# How to actually calculate the rise time for a square wave generated by an Astable Multivibrator?

If I have an astable multivibrator like the one shown in the picture, how do I actually calculate the rise time for the output square wave? I know that it's the time required to change from the low state to the high state, but how do I actually calculate it in this case? • So, is this a follow-on to this? I almost thought you got a good enough answer about the timing and which parts relate to computing it, there. Maybe I'm wrong, though. Are you looking for "within 20%"? Or what? And what power supply rail voltages? (BJTs avalanche in many cases making these things almost impossible to predict.)
– jonk
Aug 5 at 5:33
• Yeah, I did. But I'm having some problems related to the rise time. Is it determined by the time constant R1C1 or R4C2? I wanted to be sure, cuz I'm very much new to electronics. Sorry for that. Aug 5 at 5:35
• What do you understand about the term 'rise time?' How would you define it, quantitatively? How would you measure a single specific, quantitative value if I drew you a specific curve with both the x and y axis carefully labeled for you? (I'm hoping that you see this isn't a "I know it when I see it" kind of thing. This is engineering, not touchy feely. So things must have definitions. I just want to know what yours are.)
– jonk
Aug 5 at 5:49
• Yeah. If I were to give a definition of rise time, I would say that it's the time taken for the wave to change to it's high state from it's low state, since it's "rise". Also, since it's a square wave, it should have minimal rise time, due to some rounded edges. I want to calculate this time. Aug 5 at 5:53
• That's not a definition, unfortunately. In theory, it takes infinite time for an RC charging circuit to fully charge. So it never happens if you define it that way. You need a better definition. For example, $V_t=V_0\cdot\left[1-\exp\left(-\frac{t}{\tau}\right)\right]$. As $t\to\infty$ then $V_t\to V_0$. But that's a long time to wait!
– jonk
Aug 5 at 5:54

I'm going to provide some thoughts for you to consider. This won't exactly be a direct answer. But an answer will be contained here just the same.

Let's look at a simple case, using $$\1\:\text{k}\Omega\$$ collector resistor, $$\1\:\mu\text{F}\$$ caps, and $$\10\:\text{k}\Omega\$$ base resistors. At first blush, when a BJT turns off the collector resistor pulls up on and charges one end of its capacitor -- the other end being literally tied down by the base-emitter voltage of the other BJT and unable to move. So we'd compute $$\\tau=R\,C=1\:\text{k}\Omega\cdot 1\:\mu\text{F}=1\:\text{ms}\$$. This means that we expect a rise from close to $$\0\:\text{V}\$$ to about $$\5\:\text{V}\cdot\left(1-e^{^{-1}}\right)\approx 3.16\:\text{V}\$$ in $$\1\:\text{ms}\$$.

Let's take a look at two schematics: one that is as described above (case 1) and will have plenty of time to rise near the rail voltage and another one (case 2) that is changed so that it doesn't have that much time. simulate this circuit – Schematic created using CircuitLab

Let's use $$\V_{_\text{CC}}=5\:\text{V}\$$ to avoid questions of BJT avalanche at the base.

Here's an LTspice run: I've used LTspice to work out the time between the $$\\approx 0\:\text{V}\$$ and $$\\approx 3.16\:\text{V}\$$ points for the green curve as $$\103.49\:\text{ms}\$$ and $$\104.49\:\text{ms}\$$ and in the case of the red curve for the same voltage points as $$\103.66\:\text{ms}\$$ and $$\104.66\:\text{ms}\$$.

Note that in both cases the difference is as close to $$\1\:\text{ms}\$$ as is possible to achieve.

So, even though the frequency is quite different between these two designs, the "rise time" as you say is the same for both. There's no difference that matters. The key difference is that I've prevented case 2 from being able to have enough time to reach $$\V_{_\text{CC}}\$$. But they are otherwise the same.

The point I'm trying to make, aside from having a precise definition for the meaning of "rise time", is that you can actually compute the rise time if you define it, correctly. And, I suppose, that the rise time really doesn't tell you the frequency that the circuit will operate at. Other factors impinge on that. In case 2 I didn't allow it to reach $$\V_{_\text{CC}}\$$ before the next cycle started, again.

Definitions matter. That's the point.

• That's a really nice simulation. I'm starting to get the hang of it now. Thanks :) Aug 5 at 8:51
• @AmpliFire I'm glad to hear that. There is a kind of "getting the hang of it" to all this. It's really cool to realize that multiplying Ohms by Farads actually gives time (dimensional analysis is something else to learn about.) This time is called "tau." It's merely a way of saying that something approaches "this far" towards a goal in "this time." But that once that is achieved it takes the same time to get just that much closer, once again. It is often said that $5\tau$ is sufficient "for all intents and purposes." And that's pretty much true because that means 99.3% of the way there.
– jonk
Aug 5 at 8:59
• @AmpliFire But $4\tau$ is 98.2%, $3\tau$ is 95%, and $2\tau$ is 86.5%. So you get to pick your poison, so to speak.
– jonk
Aug 5 at 9:01

Time taken for collector of Q2 to go from ground (almost) to within 1% of Vcc

t = C2 * R4 * ln(Vcc/(1% of Vcc)) = 4.6 * C2 * R4

By adding the extra diodes and resistors shown in the circuit below you can obtain a square wave output. • Yeah I was having some difficulty to find out that constant in terms of ln, I knew that it would be proportional to R4C2. But then working out the equation from scratch helped me realize it. Thanks for the answer. Aug 5 at 8:53