- #1
WifesBoyfriend
- 1
- 0
Hi,
These circuits are from Forrest Mims book "Getting started in electronics"
I think I understand the first circuit, but may as well check with you guys if my explanation is ok. The first circuit is the metronome circuit.
At first, C1 will charge to 0.7v and there will be a slight current just enough to open Q1. When Q1 opens, Q2 opens, the voltage on right side of C1 rises, this in turn causes more current to flow from C1 (but from the left plate) through base of Q1 further increasing the voltage rise on the right side. When C1 right plate voltage is at ~9V, less current flows in Q1 base causing it to slowly start closing, which causes Q2 to start closing. Base current for Q1 is now only provided by R1 and R2 and it's not enough to keep it open. Voltage on right side of C1 will drop as well but I'm assuming the transistors will close before the charge on the right plate is full discharged. So, finally, C1 is still negatively charged (between -9 and -8v) and it discharges from the right plate through the battery, through the resistors finally to the left plate. The voltage across the capacitor starts rising to 0v and then to 0.7v and the cycle repeats.
The second, siren circuit is similar, but I can't really figure out what C1 does so I will write my interpretation below.
Again, at beginning, caps are at 0 charge, Q1 and Q2 are closed. Hitting the switch causes C1 left plate to start rising to Vcc - Vr1. Current will flow through C1 (or rather left to right plate through R2) to the base of Q1. Im assuming R2 and R3 are to limit the base current of Q1 since their values are kinda high. C2 left will be at VbeQ1 (assuming 0.7v). Again, Q1 will open. As it does Q2 opens, and the oscillation cycle will go as mentioned above. C2 is very small so I'm assuming it will oscillate much faster then above.
Why does charging C1 cause the frequency to rise? I'm assuming because as C1 voltage rises there is less current going through it and more current going to base of Q1 through r2. More current going through base of Q1 causes Q2 to be more open which causes C2 to charge faster. Is that it?
Opening S1 causes C1 to discharge. At first IbQ1 will be large, opening Q2 more, increasing the speed of oscillation. As C1 discharges, IbQ1 will gradually fall off until there's no more flow since S1 is open. The frequency will fall since C2 will charge more slowly. When C1 is fully discharged the circuit will stop fully until S1 is pressed again
Sorry for the wall of text, thank you for any help you able to provide.
These circuits are from Forrest Mims book "Getting started in electronics"
I think I understand the first circuit, but may as well check with you guys if my explanation is ok. The first circuit is the metronome circuit.
At first, C1 will charge to 0.7v and there will be a slight current just enough to open Q1. When Q1 opens, Q2 opens, the voltage on right side of C1 rises, this in turn causes more current to flow from C1 (but from the left plate) through base of Q1 further increasing the voltage rise on the right side. When C1 right plate voltage is at ~9V, less current flows in Q1 base causing it to slowly start closing, which causes Q2 to start closing. Base current for Q1 is now only provided by R1 and R2 and it's not enough to keep it open. Voltage on right side of C1 will drop as well but I'm assuming the transistors will close before the charge on the right plate is full discharged. So, finally, C1 is still negatively charged (between -9 and -8v) and it discharges from the right plate through the battery, through the resistors finally to the left plate. The voltage across the capacitor starts rising to 0v and then to 0.7v and the cycle repeats.
The second, siren circuit is similar, but I can't really figure out what C1 does so I will write my interpretation below.
Again, at beginning, caps are at 0 charge, Q1 and Q2 are closed. Hitting the switch causes C1 left plate to start rising to Vcc - Vr1. Current will flow through C1 (or rather left to right plate through R2) to the base of Q1. Im assuming R2 and R3 are to limit the base current of Q1 since their values are kinda high. C2 left will be at VbeQ1 (assuming 0.7v). Again, Q1 will open. As it does Q2 opens, and the oscillation cycle will go as mentioned above. C2 is very small so I'm assuming it will oscillate much faster then above.
Why does charging C1 cause the frequency to rise? I'm assuming because as C1 voltage rises there is less current going through it and more current going to base of Q1 through r2. More current going through base of Q1 causes Q2 to be more open which causes C2 to charge faster. Is that it?
Opening S1 causes C1 to discharge. At first IbQ1 will be large, opening Q2 more, increasing the speed of oscillation. As C1 discharges, IbQ1 will gradually fall off until there's no more flow since S1 is open. The frequency will fall since C2 will charge more slowly. When C1 is fully discharged the circuit will stop fully until S1 is pressed again
Sorry for the wall of text, thank you for any help you able to provide.
Last edited by a moderator:
