Electrolytic capacitor behavior in monostable multivibrator

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ApexMark

Joined May 17, 2020
1
1589853764755.png

I've been building and studying this circuit for some days now and understand most things about it quite well. There are a couple of points which baffle me however, and that I can't seem to find answers for. I appreciate that I'm a newcomer to electronics and so try to research everything as much as possible before asking questions. I have the circuit on a 9V battery so I can put probe grounds where I need them. I hook my scope channel 1 (yellow) to monitor Q2B and channel 2 (pink) is hooked up across CT to monitor its discharge and charge cycles. (Channel 1's ground shares CT's cathode and I invert the channel). Here is what I see:
1589854076266.png

Most things here make perfect sense to me. Q2B is hovering just above 0.7V and keeping Q2 open. During the trigger, the sudden grounding of Q1C takes that nodes voltage to zero, causing a subsequent drop to around -9V on Q2B which closes Q2. As C begins discharging, Q2B's negative voltage begins to rise to match it. Once Q2B reaches 0V and goes positive, RT then supplies current into Q2's base just until it reaches a bit over 0.7V, after which Q2 goes into saturation and then its base voltage is constant.
What I'm studying is the area in the circle. Channel 1 & Channel 2 intersect at 0V. Q2B (Channel 1 Yellow) continues to rise positive. Does this mean there is now a brief positive charge building on the cathode of the electrolytic capacitor, which would explain why the discharge of the anode seems to keep going past 0 into negative, because it needs to keep matching the opposite of the other plate. I have read that aluminum electrolytic capacitors are able to sustain brief, minimal reverse voltages without permanent damage. Does this qualify, is that what we're looking at here - an acceptable and necessary price?
My other question is on initial supply of voltage to the circuit: everything I've read states that "on powering on, Q2 starts open." Well - it eventually wins the race to what stays open, but they both often open when power is first supplied. I don't see anything absolutely preventing Q1 from also being open at start, even though Q2 will shut it down. This is the behavior I see at my desk and in simulators. So then my question would be: in a real-world implementation, it seems like something would have to prevent Q1's initial opening, otherwise the 'Out' will fire momentarily (in this case, my LED).

Thank you for reading and for any insights.

Kind regards,
Mark
 

MisterBill2

Joined Jan 23, 2018
6,076
View attachment 207545

I've been building and studying this circuit for some days now and understand most things about it quite well. There are a couple of points which baffle me however, and that I can't seem to find answers for. I appreciate that I'm a newcomer to electronics and so try to research everything as much as possible before asking questions. I have the circuit on a 9V battery so I can put probe grounds where I need them. I hook my scope channel 1 (yellow) to monitor Q2B and channel 2 (pink) is hooked up across CT to monitor its discharge and charge cycles. (Channel 1's ground shares CT's cathode and I invert the channel). Here is what I see:
View attachment 207546

Most things here make perfect sense to me. Q2B is hovering just above 0.7V and keeping Q2 open. During the trigger, the sudden grounding of Q1C takes that nodes voltage to zero, causing a subsequent drop to around -9V on Q2B which closes Q2. As C begins discharging, Q2B's negative voltage begins to rise to match it. Once Q2B reaches 0V and goes positive, RT then supplies current into Q2's base just until it reaches a bit over 0.7V, after which Q2 goes into saturation and then its base voltage is constant.
What I'm studying is the area in the circle. Channel 1 & Channel 2 intersect at 0V. Q2B (Channel 1 Yellow) continues to rise positive. Does this mean there is now a brief positive charge building on the cathode of the electrolytic capacitor, which would explain why the discharge of the anode seems to keep going past 0 into negative, because it needs to keep matching the opposite of the other plate. I have read that aluminum electrolytic capacitors are able to sustain brief, minimal reverse voltages without permanent damage. Does this qualify, is that what we're looking at here - an acceptable and necessary price?
My other question is on initial supply of voltage to the circuit: everything I've read states that "on powering on, Q2 starts open." Well - it eventually wins the race to what stays open, but they both often open when power is first supplied. I don't see anything absolutely preventing Q1 from also being open at start, even though Q2 will shut it down. This is the behavior I see at my desk and in simulators. So then my question would be: in a real-world implementation, it seems like something would have to prevent Q1's initial opening, otherwise the 'Out' will fire momentarily (in this case, my LED).

Thank you for reading and for any insights.

Kind regards,
Mark
When transistor Q1 conducts the current through the capacitor certainly does reverse as it is discharged, and so the voltage will also reverse. That is why polarized capacitors are generally not the best choice in such circuits. And certainly there are ways to assure that a circuit starts in a preferred condition, but none of those methods have been added to the circuit you are looking at.
 
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