Formula to Calculate exact Time period and tON time of Bistable Multivibrator - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: January 18, 2025, 05:40:42 pm
content: What is the time period(T) and on time (tON)time of this bistable multivibrator?I got the following result from Chat GPT.But after substituting the values i get,R = 10kΩC= 220μF T = -2 ×10×10^3×220×10^-6× ln ( 1 - 0.7/3.7 )VCC = 3.7 volt T= -4.4 × ln (0.8108)e = 2.718 T= -4.4 × -0.2097VBE = 0.7 voltT= 0.92268 second 922.68 milli second.But after simulating it in LTSPICE, I get the following result. Which shows the time period is nearly (532 -501) milli second.I want to know the correct formula to calculate the exact Time period and tON time?

author: ledtester
author_rank: Super Contributor
timestamp: January 18, 2025, 07:14:43 pm
content: The ChatGPT calculation assume Vbe(ON) is 0.7V, but perhaps this isn't the case for the transistors you are using the simulation. In any case, LTspice uses a much more sophisticated model of transistors than what the ChatGPT assumes.Also, why are you using a BC546B on the left and a BC547B on the right?In the real world all components will have variation and can't be assumed to have exact values, so any calculation will only be an estimate.

author: Zero999
author_rank: Super Contributor
timestamp: January 18, 2025, 07:32:17 pm
content: That's an astable, not bistable. A bistable is latch, not an oscillator.Using different transistor models in SPICE is a good idea, because the simulator will have difficulty starting the oscillation, if it's perfectly symmetrical.I don't see any mention of the LED forward voltage in the calculations.1R is far too lower value for the current limiting resistors.Post the .asc file.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 18, 2025, 10:57:47 pm
content: (Using the component numbering in the BC546 plus BC547 diagram.) When Q1 switches on the base of Q2 and the end of C1 connected to it goes negative by (Vcc-Vce1). Immediately beforehand that point was at Vbe2. So the absolute voltage at Q2 base becomes (Vbe2-Vcc+VCe1).That makes the "intial charging" voltage across R1 equal to Vcc-(Vbe2-Vcc+Vce1). That is 2Vcc-Vbe2-Vce1. Note the 2 times Vcc.And yes, 1 Ohm is much too low for the LED series resistors.

author: PGPG
author_rank: Frequent Contributor
timestamp: January 19, 2025, 12:43:36 am
content: Quote from: wasedadoc on January 18, 2025, 10:57:47 pmAre you sure?

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2025, 11:45:18 am
content: Quote from: PGPG on January 19, 2025, 12:43:36 am

author: wasedadoc
author_rank: Super Contributor
timestamp: January 19, 2025, 12:16:47 pm
content: Quote from: PGPG on January 19, 2025, 12:43:36 amYes.

author: tggzzz
author_rank: Super Contributor
timestamp: January 19, 2025, 01:19:12 pm
content: Quote from: khatus on January 18, 2025, 05:40:42 pmLLMs hallucinate, much like politicians.LLMs work by assembling sentences and paragraphs from words that are frequently seen in proximity. That's what human buzzword bullshitters do, e.g. salesmen and incompetent managers.In order to work out whether a statement is a hallucination or is buzzword bullshit, you have to know the correct answer - in which case why are you using an LLM?

author: Ian.M
author_rank: Super Contributor
timestamp: January 19, 2025, 01:47:07 pm
content: Note that LTspice and most other circuit simulators do not usually model Base Emitter junction reverse Zener or avalanche breakdown. Seehttps://ltwiki.org/LTspiceHelp/LTspiceHelp/Q_Bipolar_transistor.htmLTspice uses SPICE3 modified Gummel-Poon models for >99% of the transistors in its library. It has the capability to use VBIC models which include reverse breakdown, but only three devices in standard.bjt have VBIC models (+ one extra if using Bordodynov's library). It is also possible to simulate reverse breakdown using a subcircuit model, but they aren't in the standard library. You may encounter one as a 3rd party OEM model.In the real world, at higher supply voltages, the astable circuit's base voltage negative peaks may be clamped by the base emitter junction breakdown, shortening the off time. The clamping voltage is *NOT* a datasheet parameter and can vary considerably from device to device of the same type. If you want predictable timing, either keep the supply voltage below the datasheet abs. max. reverse Vbe (6.0V for BC546/7/8A/B/C) or modify the circuit e.g. with series base diodes so significant reverse Vbe breakdown current cannot flow.

author: PGPG
author_rank: Frequent Contributor
timestamp: January 19, 2025, 01:50:41 pm
content: Quote from: wasedadoc on January 19, 2025, 12:16:47 pmCan you explain how did you become convinced that turned off LED will provide enough current to be able to charge in time the 220uF to VCC because I still have doubts.

author: NightMoth
author_rank: Contributor
timestamp: January 19, 2025, 04:10:31 pm
content: HelloI did very same project more than year and half agoI found formula for multivibrator without LED herehttps://rayshobby.net/wordpress/learning-electronics-1-multivibrator/Then I modified formula from above link (taking into account LEDs) and got this formula:T = -(R1*C1 + R2*C2)*ln((Vcc - Vb)/(2Vcc - Vb - Vled))(TON should be 1/2 of T)My circuit behaved as predicted by the formula (for my values of R and C):Calculated with the formula: f = 1.994 Hz, T = 0.502 sReal circuit behavior: f = 2.013 Hz, T = 0.497 sAttaching my circuit with commentsThat time I tested it with multimeter and timer, so I dont have oscilloscope waveforms

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2025, 09:27:11 pm
content: Quote from: Ian.M on January 19, 2025, 01:47:07 pmThat's true, but the supply voltage in this case is 3.7V, which is well below the reverse breakdown voltage of the BC547.Quote from: NightMoth on January 19, 2025, 04:10:31 pmDoh, I wanted the original poster to figure that out for himself.Thanks for posting.Also note that the LEDs never fully turn off, although it might not be noticeable. Putting another resistor in parallel will solve that, as well as making the circuit less sensitive to the LED's forward voltage.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 20, 2025, 01:17:37 am
content: Quote from: PGPG on January 19, 2025, 01:50:41 pmThe LED doesn't have any part in the charging of the capacitors.

author: PGPG
author_rank: Frequent Contributor
timestamp: January 20, 2025, 09:00:40 am
content: Quote from: vk6zgo on January 20, 2025, 01:17:37 amSo which way they get VCC (at their ends connected with collectors) needed to allow for negative step by (vcc-Vce1)?About what is the Vce1 value in this equation I have also doubts. I suppose that at the moment of switching it is small, but shortly (when base is no longer driven by first pulse loading capacitor but only by resistor) then collector current is limited by base resistor and beta and it can be not enough to get small Vce1 (LED V/I curves have to be analysed). This move up at collector will correspond to the same move up at the other capacitor end so the process of charging this end (at base) is from different point at first and then from different point after transistor is not able to drive LED with enough current. I am all the time speaking about a circuit for which the question was asked.The question was to find equations for circuit and not circuit to satisfy assumed equations.

author: Zero999
author_rank: Super Contributor
timestamp: January 20, 2025, 07:11:57 pm
content: Of course the capacitors charge via the LEDs. Note the current through the capacitor as it charges, is equal to that through the LED, as it turns off. This is why it's a good idea to buffer the outputs, if it's being used to drive large loads.What colour are the LEDs? I presumed white, which I've modelled as 4.5 silicon diodes in series, note the N=4.5 next to the LEDs.Another way to kickstart the simulation is to declare the initial voltage on one of the capacitors as zero, hence the ic=0 statement.To make the plots easier to read, I skipped the first cycle, so the circuit is oscillating in its final state.

author: PGPG
author_rank: Frequent Contributor
timestamp: January 20, 2025, 08:17:12 pm
content: Quote from: Zero999 on January 20, 2025, 07:11:57 pmGood job to find how this circuit really works.I was using PSpice, generally long time ago. In 90s our access control system was communicating with PC using 20mA current loops (1200m, 4800 bps). Designing their circuits my idea was:- for 0..5mA all current was going through resistor,- for 5..10mA all current over 5mA shell go through transoptor LED,- for 10..20mA all current over 10mA shell go through red LED indicating that loop transmits some communication.I have measured V=f(I) characteristics of transoptor LED and red LED. If remember well to model them (in interesting region) I used voltage sources inside models. So my remember tells me that I don't have good LED models to be used in this circuit and also my PSpice version refused to work when I moved from Windows-XP to 64 bit Windows 7. Practically I don't have Spice at hand as I didn't needed it since 10+ years. And an year ago I spend one day trying how Spice in KiCad works.Those time (90s) I was reading this PSpice documentation carefully. I don't remember of having possibility to use something like N=4.5, but it can be I just missed it. I don't remember any texts (except references and values) used at PSpice schematic. When (20 years ago) I tried once LTSpice I was surprised that I have to use texts at schematic to specify this or those.Going to the beginning of this thread:- does it work (in simulation) with R3 = R4 = 1 ?- does it work with R3 = R4 = 0 ? - Chat GPT was asked to find equations for such circuit, what in my opinion is very far from a trivial task.