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Amplify current of a flickering candle LED

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ok i have a 20K pot and connected as follows.
That's not at all what I meant!

Only connect R3 to the pot wiper, and keep the 1K feeding the LED with pot ends connected across that.

The idea is not to vary the LED current, but allow the transistor base voltage to be adjusted for best effect.
 
That's not at all what I meant!
I am sorry but I did try to look up what "across from" meant but i could not figure out where you wanted me to place it. I still do not know where to place the pot. What I do understand is series and parallel. If I were asked to measure voltage across a diode or 2 points across a transistor I could do that. But when something reads place a component across from another I just cannot picture it. I do agree now though that replacing the resistor at R1 was completely wrong.


In another post I had the same problem.
Thread 'Flickering (candle) LED to trigger 555' https://www.electro-tech-online.com/threads/flickering-candle-led-to-trigger-555.163572/
 
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Across, as in parallel, so the ends of the pot are in parallel with the 1K resistor that feeds the LED.

Across is just a more generic term, as in connecting a decoupling capacitor across the power supply to something.
 
OK, try adding a pot or preset across the 1K (e.g a 4.7K or 10K), with R3 connected to the wiper.
That should allow you to find an appropriate bias point for the transistor?
OK. NICE. The 5 leds begin to flash and are their brightest at this setting.

The results are measured with the components in the bread board.
Pot R1 = 52.9K
C-B = 3.88K
C-A = 4.71K
B-D = 13.86K
A-D = 14.68K
Pot R1 measured out of the board
C-B = 4.1K
C-A = 49.1K
flicker_03.png
 
The results are measured with the components in the bread board.
Pot R1 = 52.9K
C-B = 3.88K
C-A = 4.71K
B-D = 13.86K
A-D = 14.68K
Pot R1 measured out of the board
C-B = 4.1K
C-A = 49.1K
View attachment 142201
The circuit diagram shows a 20 kOhm pot, but the readings indicate a 50 kOhm pot. Is that just a typo in the circuit diagram? It probably doesn't make any difference to the operation of the circuit.
 
The circuit diagram shows a 20 kOhm pot, but the readings indicate a 50 kOhm pot. Is that just a typo in the circuit diagram? It probably doesn't make any difference to the operation of the circuit.
Yes, a typo. I got this batch from Tydaya but a few read as 20k. I try to test every component as they arrive.

So with all the data I provided what resistance should I add to the 10k? My guess would be 4k.

I know some components cannot be read while installed but I am also curious why the pot read 3.8 & 4.7 while installed but 4.1 & 49 after I pulled it.
 
I know some components cannot be read while installed but I am also curious why the pot read 3.8 & 4.7 while installed but 4.1 & 49 after I pulled it.
The measuring device was measuring the through the 1 kOhm resistor (R9) and any other paths through the circuit as well. When you were measuring the 49 kOhm part, the reading you got was approximately the 4.1 kOhm (the other end of the potentiometer) plus the 1 kOhm.
 
The educational experience from this group is nothing less than great. I've not connected a pot that way before. Usually the wiper is connected to one of the outer pins. Regardless i am having the same problem with the Quasi-random sq wave gen circuit as it functions as documented but if the voltage rises above 11v the led at pin 3 fails to light. I will use this technique to figure out what resistor value is needed at R2.

Now the result for this circuit states that I need a 14K resistor to replace the 10k at R3. The next size up is a 20k so ill join the 10k in series with a 3.7k

Thank you
 
I really did not think I'd have a problem with this but when i replace the POT with a resistor (4.7K) the result is not the same as when the pot was in place. Meaning with pot in place and adjusted to 4.1K the 5 leds flicker but when i replace the pot with a 4.7K they are lit but not flickering any longer. im connecting the additional resistor at points A+C. What have I done wrong this time?
breadB.jpg
 
Your using a real breadboard, make sure your parts are in nice and snug. Then wiggle the part your working with until you get the suspected value.
 
You need to replace both ends of the potentiometer with suitable resistors. You need A-C and B-C
Im understanding but not following. Meaning I'm a bit confused. With the data i provided earlier can you tell me what values and where to place them? I think your saying to add 4.7K to R3 AND 3.7K to R1 OR to add the total of A-C & B-C (8.4K) to R3?
 
Yes, like in post 36.
i am not following you. Post 36 is what I did and the circuit behaves differently then it did with the pot in place.
Mr. D300 says in post 35 that i need to do something else but i did not understand what he meant. What is confusing me is that a POT is one resistor. I thought i was implementing this to find the resistance needed for R3. AS it has done successfully. So i measured A-C and added the calculated resistance to the circuit but that did not work. So i understand i need a second resistor for B-C. So now i do not know what to do. Do i simply add AC & BC together and that's the resistance i need to add?
 
You have not been consistent with the labelling of the potentiometer. In post #36, A is the wiper. In post #26 and #31, C is the wiper.

A potentiometer is two resistances. Each resistances connects the wiper to one end. The total value of the two resistances is (approximately) the value of the potentiometer. Adjusting the potentiometer makes one resistance larger and the other smaller.

To replace a potentiometer with fixed resistors, you need to replace both of the resistances with fixed resistors.

Example. You have a 10 kOhm potentiometer. When it is adjusted where you want, and remove it from the circuit, it measures 3 kOhm one end and 7 kOhm the other. You replace that with a 3 kOhm resistor from one end to the wiper, and a 7 kOhm resistor from the other end to the wiper.

You do not want a 10 kOhm resistor for the example.
 
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