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This is the schematic for a modified astable multivibrator I've built. I incorporated the diodes in order to get a sharper rise time, and therefore produce actual square waves.

Principle of Operation (As I Understand It):

One transistor will turn on first due to minute differences in the gain of the transistor. Assume Q1 turns on first.

1) Q1 Turns on. 2) The left side of C1 is at .7V. The right side of C1 is at 0V. 3) C1 charges through R3 at \$T = R_3C_1\$. 4) When C1's right plate achieves .7V Q2 turns on. Because Q2 was previously off, the right side of C2 was at rail voltage during the previous state. Since the right plate has dropped 5V the left plate also 5V. The negative voltage at C2's left plate is coupled to Q1 holding it off. 5) R4 charges C2 from negative rail voltage to +.7V at which point the states switch and the process repeats.

Unfortunately, this oscillator is producing a standard voltage. This is confirmed by a simulation.

What's the problem? What are the errors in my thinking?

EDIT

See related question here.

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The basic circuit is OK but the resistors R3 and R4 are much too low in value. I would expect them to be about about 10 times the collector resistor (500 ohms as it consists of 2 1K resistors in parallel) so 5 k or so. The transistors are so heavily biased into conduction that the cross-coupling (C1, C2) cannot turn them off.

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  • \$\begingroup\$ Meaning that the equivalent negative voltage that builds up is not at a high enough current to turn the base off? So, the collectors must have a lower value than the base resistors? \$\endgroup\$
    – Allenph
    Jun 10, 2015 at 2:27
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The diodes are the problem. As modified, the circuit has a stable state, with both transistors turned on, and about 0.7 volts on both sides of both capacitors.

Consider that the circuit was turned off for "long enough" for the caps to discharge to zero. When you apply power, the caps start out at zero, and R2, R3, R4, and R5 all follow the power rail as it starts to go up. When the transistor bases get up to about 0.7 Volts, the transistors turn on with the current through those resistors. That clamps the collectors near zero. The anodes of the diodes can get pulled up to 0.7 above the transistor collectors, and that's enough to turn on the transistor bases.

You could replace your 1N4007 diodes with lower-voltage Schottkey diodes (maybe a BAT48, or 1N5817, 18, or 19), and that might allow the circuit to oscillate, but I'm not sure you'd really achieve your goal.

Here's a selection of likely Schottky diodes at Digikey: http://tinyurl.com/qat53yk

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There are two problems with your circuit, the diodes and the base bias resistors R15 and R16, the diode drops keeping the transistor bases above cutoff, no matter what, and the 330 ohm resistors putting too much current into the transistors' bases for each other's collectors to suck out and turn the transistors off.

Here's your circuit and one that works, so you can compare them, and here's the LTspice .asc file so you can play with the circuits if you want to.

enter image description here

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  • \$\begingroup\$ The diodes are a common way of speeding up the positive edge on the collectors. Without the diodes the positive edge rate is limited by having to charge the capacitors. The diodes shouldn't stop it oscillating. On a side note your design has the 10:1 ratio between the collector and base resistors I was suggesting. kevin \$\endgroup\$
    – Kevin White
    Jun 10, 2015 at 3:46
  • \$\begingroup\$ @KevinWhite: So post a circuit with diodes to prove your point. So you think I plagiarized your answer? What gall! The ratio of the collector resistance to the base bias resistance isn't critical; in my circuit, with 1k of collector resistance, anything from about 5k to >100k works for base bias. What really matters is that the edge rate of the collector going low be high enough that the reactance of he coupling cap look small enough that Vce of the transistor turning on plus the drop through the cross-coupling cap be small enough to turn the other transistor off. \$\endgroup\$
    – EM Fields
    Jun 10, 2015 at 4:29
  • \$\begingroup\$ The posters original topology is correct. The value of the resistors is the only issue. Here is an example with the diodes learnabout-electronics.org/Oscillators/osc41.php. As you say the ratio of the collector to base resistors is not critical, however there must be enough bias current to saturate the transistor and commonly an HFE of 10-20 is assumed. If you used 100k in your circuit it would require a transistor that guaranteed an HFE of >100 which even these days cannot be assumed. \$\endgroup\$
    – Kevin White
    Jun 10, 2015 at 19:10

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