP faster than it would with just the resistor, saving a modicum of power during the switching transition.
Correct - the base current of the NPN transistor is why, and because that base current is not constant (at high temperature, it'll go backward), this is not a particularly good circuit.
Usually, a blocking oscillator is done with an inductor and ONE transistor; this uses two transistors and a capacitor (why?). With two transistors, you can make a real astable multivibrator, which is how I'd do it. Or, you can use a '555 variant.
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