I was trying to understand flasher circuit from following link http://www.talkingelectronics.com/projects/FlasherCircuits/Page83FlasherCircuitsP1.html. Author wrote following: "

The negative end of the electrolytic is `jacked up' by this voltage and the positive end pushes the charge on the electrolytic into the base of Q1 to turn it on even harder

."

I am haven't understood what is meant by jacked up. Would someone be able to explain me what author is trying to say.

 

Jacked up is usually slang for "increased."

That web site is run by Colin Mitchell, who is mentioned in this thread. I suggest you read the thread before you put too much stock in what you find on "Talking Electronics."

http://forum.allaboutcircuits.com/showpost.php?p=427432&postcount=2

SgtWookie is one of our most experienced and trusted members.

 

I'd suggest looking at other sites on the net to learn electronics from, the e-book at the top of the page is a good start.

That aussie site is a guy who tries to look good by bringing others down. Sometimes he is right, but not all the time. Some of the circuits he has posted as projects don't work well.

 

Thanks tracecom

Was his explanation about flasher circuit wrong? I am trying to understand few things, and if talkingelectronics is not correct source, then I should discard, and move on to some other source, otherwise I will un-necessarily confuse myself.

The biggest bit confused me was how capacitor was jacked up (increased). Your help will be appreciated.

 

The voltage on the capacitor isn't 'jacked up.' That voltage never gets above ~.7V due to the B-E junction of the NPN transistor. I can tell you how it works, but it's very late and I'm heading to bed. We'll chat again tomorrow unless someone else beats me to it.

In my experience, Colon is knowledgable, but this explanation misses the mark.

 

The voltage on the capacitor isn't 'jacked up.' That voltage never gets above ~.7V due to the B-E junction of the NPN transistor.

This is what I was thinking. Will definitely wait for an answer.

 

I was going to post a simulation that plotted the voltages, but the LED just stays on in LTSpice simulation....

 

Yes "jacked up" just means the voltage appearing across 22R, when the PNP turns on, raises the (+) side of the 10u capacitor. This pushes charge from the capacitor into the base of the NPN, turning it further on.

 

The origin of the phrase comes from the mechanical device known as a jackscrew. It uses the mechanical advantage of a screw thread to turn rotational motion into linear motion. The device, in several configurations, is used to jack up cars, buildings, and other large loads. Here is the wiki on the device

http://en.wikipedia.org/wiki/Jackscrew

 

raises the (+) side of the 10u capacitor

Please ignore my ignorance. Can this be elaborated bit more.

 

I still owe you an explanation. But first, a disclaimer. I have not built nor simulated this particular curcuit, and so I cannot guarentee that it actually works. My explanation comes from similar circuits I have built and or simulated, but there are differences, perhaps major ones. To the best of my limited knowledge:

The circuit is a form of an astable multivibrator. It has two unstable states:

A) both transistors conducting and saturated
b) both transistors cut off

The circuit works in 4 statges. In state A, B and the transistions from A-> B and B-> A. It is during these transistions that the real action takes place. First, let us assume we are in state A, and both transistors are conducting in saturation. The circuit is in a state of regenerative feedback, where transistor Q1 is turning on Q2 and visa-versa. The feedback is provided by the 10u capacitor. The current from Q2 ( the PNP,sorry I just realized those transistors aren't numbered) flows through the capacitor and into the base of Q1(the NPN) keeping Q1 turned on and in satruation. Likewise the current from Q1 collector flows into Q2, keeping it turned on and saturated. Eventually, the capacitor charges up and the current through it diminishes. That current, which was keeping Q1 in saturation now is decreasing. It eventually decreases to the point that Q1 comes out of saturation. This causes a drop in the base current of Q2, which also comes out of saturation, which in turn, causes Q1 to turn off even more. Now both transistors are nearing cutoff, and the capacitor is charged almost to the battery voltage. Once Q2 turns off, the + plate of the capacitor rapidly falls to nearly 0V. Because a voltage cannot change across a capacitor instantainiously, the - plate, attached to Q1 is now at a negitive voltage. Remember, there was alomost 6V across the capacitor, and when the + palte falls to near zero, the - plate must be at nearly -6V. Both transistors are cut off, and the charge in the capacitor bleads out through the 22R resistor. This is the charge that is holding Q1 in cutoff. Once it bleads down enough, current through the 330K resistor begins to turn Q1 on again. Once Q1 turns on, it turns Q2 on , which turns on Q1 even more ( as explained above ) and the whole cycle repeats. The cycle is self-perpetuating, and thus, you have pulses form both transistors. The pulse from Q2 is used to strobe the LED.

 

Thanks Brownout

the - plate, attached to Q1 is now at a negitive voltage

Shouldn't plate connected to Q1 be + and plate connected to 22R be -ve? For example, when circuit is initially started, all the current will flow from 22R towards right-hand side plate of capacitor thereby accumulating more electrons, hence, negative charge? Am I fundamentally wrong?

 

Have you got the circuit assembled and working? A 10MΩ Impedance DMM can show you the voltage across the capacitor if it is working good.

 

Thanks Brownout

Shouldn't plate connected to Q1 be + and plate connected to 22R be -ve? For example, when circuit is initially started, all the current will flow from 22R towards right-hand side plate of capacitor thereby accumulating more electrons, hence, negative charge? Am I fundamentally wrong?

That's a good point. Consider when Q2 is on, positive current flows in the capacitor from the right plate to the left plate, charging the capacitor. When Q2 is off, current flows in the opposite direction. However, the capacitor doesn't really charge up in the opposite direction, because as soon as the voltage rises to ~.7v, Q1 turns on and the capacitor again charges in the forward direction. It's probably not good to use an electrolytic capacitor in this manner, but since this is just a demonstration circuit, it's OK. You wouldn't want to design a production circuit like that.

 

Hi Brownout,

Thanks once again for your great explanation. Like every beginner, head is exploding with lot of questions .

Please see attached screenshot so that I get a better understanding of capacitor working.

Figure 1: A simple capacitor with Vcap of 10V.

After sometime, voltage at point A is risen to 3V. Question I have is what will be Vcap at that instance. May be this is a silly question; I am trying to understand what happens when voltage at one end is pulled up, for example, to 3V.

 

It's not possible to know what the capacitor voltage is. It will be 3V - VB, but what is VB? If you ground point B, then Vcap = 3V.

 

Sorry, A is connected to ground point whereas B is pointed to 10V battery source.

 

Never mind

 

Remember, there was alomost 6V across the capacitor, and when the + palte falls to near zero, the - plate must be at nearly -6V.

I was trying to understand above line that is, how voltages are affected at points A & B which are connected to + & -ve plates respectively.

 

Let's say point A is connected to 3V,and point B is connected to a resistor 1K ohm. The other end of the reisistor is connected to ground. Now, quickly switch point A to ground. Since the voltage across a capacitor cannot change instantantainiously, 3V must still exist across the capacitor. So, since piont A is now at ground, point must be -3V until the charge drains off the capacitor.

Two thins to remember about capacitors,

1) voltage cannot change across a capacitor instantainiously
2) current through a capacitor is proportional to how fast the voltage changes, mathetically this is Ic=C*dV/dT.