Issue 262: EQ Answers

Problem 1—The classic two-transistor astable multivibrator is shown below. Typically, R2 and R3 have at least 10 times the value of R1 and R4. This circuit oscillates, with Q1 and Q2 turning on alternately. From the point in time in a cycle where Q1 first switches on, describe what happens until Q2 switches on.

Source: D. Tweed, CC262

Answer 1—Right before the moment Q1 switches on, C1 is charged to VCC – VBE, with its left end positive, and the left end of C2 has just reached +VBE. The right end of C2 is being held at VCE(SAT) by Q2.

Source: D. Tweed, CC262

So, as Q1 begins to switch on, it pulls the left end of C1 low, and this also pulls the right end of C1 low, cutting off Q2. This in turn allows the right end of C2 to rise, emphasizing the turn-on of Q1 by increasing the voltage (and current) at the base of Q1.

Once Q1 is fully on, the right end of C1 is now at VCE(SAT)– (VCC – VBE) (a fairly substantial negative voltage), and C1 now begins to charge in the other direction, through R2. Once the right end of C1 reaches +VBE, Q2 begins to turn on, starting the second half of the cycle.

Problem 2—What determines the time of one half-cycle of the oscillation? Does this depend on VCC?

Answer 2—The time of the half-cycle described previously is the time that it takes the right end of C1 to charge from –(VCC – (VBE + VCE(SAT))) to +VBE.


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Now, keep in mind that the capacitor is charging “toward” +VCC, but it gets halted by the B-E junction of Q2 at +VBE. This charging is occuring at a rate determined by the time constant C1 × R2, and we’re basically interested in the time that it takes to move halfway from its starting value to its final value. This works out to –ln(0.5), or 0.693 times the R-C time constant.

As long as VCC >> VBE, the time does not depend on VCC. That isn’t to say, however, that VCC can be arbitrarily large. If it exceeds the reverse-breakdown voltage of the transistors’ B-E junctions, current will flow and perturb the timing.

Problem 3—Recently, a different circuit appeared on the web, shown below. Again, R2 and R3 are significantly larger than R1 and R4. The initial reaction of one observer was that this circuit can’t work, because there’s no DC bias path for either transistor. Is this assessment correct?

Source: D. Tweed, CC262

Answer 3—No, it isn’t. This circuit can oscillate just fine. Again, look at how C1 charges and discharges.

Source: D. Tweed, CC262

If C1 starts out discharged, it will charge through R1 and the B-E junction of Q2. This current will turn on Q2, holding its collector at ground (really VCE(SAT)) and preventing Q1 from turning on.

However, as C1 reaches full charge, the current through it decays below a level that will keep Q2 turned on. When it starts to turn off, its collector voltage rises, which also forces current into Q1’s base through C2. As Q1 begins to turn on, it pulls its collector low, which also pulls the base of Q2 lower, emphasizing its turn-off. The circuit quickly “snaps” to the other state, with Q1 on and Q2 off. C1 is discharged through Q1 and D2 at the same time that C2 begins charging through R4 and Q1’s B-E junction.

Problem 4—What role do R2 and R3 play in this circuit?

Answer 4—R2 and R3 never have more than ±VBE across them; as a result, the current through them is negligible relative to the current through the capacitors. In other words, they’re superfluous.

Question 5—Does the timing of this circuit depend on VCC? If not, what does it depend on?


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Answer 5—The time from when one of the transistors turns on to when it turns off is determined by the currents flowing into its base and collector. When the current into the base drops below the value needed to sustain the current into the collector, the transistor begins to turn off, and the circuit feedback then insures that this happens quickly.

Looking at Q2, and ignoring the transient associated with discharging C2 for now, the collector current is set by R4. The initial base current is set by R1, but this decays exponentially with a time constant of R1 × C1.

Therefore, the primary determinant of the half-cycle time period (in addition to the R-C time constant) is the current transfer ratio, or hFE of each transistor. When the base current drops to a value of 1/hFE of the collector current, the transistor begins to turn off.

Since both currents scale in the same way with VCC, it has no direct effect on the timing. There is only a secondary effect if the value of hFE changes with the value of the collector current.

Contributor:  David Tweed



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Issue 262: EQ Answers

by Circuit Cellar Staff time to read: 3 min