fig. 1enter image description here

I've witnessed that the output square wave from an astable multivibrator using two BJTs has its edges rounded during the rise up time(just as it is seen in fig. 1). Across resources, I've come to know that this is due to the initial current drawn in by the capacitor, or precisely the way the capacitors get charged up. I'm aware of the way an astable multivibrator works, but I'm unable to understand how does the charging of the capacitor lead to a change in the rise up time of the square wave. Which part of the multivibrator is actually causing the nuisance here? Also, how can we actually show the error dependence in the rise up time with an error in that part creating nuisance, mathematically? It would be really helpful if someone could give a nice explanation with the mathematical workout in the dependence.

  • \$\begingroup\$ There is more than one way to make an astable multivibrator (aka oscillator) with two BJTs and capacitors (don't forget the resistors). Care to post a schematic of the circuit you are asking us to analyze? \$\endgroup\$
    – Aaron
    Aug 4, 2021 at 17:35
  • \$\begingroup\$ Sure. I'm doing it right now. \$\endgroup\$ Aug 4, 2021 at 17:36
  • \$\begingroup\$ I've added now. \$\endgroup\$ Aug 4, 2021 at 17:39
  • 1
    \$\begingroup\$ The rising output at collector Q2 is when Q2 is off. So take that out of the circuit and you are left with C2 charging through R4. That is an RC circuit. \$\endgroup\$
    – Aaron
    Aug 4, 2021 at 18:11
  • \$\begingroup\$ You might want to sho the waveform and highlight the feature you're asking about. \$\endgroup\$
    – user16324
    Aug 4, 2021 at 18:11

2 Answers 2


Let's assume that the output is Vout, at the junction of R4, C2 and the collector of Q2.

Start by understanding why the falling edge of Vout is so rapid. There's a relatively slowly rising voltage at Q2's base, as C1 charges, but the transition from off to on occurs only within a very small "window", perhaps between 0.6V and 0.7V. Thus in spite of the sedate pace of the rising base voltage, the transistor switches on comparatively quickly, as its base reaches 0.6V or so.

At that point, Q2 "instantly" becomes a dead short between collector and emitter, pulling Vout immediately to 0V.

However, when Q2 eventually switches off again, similarly rapidly, the transition of Vout is not rapid at all. This is because there's no such "dead short" to Vcc like there was to ground.

In the absence of a transistor to pull Vout hard and quickly towards Vcc, as Q2 did to pull it low, there's only R4 to perfom that function. In comparison to the "dead short yank" of Q2 to ground, R4 is merely a "gentle tug" back up!

The voltage across C2 changes only as fast as R4 permits, because together they form the familiar resistor capacitor series pair (don't forget, Q2 is off, and can be treated as if it were not even there). The voltage Vout, at their junction, will rise exactly as you would expect from such a pair, exponentially, with a time constant of R4 x C2, which is what you see in the graph.

I'd like to point out that this is somewhat simplified, because there is also Q1's base and R3 to consider, but the dominant players in the rise of potential Vout, at Q2's collector, are C2 and R4.

In summary then, Q2 switches on rapidly, dropping Vout to zero more-or-less instantly. But when Q2 switches off, it is the combination of C2 and R4 that determines the rise of Vout, which is approximately the classic exponential curve with time constant R4 x C2.

  • \$\begingroup\$ That's a great answer. Thanks for that. Also, for nicer situations, which one would be easier to change, the resistor or the capacitor(value) so as to somewhat adjust the round edges? \$\endgroup\$ Aug 4, 2021 at 19:30
  • \$\begingroup\$ I know that adding diodes help, cause it'd be as fast as the BJT to make a transition, but if we had to only choose between R4 and C2, which one would've been a better gamble? \$\endgroup\$ Aug 4, 2021 at 19:32
  • \$\begingroup\$ @AmpliFire: Well, capacitor C2 and R3 determine (almost exclusively) the interval t1 (or t2) in your graph. Since you've already determined those values to set this interval, C2 isn't really an option. You're only really left with R4 to play with, so to get C2 to charge faster, and get Vout up to Vcc as quickly as possible, make R4 as small as you can without blasting anything into orbit. \$\endgroup\$ Aug 4, 2021 at 19:44
  • \$\begingroup\$ @AmpliFire: The best you can do with this circuit is to mininise the rise time in comparison to t1 (or t2). You achieve that by making the ratio of R3 to R4 as large as possible (meaning make R3 much, much greater than R4), while keeping their values sane. \$\endgroup\$ Aug 4, 2021 at 19:46
  • \$\begingroup\$ I see that now. That's nice. Thanks. \$\endgroup\$ Aug 4, 2021 at 19:47

This circuit is little tricky means if you not design it properly it will not oscilate, roundings on outputs can become extreme or another thing even small ripple in Vcc can cause false oscilation.

Regarding the roundings you can avoid them with high ratio R1:R2, lets say if you choose R1=1k you must use R2 about 47k at least. The reason is R1 charges C1 and you want to do it as fast as possible to roundings be a minimal in comparison to all signal waveform. But remember this high R1:R2 ratio require high beta transistors (I woulds say 200 at least).

  • \$\begingroup\$ Say, I would like to calculate a discrepancy in the rise up time of the output wave. So, this discrepancy is caused due to the charging resistor, or the capacitor that gets charged itself? \$\endgroup\$ Aug 4, 2021 at 18:29
  • \$\begingroup\$ This discrepancy would be required to know by how much I should change the resistor's or capacitor's value, whichever one is at fault, to correct the waveform. \$\endgroup\$ Aug 4, 2021 at 18:31
  • \$\begingroup\$ C1 is charged thru path Vcc->R1->C1->Q2_base so the waveform of Q1 collector voltage is shaped by R1,C1 time constant. Lower R1 or C1 quicks the charging. \$\endgroup\$
    – user208862
    Aug 4, 2021 at 18:41
  • \$\begingroup\$ Thanks, I would like to know which is more flexible in this case, changing R1 or C1? Capacitors might have residual charges so there is a chance for a sudden spike. Is it better to deal with the resistor? \$\endgroup\$ Aug 4, 2021 at 18:48
  • \$\begingroup\$ In real application the R1 affects the power consumption because half of period is one of transistor fully open. So you dont much to do with R1 value (if you care about power draw). Changing C1 affect the frequency on other hand. There is some place to variable R2. \$\endgroup\$
    – user208862
    Aug 4, 2021 at 19:06

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