# Issues with a modified BJT multivibrator?

Discussion in 'General Electronics Chat' started by Allenph, Jun 9, 2015.

1. ### Allenph Thread Starter Member

May 27, 2015
76
2

but I didn't want to resurrect an older thread.

This schematic is a BJT astable multivibrator with diodes in the feedback channel to decrease rise time. (I.E. get actual square waves.)

I think I have a handle on how it works, but let's make sure. (Assume Q1 turns on first.)

1) Q1 is ON, so all the current through R1 and R2 is diverted to Q1's collector and into ground. The left plate of C1 is at 0V.
2) C1 charges through R3, dictated by T = R3(C1). C2 is at rail voltage on both plates.
3) When the right plate of C1 achieves +.7V, Q2 turns on. C2's right plate suddenly becomes 0V, and the left side decreases from +5V to 0V turning Q1 off.
4) The process repeats in the opposite direction.

I've built this circuit, but unfortunately it's not working.

The 1k resistors limit the current to the collector of the transistors. According to the datasheet, 2N3904's are able to handle 200mA. At 5V and 1kΩ, using Ohm's Law the current shunt should only allow 5mA which is more than safe.

I tried my best to get the frequency of the output. What I came up with is...

RC = T

T5 = 5V

T = 1V

T/1V = X(.6V)

1/X = Ψ

In the same order substituting my values:

(3.3 * 10^2)(2.2 * 10^[-9]) = 7.26 * 10^[-7]

(7.26 * 10^[-7])/1 = X(.6)

X = 1.21 * 10^[-6]

1/(1.21* 10^[-6]) = 8.2644628099 * 10^5

If my process and math is correct, I should get a frequency of about 826446 Hz, or about 826kHz.

However, the output of my circuit is nothing. Not even a voltage. It just sits at 0V on my oscilloscope.

Is my explanation of the circuit correct? Is my process correct? Is my math correct? Is this component failure, or a circuit design flaw?

Thanks!

2. ### upand_at_them Active Member

May 15, 2010
246
29
How would C2's right plate ever be zero volts? The diode prevents it from ever getting below 0.7V.

3. ### Allenph Thread Starter Member

May 27, 2015
76
2
That's true, but for all intents and purposes it can be considered to be at 0V. The only thing that matters in this situation is that the potential difference be or not be .7V on the capacitor's plates. The explanation was simplified to omit the diode's voltage drop.

I.E. the capacitor will actually discharge and charge at 1.4V, but since the left plate will never be below .7V, you can just consider a difference of .7V.

Right?

4. ### upand_at_them Active Member

May 15, 2010
246
29
Last edited: Jun 9, 2015
5. ### Allenph Thread Starter Member

May 27, 2015
76
2
Thanks!

A couple questions...

1) Why are plates of the capacitors ever going negative?
2) Isn't the time constant dictated by R3 and R4? Changing them to 10K could cause issues with frequency.
2) Why does it matter what the base current is?

6. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,698
2,752
Why don't you just sym it in LTspice?

7. ### Allenph Thread Starter Member

May 27, 2015
76
2
He supplied a sim.

These are more theoretical questions, aren't they? Questions about WHY that would be happening.

For instance my first question, "Why are plates of the capacitors ever going negative?"

In an instance where Q1 has just turned on, I'm not sure why the right hand plate of C1 should be negative.

I see no reason why the right hand of C1 should be at 0V when the change in state occurs. I understand that when that change in state does occur, the left hand plate of C1 will be dropped from +V to GND, therefore the right plate must undergo a proportional drop in voltage. The problem is that, as far as I can tell, the voltage at the right hand plate of C1 should not be 0V before the change of state occurs.

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8. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,698
2,752
You're right... simulating something is not equal to understanding it... now you've got me curious... I think I'll sim it myself and see if I can come up with an answer.

9. ### Allenph Thread Starter Member

May 27, 2015
76
2
I think I might have figured out my own question.

The aforementioned plate on the capacitor is at 0V during the switch because that plate is also connected to the base of a transistor which is currently on. Since that is the case, all the current that would have brought that plate of the capacitor to +5V is flowing to ground through the base.

This messes up my process a little bit. Since it would be at -.7V during the switch, this portion of my process:

(7.26 * 10^[-7])/1 = X(.6)

Will actually be:

(7.26 * 10^[-7])/1 = X(1.4)

Because the capacitor has to overcome the negative voltage as well as get to +.7V.

Correct?

10. ### Allenph Thread Starter Member

May 27, 2015
76
2

2) Isn't the time constant dictated by R3 and R4? Changing them to 10K could cause issues with frequency.
3) Why does it matter what the base current is?

Yes. R3 and R4 dictate the constant. So again, I don't know why we needed to change these to make it work, or why it helped. Using 330 for the other resistors is not really doing anything logically. It's just changing our output current.

The base current matters because the output current is dictated by the current at the base. But, here, we're not really concerned with that, so in essence as long as we're within safe ranges it doesn't* matter.

Yes?

11. ### Allenph Thread Starter Member

May 27, 2015
76
2
I think I figured it out.

The answers to both question 1 and 2 are similar.

The reason it wasn't working is because I was just thinking about voltage, not current. In this configuration the base resistors are smaller than the collector resistors. This is a problem, because when the capacitors couple the voltage drop to the base of the other transistors, their drop is not only proportional in voltage, but also current.

This means that when the drop occurred, it wasn't enough to turn the bases off. That effectively halted oscillation.

12. ### DickCappels Moderator

Aug 21, 2008
2,753
665
This is probably not what is stopping your circuit from working, but it will probably keep the rise times from being as fast as you expect.

1N4007's typically have a reverse recovery time of 30 us, meaning that they aren't good at switching at 900 kHz. You might want to try a faster small signal diode such as 1N914, 1N4148, 1N915, iN456. Even the collector-base junction of a 2N3904 would work well if you don't have any fast small signal diodes.

Allenph likes this.
13. ### Allenph Thread Starter Member

May 27, 2015
76
2
This is something I was going to add. I was getting a very strange rise in my waveform. I kind of thought I needed a shotkey, now I know I'm right! Thanks.

14. ### Bordodynov Active Member

May 20, 2015
670
194
I'm a little bit modified multivibrator to increase frequency and reliable start.
View attachment 86984

Last edited: Jun 11, 2015
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15. ### Bordodynov Active Member

May 20, 2015
670
194
All of these calculations may not be accurate at high frequencies.
For example, they cannot consider turn-off time bipolar transistor.
I tried to fix the known limitations of the simple schema multivibrator.
I applied the Schottky diodes. I have turned into 1.8 Mhz switching frequency.

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