Not sure why current doesn't flow through these paths

Thread Starter

Tp86

Joined Sep 13, 2023
41
Hi all. I am looking at a new circuit which uses a switch to activate a fast switching astable multivibrator oscillator and when the switch is removed it stops the oscillating processes and only 1 or the transistors will remain on. (Its like a decision maker machine)

I understand mostly how the astable multivibrator works but only it its form without the human-operated switch. With the switch labeled "GO" and extra 2 paths with resisters R5 and R6 a have a question

Why does the astable multivibrator not continue to oscillate when the GO switch is either open or completely removed from the circuit due to the paths which I have colored in red and blue that seem to be a complete circuit to allow current continue charging the capacitors and allow current through the Base Emitter?

1701222666166.png
 

panic mode

Joined Oct 10, 2011
2,670
pick state and see what is going on...

since it is symmetrical lets assume Q1 is on and Q2 is off...
that means Vce for Q1 is low (about 0.1V) and Vbe is "high" (0.7V)
opposite is on Q2, Vce is high (couple of volts) and Vbe is low (close to 0V) since supplied from 0.1V on /q1 collector through R5.

now for change to take place, C1 need to be charged high enough so that Q2 can turn on.
but that never happens unless base of Q2 is connected to sufficiently high voltage.
in this case it is only connected to 0.1V which is too low to turn it on (need to be 0.6V or more)
 

MrChips

Joined Oct 2, 2009
30,473
With the switch not closed, the circuit is a bi-stable.
Let us assume that Q1 is OFF and Q2 is ON.
Let us look at what is happening at Q2.
Q2 base is pulled high.
Q2 collector is close to 0V.
C2 is discharged through R6.

At Q1, its collector is close to 6V.
C1 is also discharged through R5.

The circuit is stable and does not oscillate.
 

Thread Starter

Tp86

Joined Sep 13, 2023
41
since it is symmetrical lets assume Q1 is on and Q2 is off...
that means Vce for Q1 is low (about 0.1V) and Vbe is "high" (0.7V)
opposite is on Q2, Vce is high (couple of volts) and Vbe is low (close to 0V) since supplied from 0.1V on /q1 collector through R5.
I'll use your example cause you posted first. I understand what you mean now. I believe the problem I was having was that I envisioned 2 seperate voltage potentials within the same wire. (I calculated the series path as parallel) For example below I assumed the orange path would be at 6V and the blue at 0.1V. Im pretty sure thats where Ive gone wrong
1701229248970.png
 

Thread Starter

Tp86

Joined Sep 13, 2023
41
With the switch not closed, the circuit is a bi-stable.
Let us assume that Q1 is OFF and Q2 is ON.
Let us look at what is happening at Q2.
Q2 base is pulled high.
Q2 collector is close to 0V.
C2 is discharged through R6.

At Q1, its collector is close to 6V.
C1 is also discharged through R5.

The circuit is stable and does not oscillate.
With the switch not closed and after the capacitors discharge through the resisters you mentioned, they won't charge up again as there is no potential difference between each end of the individual capacitors. Is this correct?
 

WBahn

Joined Mar 31, 2012
29,857
I'll use your example cause you posted first. I understand what you mean now. I believe the problem I was having was that I envisioned 2 seperate voltage potentials within the same wire. (I calculated the series path as parallel) For example below I assumed the orange path would be at 6V and the blue at 0.1V. Im pretty sure thats where Ive gone wrong
1701232379141.png

What basis do you have for assuming 6 V at that node?

It means that the base-emitter voltage of Q2 is 6 V (assuming that your "ground" is the negative terminal of the battery, which you need to indicate the moment you start talking about specific voltage at a point instead of voltage differences between points). Does it make sense for the Vbe of an NPN BJT to be 6 V?

IF the voltage on that node was somehow 6 V, what direction would current be flowing in R3, R5, and the base of Q2? Does that make sense in like of Kirchhoff's Current Law (do you know about that law)?
 

k1ng 1337

Joined Sep 11, 2020
910
I'll use your example cause you posted first. I understand what you mean now. I believe the problem I was having was that I envisioned 2 seperate voltage potentials within the same wire. (I calculated the series path as parallel) For example below I assumed the orange path would be at 6V and the blue at 0.1V. Im pretty sure thats where Ive gone wrong
View attachment 308627
Try "Falstad" and "LTspice" circuit simulators. Falstad has a LOT of examples for beginners but overall it's not accurate. Play with it a bit then move on to LTspice.

To show you the power of spice, I simulated your circuit and plotted current through every component on the graphs. You can clearly see when current flows and in what direction as well as the voltage at the collector of each BJT. Try simulating the circuit without connecting R5 and R6 to the positive rail to see why it doesn't oscillate.

TP83 2.png


TP83 1.png
 
Last edited:

Thread Starter

Tp86

Joined Sep 13, 2023
41
What basis do you have for assuming 6 V at that node?

It means that the base-emitter voltage of Q2 is 6 V (assuming that your "ground" is the negative terminal of the battery, which you need to indicate the moment you start talking about specific voltage at a point instead of voltage differences between points). Does it make sense for the Vbe of an NPN BJT to be 6 V?

IF the voltage on that node was somehow 6 V, what direction would current be flowing in R3, R5, and the base of Q2? Does that make sense in like of Kirchhoff's Current Law (do you know about that law)?
I didnt explain this well. Imagine when the human operated switch is closed. The capacitor C1 will charge up to around 6V - 0.2V via the R3. But before it could theoretically get to 5.8V it would stop at 0.7V and the Q2 transistor would switch on thus causing the Q1 transistor to switch off. C1 will now build up a negative charge.

The path in orange was my mistake. I'm still getting used to electrical circuits. Completed defied the laws of electrical circuits
 

Thread Starter

Tp86

Joined Sep 13, 2023
41
Try "Falstad" and "LTspice" circuit simulators. Falstad has a LOT of examples for beginners but overall it's not accurate. Play with it a bit then move on to LTspice.

To show you the power of spice, I simulated your circuit and plotted current through every component on the graphs. You can clearly see when current flows and in what direction as well as the voltage at the collector of each BJT. Try simulating the circuit without connecting R5 and R6 to the positive rail to see why it doesn't oscillate.

View attachment 308633


View attachment 308635
Hi mate. Can you please send me the file for your simulation if you still have it?
 

k1ng 1337

Joined Sep 11, 2020
910

Thread Starter

Tp86

Joined Sep 13, 2023
41
It does have a steep learning curve but it is worth it. Also, get yourself one of these scopes. I have the same model and it works great for hobby projects. With the proper equations, a scope and a simulator, you'll have what you need to understand oscillators.

DSO Shell DSO150 Oscilloscope Full Assembled with P6020 BNC Standard Probe: Amazon.com: Industrial & Scientific
That scope looks cool and is great value for money. The only thing is the record length of 1024 is not great compared to 14 Mpts on the A$500 Siglent scopes for instance. I have an analog scope at home I could probably play around with on this circuit though. I need to adjust the voltage on CH1 as it is incorrect
-2730817676128995117.jpg

By the way how does Phet and Multisim compare to LTspice?
 

WBahn

Joined Mar 31, 2012
29,857
It does have a steep learning curve but it is worth it. Also, get yourself one of these scopes. I have the same model and it works great for hobby projects. With the proper equations, a scope and a simulator, you'll have what you need to understand oscillators.

DSO Shell DSO150 Oscilloscope Full Assembled with P6020 BNC Standard Probe: Amazon.com: Industrial & Scientific
The specs seem to have some errors. Perhaps you could look at yours and say what they really are.

It says that the sensitivity range only goes from 5 to 20 mV/div. That's got to be a mistake.

It doesn't say how it is powered and the pictures don't show any kind of power adaptor, but the specs say that it includes just the scope and none of the accessories shown in the pictures (which implies that the BNC probe is not included, either, even though it is explicitly stated in the product title). But it does look like the bottom of the meter might have a power jack there next to the switch (which is probably the power switch). They don't have a picture of the back, so I can't tell whether there is a battery door there, but if it's battery powered the batteries can't last long at the spec'ed draw of 120 mA.

@Tp86: Before getting this scope, think a bit about what your immediate needs are. It only has a 200 kHz bandwidth, which is going to make looking at digital signals much above 20 kHz to 50 kHz on the more difficult side. But if you are only doing audio-type stuff for the near future, it's likely adequate -- and at the price it's hard to go wrong, even if it turns out being inadequate for your needs.
 

Thread Starter

Tp86

Joined Sep 13, 2023
41
Yeah I didnt realise it was only 200khz bandwidth. See my post above I have a 60Mhz analog oscilloscope which would perform much better for the signal display however my issue with the Amazon DSO scope was the lack of record length points. I think I should just save up for one with more recording and higher bandwidth in the future. My dicksmith (famous Australian brand) which is actually a "Pintek" should work okay I will have a look tonight if I can calibrate the voltage. I believe its a matter of finding the correct potentiometer
 

k1ng 1337

Joined Sep 11, 2020
910
The specs seem to have some errors. Perhaps you could look at yours and say what they really are.

It says that the sensitivity range only goes from 5 to 20 mV/div. That's got to be a mistake.

It doesn't say how it is powered and the pictures don't show any kind of power adaptor, but the specs say that it includes just the scope and none of the accessories shown in the pictures (which implies that the BNC probe is not included, either, even though it is explicitly stated in the product title). But it does look like the bottom of the meter might have a power jack there next to the switch (which is probably the power switch). They don't have a picture of the back, so I can't tell whether there is a battery door there, but if it's battery powered the batteries can't last long at the spec'ed draw of 120 mA.

@Tp86: Before getting this scope, think a bit about what your immediate needs are. It only has a 200 kHz bandwidth, which is going to make looking at digital signals much above 20 kHz to 50 kHz on the more difficult side. But if you are only doing audio-type stuff for the near future, it's likely adequate -- and at the price it's hard to go wrong, even if it turns out being inadequate for your needs.
Its 20V not 20mV. Mine came up alligator style probes and a 9V barrel adapter. It was recommended to me by Yaakov a few years ago and turned out to pretty handy for the price at $50 CAD with shipping. My brand is "Starto600" which doesn't seem to be listed anymore.

Yeah I didnt realise it was only 200khz bandwidth. See my post above I have a 60Mhz analog oscilloscope which would perform much better for the signal display however my issue with the Amazon DSO scope was the lack of record length points. I think I should just save up for one with more recording and higher bandwidth in the future. My dicksmith (famous Australian brand) which is actually a "Pintek" should work okay I will have a look tonight if I can calibrate the voltage. I believe its a matter of finding the correct potentiometer
Looks like a cool unit. What kind of signals do you want to look at?
 

WBahn

Joined Mar 31, 2012
29,857
Most of those transistor multivibrator circuits do not have the R5 and R6 in your circuit:
They also don't disconnect R2 and R3 (in your schematic) from the positive rail.

My guess is that those are there to ensure that the circuit can't stall.

In the traditional circuit, there is a stable state where both LEDs are on, which happens if both capacitors are charged at the same time. The high-value capacitors across the caps ensures that if this does happen, it won't persist and the circuit can recover.
 

k1ng 1337

Joined Sep 11, 2020
910
They also don't disconnect R2 and R3 (in your schematic) from the positive rail.

My guess is that those are there to ensure that the circuit can't stall.

In the traditional circuit, there is a stable state where both LEDs are on, which happens if both capacitors are charged at the same time. The high-value capacitors across the caps ensures that if this does happen, it won't persist and the circuit can recover.
Indeed this is what happens in my circuit when R5 and R6 are disconnected. There is no feedback so each base rises to 700mV on startup and that's it.

Without R5 and R6, the circuit is simply two common collector BJTs which are effectively uncoupled. Turning on one BJT does not turn off the other therefore the circuit cannot oscillate.
 

MrAl

Joined Jun 17, 2014
11,268
Hi all. I am looking at a new circuit which uses a switch to activate a fast switching astable multivibrator oscillator and when the switch is removed it stops the oscillating processes and only 1 or the transistors will remain on. (Its like a decision maker machine)

I understand mostly how the astable multivibrator works but only it its form without the human-operated switch. With the switch labeled "GO" and extra 2 paths with resisters R5 and R6 a have a question

Why does the astable multivibrator not continue to oscillate when the GO switch is either open or completely removed from the circuit due to the paths which I have colored in red and blue that seem to be a complete circuit to allow current continue charging the capacitors and allow current through the Base Emitter?

View attachment 308625
Hi,

The extra resistors R3 and R4 might be there to prevent the caps from charging up too much. Note the 6v power supply (battery), but they may have been planning for an even higher voltage power source like 12v.

If the caps charge too high in voltage, when a transistor turns on it could bang the other transistor base emitter with a reverse voltage of more than -5 volts, which could damage the base emitter diode. Most of these transistors have a -5v limit on that.

You could do a simulation and see if the caps ever charge up to more than 5v with different power supply voltages.
For this exact design, it doesn't seem to be a problem because the LED's will drop some of that 6v as well as the base emitter diodes of the transistors, but they may have been planning for a higher voltage. A simulation would tell us for sure.

This is always a consideration with multivibrators and various other simple oscillators. This next circuit is interesting too as it only has a max voltage input of 5v which should always be ok (5v from the external jack is allowed although battery operation has it at a bit over 4.5v with fresh batteries).

This circuit is a multi-multivibrator, with three distinct states rather than just two like a regular multivibrator (It looks like three anyway).
Notice it is set up the same way just with an extra transistor, and as mentioned the power supply is limited to 5v or lower.

Another way of looking at this is that the transistors in either circuit have to be biased correctly in order to act like inverters that can constantly interact.
Another side issue is that if the circuit was perfect (all like components exactly the same and the same value and same exact specs) it would never oscillate. To start the oscillation there has to be an imbalance somewhere and that usually happens in real life because the parts are always slightly different in value or other specs.
 

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