5
\$\begingroup\$

I am an avid electronics hobbyist, not an electronics engineer.

I built a transistor-based astable multivibrator circuit and then fed its output to an op amp buffer amplifier which then drove an 8 ohm speaker. Here is an image of the circuit schematic:

schematic

I wanted the square wave output of the BJT transistor astable multivibrator to have "harder" edges, so I added two diodes and two resistors to the circuit so that each capacitor in the astable multivibrator, when charging, would not affect the nearby transistor.

The circuit I based my astable multivibrator on was discussed in this post to StackExchange:

Astable multivibrator diode improvement

Anyway, here is the deal. As a user pointed out in the other post, with the diodes in the circuit (the original circuit schematic used 1N4148 small signal diodes), sometimes the circuit does not start oscillating when you apply power to the circuit.

That is what happened to me. With the two 1N4148 didoes in the circuit, I could not get it to oscillate. I tried various approaches to "imbalancing" the circuit, such as adding a small value capacitor from the base of one transistor to ground, to get it to oscillate, but it would not oscillate.

On a whim, I replaced the two 1N4148 diodes with two 1N5818 Schottky diodes. Voila! The circuit started up and oscillated just fine! The edges of the square wave were much more vertical. I was happy.

Here is what I do not understand: Why would it matter whether the diode used in the circuit is a 1N4148 or a 1N5818 (Schottky) type diode? I know the forward voltage of the Schottky diode is lower, but I could not figure out how that would matter.

From what I can tell, all the diode does is force the capacitor in question to charge up through the new resistor (R2 and R5 in my schematic). Why would it matter whether the diode I use is a Schottky type with a low forward voltage, or a regular small signal diode with a larger forward voltage?

I am sorry if this is a newbie sort of question! Thank you in advance for your mercy.

By the way, the circuit works great. But I hate not knowing WHY.

Thanks in advance!


I built experiment circuits using the helpful guidance from the commenters and answerers. Much gratitude!

What I ended up with is this circuit:

astable multivibrator bjt transistors with op amp buffer amplifier on the output driving 8 ohm speaker

The original problem I had was that the circuit would not start up and start oscillating if I used 1N4148 small signal diodes as the "wave form correction" diodes. This version of the circuit solves that problem. This version of the circuit starts up and starts oscillating, just fine, even when I use 1N4148 diodes instead of 1N5818 Schottky diodes for the "wave form correction" diodes. I replaced the 1N5818 Schottky diodes with the 1N4148 small signal diodes for the wave form correction part of the circuit. See D1 and D2.

What is funny is that I was not trying to solve the "won't start up and start oscillating upon applying power to the circuit" problem when I added the diodes in series with the base of each transistor. I was trying to solve the "keep the base-emitter junction of each transistor from zenering because of the negative voltage applied to the base-emitter junction of each transistor on each cycle" problem. Adding a diode in series with the base of each transistor seems to have solved BOTH problems!

In addition to the problem of not oscillating when power was applied, a problem with the first version of the circuit, as an answerer pointed out, was that base-emitter junction of each transistor was being forced to endure a negative reverse voltage of up to -7 volts on each cycle. This is bad as it might cause the base-emitter junction to "zener" as the reverse breakdown voltage of the base-emitter junction of the transistor can be as low as -5 or -6 volts.

The solution I ended up with was in a magazine article by Ray Marston. Here is the link to Ray Marston's magazine article: "Bipolar Transistor Cookbook - Part 6". See https://www.nutsvolts.com/magazine/article/bipolar_transistor_cookbook_part_6 for the article. See figure 2.

In the magazine article, Ray Marston's solution to the "zenering" of the base-emitter junction problem is to put a diode in series on the base of each transistor. That way, the very large reverse breakdown voltage of the diode (about -100 volts for the 1N4148 diodes I used), would have to be overcome before the base-emitter junction will zener. See D3 and D4.

That is what I did. Now the base-emitter junction of each transistor is protected from zenering due to any reverse voltage up to -100 volts.

What I discovered, quite by accident, is that adding a diode in series with the base of each transistor solved the "circuit won't start oscillating when power is applied" problem. I am still analyzing the circuit to figure out why this is the case. Any answers to why the diode on the base of each transistor solves the "won't start oscillating on power up" problem are appreciated.

Ray also has a proposed solution to the "circuit won't start oscillating on power up" problem in his article. See Figure 6. I have not yet tried his solution in a test circuit.

Another change I made to the circuit: I replaced the single, 20K ohms potentiometer with a dual, ganged pot, for the RC circuit. The dual, ganged pot is in the circuit to allow the frequency output of the oscillator to be varied within a certain range. For circuits that only need one frequency to be produced, the pot could be replaced with two fixed resistors. Unfortunately, the only dual, ganged pot I had handy was a 100K ohms, dual, ganged pot, which I know, per one of the helpful answers, is too large a value when compared to the 2.2K ohms resistor connected to the collector of each transistor. I will have to be careful not to set the pot to too large of a value.

If I can figure out WHY adding a diode in series with the base of each transistor solves the "won't start oscillating on power up" problem, I will post that answer here.


It occurs to me, after studying the schematic, that I should put a fixed resistor of at least 2.2K ohms in series with each side of the 100K pot. That way, if the pot is rotated to its minimum value, there will still be SOME significant resistance for each of the two RC networks in the circuit.


* Update of March 1, 2019 *

Here is the newest version of this circuit as built on my breadboard:

bjt transistor astable multivibrator with op amp buffer and push pull transistors emitter follower on output

I added two new diodes (D5 and D6). These seem to have little practical effect on the circuit.

What did have an effect on the circuit was adding the two output transistors in an emitter follower "push pull" configuration. I did this at the suggestion of an answerer.

These two transistors allow the circuit to send a lot more current to the speaker as I can reduce the final variable resistor and hump about 180 mA of current (max) to the speaker.

The problem is that, when I do send a lot of current to the speaker, the op amp starts to draw about its maximum current on pin 7 - about 2.5 mA.

I tried using a 1K resistor on the output pin (6), but that also reduced the speaker output a little.

Any advice on how to crank the current to the speaker while protecting the op amp?

* Update of March 6, 2019 *

After reviewing the schematic drawing I posted on March 1, 2019, I realized that diode D6 was in the wrong place. It was connected to R6 and the cathode of D2. It should have been connected to R5 and the anode of D2. The schematic has been corrected and is posted below:

bjt transistor astable multivibrator with op amp buffer and push pull transistors emitter follower on output location of diode D6 corrected

\$\endgroup\$
  • 3
    \$\begingroup\$ apply power from a vdd that is,already fully up; not a vdd that slowly rises. \$\endgroup\$ – analogsystemsrf Feb 22 at 2:52
  • \$\begingroup\$ Analog, how should I modify the circuit to make sure that Vdd is fully when power is applied to the circuit? Would putting a capacitor across the power rails before the push button switch do that? \$\endgroup\$ – Brock R. Wood Feb 24 at 17:30
  • 2
    \$\begingroup\$ In general, when you have updates to your question, edit the question itself. Do not post them as answers! \$\endgroup\$ – Dave Tweed Mar 1 at 22:31
  • \$\begingroup\$ Understood. Gracias. Will edit original question from now on. \$\endgroup\$ – Brock R. Wood Mar 1 at 22:45
  • 1
    \$\begingroup\$ 2.5 mA is the max current the opamp will draw as static current to operate its internal circuits. It is not the max current that can flow in/out of the power pins to supply the output pin. I looked through the datasheet and did not find a max output current spec, but the part is characterized with a 2K load. That's over 6 mA of load current. I would keep the Q3 Q4 bases connected directly to pin 6. \$\endgroup\$ – AnalogKid Mar 1 at 22:51
1
\$\begingroup\$

The design has a few problems illustrated in marked up schematic.

1) hFE is only 10% when saturated. so Rb/Rc' should not exceed 20:1
2) Vbe goes negative below -5V which exceeds Absolute Max rating. (on most transistors) -6V in this case. meanwhile 9Vdc can produce ~>-7V pulse to base.

The Schottky fix is a bandaid because it overdrives the Vbe in reverse below the typical limit of -5V "Abs. Max rating"

Since this reduces the ramp -ve peak from -7V to -0.7V it also raises the frequency towards 2x. and not the 10x you would expect because the rise time starts linear then rapidly rises exponentially due to large of pulse current on cap fixed by the reverse diode load.

enter image description here

If you search me for Astable CMOS Schmitt trigger, you may find a much more reliable low power design then use the transistors on the Op Amp output as complementary emitter followers for more power.

\$\endgroup\$
  • 1
    \$\begingroup\$ freq doubles but more perfect square wave ANYTIME \$\endgroup\$ – Sunnyskyguy EE75 Feb 22 at 5:10
  • 1
    \$\begingroup\$ dar la bienvenida \$\endgroup\$ – Sunnyskyguy EE75 Feb 22 at 5:18
  • 1
    \$\begingroup\$ I think that's due to insufficient gain " Rb/Rc' should not exceed 20:1 " and you have (20k+3.3k) / 2k2//2k2 = 21 so its marginal. Increasing 2k2 should improve that. try 10k and work down. Loop gain needs to be >1 and hFE drops to 10% or so, another trick is small cap // Rb \$\endgroup\$ – Sunnyskyguy EE75 Feb 24 at 0:00
  • 1
    \$\begingroup\$ RV1 is the biggest Rb and use a cap ~5% of other caps after all this the CMOS Schmitt Trigger Astable is perfect with R negative feedback , using RV1 and 1 series R to limit max f otherwise you get a >30:1 frequency range with the pot and cap on input to gnd .. that's all. \$\endgroup\$ – Sunnyskyguy EE75 Feb 24 at 0:09
  • 1
    \$\begingroup\$ e.g. electronics.stackexchange.com/questions/277696/… \$\endgroup\$ – Sunnyskyguy EE75 Feb 24 at 0:15
1
\$\begingroup\$

The other thing the diode does is increase the apparent transistor saturation voltage as seen by the other transistor's base. You can recalculate the waveforms on both sides of C2 with diodes of 0.2 V and 0.7 V and see the effect.

\$\endgroup\$
1
\$\begingroup\$

You may have several problems going on here. By their nature multivibrator circuits tend to NOT oscillate if both sides are somewhat balanced in transistor beta, temperature, and resistor 'true' values. Also it helps to have a fast rising supply voltage so slight differences in capacitive values can start the oscillator.

Odd that it will start with Schottky diodes and not the ultra-fast 1N4148. Speed and forward Vdrop are not the issue. Yes the Schottky has a lower forward Vdrop, but is ten times slower than the 4ns 1N4148. I suspect the Schottky diodes are not as balanced in terms of Vdrop, and it takes imbalance to start the oscillator.

NOTES:

  1. You very much need to add decoupling capacitors. A 100nF across the IC power pins will stabilise it. Also a 100uF 25V aluminum capacitor across the supply rails (after the power switch) will stabilise the voltage on the power rails. Yes, it could change frequency some and the way it starts could change.

  2. Ideally both types of diodes should work as expected. The 1N4148 should actually work better for rise and fall times.

  3. The voltage is too low for the TL0xx series. They expect +/- 6 volts minimum. If it is distorted you can add a 10K resistor from pin 6 to Vcc. The offset current will stabilise the op-amp on single-ended power supplies.

  4. Another big help would be to use two 9 volt batteries to get 18 volts. The rise in current will not hurt anything as it is so small to begin with, but it will help the op-amp stabilise and crank out extra volume with no distortion.

\$\endgroup\$
  • \$\begingroup\$ You know, when I was building the circuit, I DID use very close tolerance parts, such as 1 percent resistors from the same batch, 2N3904 transistors from the same batch, and 1N4148 diodes from the same batch. Even the capacitors were nice, military surplus MLCC types, from the same batch. I may have, unintentionally, made the oscillator circuit really, really balanced on each side! I will do some experimenting, per your suggestions, and see what happens! \$\endgroup\$ – Brock R. Wood Feb 22 at 4:11
  • 1
    \$\begingroup\$ You seem to know what you are doing, and just need a few pointers to get it working predictably. Good luck. \$\endgroup\$ – Sparky256 Feb 22 at 4:27
  • \$\begingroup\$ Sparky, get this. When I put the 1N4148 diodes back into the circuit again, alas, no oscillation. Then I added the 100uF 25V capacitor across the power rails as you suggested. Voila! Oscillation with the 1N4148's! Downside: With the capacitor across the rails, I got an annoying slow decrease in pitch when I released the button. Solution: Even with a mere .1 uF capacitor, the oscillation would start. With the small value capacitor the oscillator stopped quickly upon releasing the push button with no annoying pitch decrease. \$\endgroup\$ – Brock R. Wood Feb 22 at 5:11
  • 1
    \$\begingroup\$ I am glad you are working the problem, and results are predictable. I may not have the best answer, but I hope I gave you some useful tips. \$\endgroup\$ – Sparky256 Feb 22 at 5:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.