# Confused Flip-Flop Circuit (multivibrator)

#### Thoms_S

Joined Aug 23, 2014
13
Greetings.
I built this circuit on a bread board and it worked with my power supply (no battery test as I assumed it would work). So then I built 3 identical circuits on solder board, see picture attached, and they all work with my little bench top power supply but not with a battery. Each unit does the same thing [including the bread board now that I've tried it], it works as desired on the P.S. (5-9VDC) continuously... but when I put it on a battery the flashes speed up until they are constant on.
(1) why my circuit works with a 9VDC power supply and not with a 9VDC battery, and
(2) how can I make this work with a standard 9VDC battery?
Thank you

#### SamR

Joined Mar 19, 2019
3,449
How about a photo of the PCB front and back. What did you use for caps?

#### AlbertHall

Joined Jun 4, 2014
11,392
Can you measure the battery voltage as it runs?
It may be that the fairly high current for the LEDs is pulling down the battery voltage.

#### dl324

Joined Mar 30, 2015
12,917
but when I put it on a battery the flashes speed up until they are constant on.
I don't see how they can both be on at the same time. Have you used an oscilloscope to look at the voltages?

#### BobTPH

Joined Jun 5, 2013
3,475
At 9V that circuit will exceed the max reverse voltage on the bases of most transistors. At 5V it should work okay.

Bob

#### dl324

Joined Mar 30, 2015
12,917
You can put diodes across the BE junctions to prevent them from breaking down.

#### Audioguru again

Joined Oct 21, 2019
3,304
You can put diodes across the BE junctions to prevent them from breaking down.
Diodes ACROSS the base-emitter will prevent the capacitors from discharging slowly. The diodes must be in series with each base.

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#### ci139

Joined Jul 11, 2016
1,696
circuit works with a 9VDC power supply and not with a 9VDC battery
because the supply outputs near perfect +9Vdc
but the battery if fresh has 2Ω Internal_resistance (Wiki) from where we can estimate ::
• the constanly on LED current (9 - 2.14)·1V / 330Ω ≈ 21mA
• the intermittent LED-s current (9 - 2.14 - 0.38)·1V / 220Ω ≈ 29mA
• so the approximate dynamic external load resistance (neglecting the drive/base current) for 6LF22 9V battery is optimistically ::
.. 9V / (21mA + 29mA) = 9 / 50mA ≈ 0.18 kΩ = 180ohms
so adding an aging/discharge factor of √(10)× to the 6LF22's internal resistance we got U = ε - ir = ε (1 - r/(r + R)) =
= ε · (1 - 1/(1 + R/r) ) =ε (1 + R/r - 1) / (1 + R/r) = ε·R/(r·(1+R/r)) = ε·R/((r+R)) = ε·1/(1+r/R) =9 / (1+ 2·√(10)/180 ) ≈ 8.69V ←
, chk U = ε (1 - r/(r + R)) = 9 · (1 - 2 · √(10) / (2 · √(10) + 180) ) ≈ 8.69V ← that may be enough to keep the oscillator from starting ↑
________________________

the fast fix may be introducing the emitter feedback resistors from 1 to 27 Ω and/or trimming the 100kΩ pair by 2k2 ohm potentiometer

. . . and perhaps 4k7 to 27k to bases of both transistors . . . Fig.1 (falstad simulator) . . . comes out the current to LED-s won't be sufficient with the last suggested mod so an "alternate compile" - - Fig.2 (falstad simulator) . . .

what i'm not telling here is :

the type of oscillator has actually quite narrow range of tuning for the relative name values of the resistors​
actually you try to make an "all in one" type design where variable duty oscillator and the power amplifier (the LED driver) have been merged​
// both of which actually require a specific setup on their own to perform their function best​
________________________

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#### Thoms_S

Joined Aug 23, 2014
13
How about a photo of the PCB front and back. What did you use for caps?
Hi Sam.
Did you see my attached image? Not giving you grief, I'm new so I want to know if you see it (I do) to make sure I'm doing things correctly here that's all.
They are 22uF 50v Caps with 100k resistors for the RC ckt - I used the 'Digi-Key Resistor-Capacitor (RC) Time Constant Calculator' to find out what I should use to get the delay.
Front/Back pictures are attached.

#### Thoms_S

Joined Aug 23, 2014
13
I don't see how they can both be on at the same time. Have you used an oscilloscope to look at the voltages?
Hi there, thank you for looking at my note.
I don't have an oscilloscope.
I made a video so you can see the "run away" blink rate when it is on the battery.
The blink rate speeds up so fast it looks like a constant on.

#### Thoms_S

Joined Aug 23, 2014
13
At 9V that circuit will exceed the max reverse voltage on the bases of most transistors. At 5V it should work okay.

Bob
Hi there BobTPH.
I looked up my transistor BC337 here:
-- components101.com/transistors/bc337-transistor-datasheet-pinout-equivalent --
and found out...
' ' The BC337 is an NPN transistor with a maximum gain of 630, commonly used in low power audio applications. It can also switch loads upto 45V and 800mA hence also used as general purpose transistor. ' '
So I'm guessing 9v will be fine.
Thanks.

#### Thoms_S

Joined Aug 23, 2014
13
because the supply outputs near perfect +9Vdc
but the battery if fresh has 2Ω Internal_resistance (Wiki) from where we can estimate ::
• the constanly on LED current (9 - 2.14)·1V / 330Ω ≈ 21mA
• the intermittent LED-s current (9 - 2.14 - 0.38)·1V / 220Ω ≈ 29mA
• so the approximate dynamic external load resistance (neglecting the drive/base current) for 6LF22 9V battery is optimistically ::
.. 9V / (21mA + 29mA) = 9 / 50mA ≈ 0.18 kΩ = 180ohms
so adding an aging/discharge factor of √(10)× to the 6LF22's internal resistance we got U = ε - ir = ε (1 - r/(r + R)) =
= ε · (1 - 1/(1 + R/r) ) =ε (1 + R/r - 1) / (1 + R/r) = ε·R/(r·(1+R/r)) = ε·R/((r+R)) = ε·1/(1+r/R) =9 / (1+ 2·√(10)/180 ) ≈ 8.69V ←
, chk U = ε (1 - r/(r + R)) = 9 · (1 - 2 · √(10) / (2 · √(10) + 180) ) ≈ 8.69V ← that may be enough to keep the oscillator from starting ↑
________________________

the fast fix may be introducing the emitter feedback resistors from 1 to 27 Ω and/or trimming the 100kΩ pair by 2k2 ohm potentiometer

. . . and perhaps 4k7 to 27k to bases of both transistors . . . Fig.1 (falstad simulator) . . . comes out the current to LED-s won't be sufficient with the last suggested mod so an "alternate compile" - - Fig.2 (falstad simulator) . . .

what i'm not telling here is :

the type of oscillator has actually quite narrow range of tuning for the relative name values of the resistors​
actually you try to make an "all in one" type design where variable duty oscillator and the power amplifier (the LED driver) have been merged​
// both of which actually require a specific setup on their own to perform their function best​
________________________
Errrrr this post is intimidating at best.
Not being snarkey, it's just like --- wow! I'm just doing a copy and build the thing for a friend.
All of that is really way above me, but I sincerely do appreciate you looking at my post.
Thank you.

#### Audioguru again

Joined Oct 21, 2019
3,304
With a 9V battery, the capacitors charge to about 6.3V but the maximum allowed emitter-base reverse voltage is only 5V so the circuit will not work or work poorly with some transistors. The circuit will work perfectly with all transistors if the protection diodes are added.

A BC337 transistor has a current gain range from 100 to 600 and you do not know how much unless you order BC337-16 for around 160, a BC337-25 for around 250 or a BC337-40 for a current gain around 400.

#### ci139

Joined Jul 11, 2016
1,696
perfectly with all transistors if the protection diodes are added.
? forward conducting to the bases
(( PS! -- a note. most simulators DO NOT simulate zener-effect of the reverse biased BE BC junctions . . . ))
______________________________________
wow! I'm just doing a copy and build the thing for a friend
i hope i got it wrong -- (if not then) don't copy me ← it's because the real circuits need higher control currents and are less sensitive to differential voltage changes -- in one word -- you need to build test and tune them from scratch or from blank and perhaps modify/re-design/replace the insensitive control chains ←← it's also because i didn't had time for "in depth" analysis of the target design
_________________
other than that :: below is the module level design
it's more easily done with std. parts ( iC-s ) intended for shown function
i only provide it as a theoretical example of into how many parts the simple LED - flip-flop can be divided - to be able to configure 'em separately
, ,
No-one usually has time/nerve/parts to go this complex -- unless you need a more complete and reliable product.
Note: the shown bi-polar pass-gate is "on a run 'compile' / experiment" -- should be replaced --or-- left off if CD4013 or other divider is used
_____________________________
+ now i got some more time to test the "odds" of the circuit ( also included the suggestion by Audioguru again ) /!\ but it's still Spice experiment with somewhat "idealistic" components ← what that means is - that it is highly unlikely the circuit works in real without adjustments as it does simulate /// the D3 & D4 were put inplace of the 33MΩ resistors (they function as resistors and somewhat sharpen the closing of Q3 & Q2)

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#### Thoms_S

Joined Aug 23, 2014
13
At 9V that circuit will exceed the max reverse voltage on the bases of most transistors. At 5V it should work okay.

Bob
Hi Bob.
So I got a 4-AA battery holder and hooked it up..... U-R-GENIUS ….. it's been running correctly for about 10 minutes now.
I need to leave the 9v battery in, due to room for parts, so I'm going to see if I can fit a LM7805 on the board and then test again.
Thank you!
T

#### crutschow

Joined Mar 14, 2008
27,395
I need to leave the 9v battery in, due to room for parts, so I'm going to see if I can fit a LM7805 on the board and then test again.
You do realize that will waste over 40% of the battery's energy?

#### Tonyr1084

Joined Sep 24, 2015
6,060
I'm going to see if I can fit a LM7805 on the board
BAD IDEA! Will kill your batteries in no time. Will turn a lot of energy into heat. 9v - 5v = 4V. You're converting four volts at whatever amperage into heat energy. The regulator has to dissipate all that wasted power. A buck converter would be a better idea. Better yet, if it works fine on 4.5 volts then why not just build it that way?!

#### ci139

Joined Jul 11, 2016
1,696
Re - #15

i analyzed the circuit and it seems to dislike the LED-s in the Collector nodes -- i didn't quite got the reason why -- though it may be because the LED-s have a "sharp" forward "turn on" -- e.g. -- their FW U(I) graph is much like the Zeners' reverse conductance one . . . ← what it may mean is that at low bias their resistance is getting relatively higher than near the nominal ...
the resistance is shown below as log₁₀R -- the R Y graphs -- it goes over the MΩ-s near zero current ←
← to improve this set around 20kΩ resistors in parallel with your LED-s - so the Collectors won't be seeing higher than this resistance to Vs ( ← is the supply voltage , might also be labelled as Vdd or Vcc or other )
.
,

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#### Thoms_S

Joined Aug 23, 2014
13
BAD IDEA! Will kill your batteries in no time. Will turn a lot of energy into heat. 9v - 5v = 4V. You're converting four volts at whatever amperage into heat energy. The regulator has to dissipate all that wasted power. A buck converter would be a better idea. Better yet, if it works fine on 4.5 volts then why not just build it that way?!
HI there and thanks for the note!
The reason why I don't build it that way is I don't know how. I am a novice at this stuff. I read up what I can and try to apply it - and I don't always catch everything I should in my readings.

I found the circuit I'm using because it does what I want, but... it did not have all the parts I have. To save some  I googled for equivalents and then used what I had. Then the flash rate was not correct so I found out about R-C time and got that correct. I know what a LM7805 will do and if that will make it work I was gonna try it.

I sincerely appreciate everyone here who is trying to help, and I'm trying to apply what I've learned but I'm still stuck on 9v is 9v! Why will it run on a 9vdc power supply and not a 9vdc Duracell?

#### Audioguru again

Joined Oct 21, 2019
3,304
Why will it run on a 9vdc power supply and not a 9vdc Duracell?
The 9V power supply has a low impedance output capacitor but the battery does not. Add a 100uF capacitor on the circuit board from +9V to 0V.