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Need TTL Cookbook page 171 & 172. 555 timer 2N3055 vintage Electric Fence Charger

gary350

Well-Known Member
Math according to the TTL book does not work.

There are 2 formulas, one is to find R2 first, the other uses R2 and R1.

According to fig 4-16 page 173 the graph shows 1uf & 10M ohm = about .2Hz. Also 10uf & 1M = .2Hz. Also 100uf & 100K = .2Hz

Doing the math LOW = .685(R2)C = .685 seconds no matter which combination of cap resistor you use.

Now I see bottom left corner of graph, R1+2R2 so what is this?

Charge time LOW is R2 only so mayby graph does not apply to R2 only applies to HIGH for R1+R2

Limits for R1+R2 is 3.3M but graph shows 10M

That means capacitor has to be 10uf

I am stuck. I don't remember how to transpose so I can work the formula backwards starting with the answer .1 to find R2




100_2877.JPG
 

gary350

Well-Known Member
These programs counterdick them self. Low + high = period = Hz. Trial and error 1 hour later nothing works. When period says 34 then Hz is .029 If period = Hz the 2 numbers have to be the same. No matter what I try HIGH is never less than 3 second and LOW is never less than 6 seconds. Circuit drawing is wrong we need the circuit with the 2 diodes. Even with one resistor at 0 ohms it still does not work. Book say, R1+R2 never less than 1000 ohms, never more than 3.3M ohms.

A different program someone listed shows R1=50 ohm R2=950 ohm works but do the math or put those number in a different online program they do not work.
 
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audioguru

Well-Known Member
Most Helpful Member
Gary, you said you have NE555 which is the original 555 if it is not Chinese and you have LMA555 which has a wrong letter. I said that an LMC555, TLC555 and ICM7555 are weak Cmos.
 

crutschow

Well-Known Member
Most Helpful Member
rial and error 1 hour later nothing works.
Exactly which circuit schematic are you using?
The ones I posted with diodes should work.
 

jjw

Member
These programs counterdick them self. Low + high = period = Hz. Trial and error 1 hour later nothing works. When period says 34 then Hz is .029 If period = Hz the 2 numbers have to be the same. No matter what I try HIGH is never less than 3 second and LOW is never less than 6 seconds. Circuit drawing is wrong we need the circuit with the 2 diodes. Even with one resistor at 0 ohms it still does not work. Book say, R1+R2 never less than 1000 ohms, never more than 3.3M ohms.

A different program someone listed shows R1=50 ohm R2=950 ohm works but do the math or put those number in a different online program they do not work.
Period is time in seconds.
Frequency is 1 / time (Hz)
 

gary350

Well-Known Member
Exactly which circuit schematic are you using?
The ones I posted with diodes should work.
YES I think your right. I did the math for that circuit and get 50 ohms & 950 ohms. I need to build it with an LED output so I can see it blink so I know if times are right.
 

crutschow

Well-Known Member
Most Helpful Member

gary350

Well-Known Member
I have an idea that might work better then the 555. Maybe I can use my LED flasher circuit to power a 2N3055 to power the ignition coil. The flasher blinks 1 second on each side with 100 uf cap on each side. If I change a 100uf to 10uf it will flash a .1 second. But voltage across Red LED is about 2v that is not much to base of the 2N3055. If I put 3 or 4 of the higher voltage LEDs in series I can get 10v across them. R1 to R4 need to be changes for higher voltage 11 extra volts. Will this work?

100_2879.JPG
 
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crutschow

Well-Known Member
Most Helpful Member
Add a resistor in series with the 2N3055 base and connect to the collector of Q2.
 

AnalogKid

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voltage across Red LED is about 2v that is not much to base of the 2N3055. If I put 3 or 4 of the higher voltage LEDs in series I can get 10v across them.
Your understanding of how to bias a transistor to act as a saturated switch is dangerously incorrect. As Wally (and many others) said, you need to have a resistor in series with the base. The resistor value is calculated based on the desired base current, the available drive voltage, and Ohm's Law.

ak
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Your understanding of how to bias a transistor to act as a saturated switch is dangerously incorrect. As Wally (and many others) said, you need to have a resistor in series with the base. The resistor value is calculated based on the desired base current, the available drive voltage, and Ohm's Law.
I'm in two minds whether he's just a troll or what?.

He repeatedly posts long running threads, makes up random obviously incorrect circuits all the time, ignores all the answers and just keeps on posting more and more utterly incorrect circuits, none of which ever make any sense.
 

gary350

Well-Known Member
Your understanding of how to bias a transistor to act as a saturated switch is dangerously incorrect. As Wally (and many others) said, you need to have a resistor in series with the base. The resistor value is calculated based on the desired base current, the available drive voltage, and Ohm's Law.

ak

2N3055 datasheet says, base current 7a, collector current 15a. PS voltage 14v divided by 7 = 2 ohms. Is this right? If I knew what I am doing I would not be asking questions.

100_2880.JPG
 

ronsimpson

Well-Known Member
Most Helpful Member
2N3055 datasheet says, base current 7a,
No! Base current of 7A is the max just before the part breaks. Do not go there!

Here is a graph of collector-emitter voltage, collector current and base current.
I don't know what your collector current is but lets say 4A. You need to keep the C-E voltage down low. With 0.5A of base current the transistor seems to close well.
1553375865625.png
If your collector current is 1A then a base current of 20 to 100mA is good.
 

crutschow

Well-Known Member
Most Helpful Member
If your collector current is 1A then a base current of 20 to 100mA is good.
That is a nominal, not worst-case characteristic graph. Designing to nominal values is problematic.
For the 2N3055 I would never use less than 10% of the collector current for the base current, to fully saturate it.
 

AnalogKid

Well-Known Member
Most Helpful Member
2N3055 datasheet says, base current 7a, collector current 15a. PS voltage 14v divided by 7 = 2 ohms. Is this right?
You tell me. Look at your schematic. Then t.h.i.n.k. Do you really want to pull 7 A through an LED?

ak
 

eTech

Active Member
I have an idea that might work better then the 555. Maybe I can use my LED flasher circuit to power a 2N3055 to power the ignition coil. The flasher blinks 1 second on each side with 100 uf cap on each side. If I change a 100uf to 10uf it will flash a .1 second. But voltage across Red LED is about 2v that is not much to base of the 2N3055. If I put 3 or 4 of the higher voltage LEDs in series I can get 10v across them. R1 to R4 need to be changes for higher voltage 11 extra volts.
I've breadboarded 555 circuit I posted in #38 and it works.

eT
 

gary350

Well-Known Member
I finally figured this circuit out. Book shows R1 & R2 in series but this circuit below has R1 & R2 in parallel, each resistor in series with a diode. Reason answers to formulas are wrong is because they do not apply to this circuit with resistors in parallel. The only formula that applies to this circuit is Time=.685(R)C Same formula applies to both resistor R1 & R2. With R1=15k HIGH time = .1 seconds. With R2=150K LOW time = 1 second. If I chance R2 to 300K LOW time = 2 seconds. I have not removed the 7812 voltage regulator maybe the 555 can use a little over voltage protection. Maybe 2N3055 base resistor chould be lower for more power to ignition coil.100_2883.JPG
 
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crutschow

Well-Known Member
Most Helpful Member
Maybe 2N3055 base resistor chould be lower for more power to ignition coil.
You are limited by how much output current the 555 can deliver.
Also if you pull down the output of the 555 too much, it will mess up the timing.
That's why I prefer the second design in post #16 (below) where the timing circuit is independent of the output voltage.

 

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