Guest
Hello,
I have questions about this version of an Astable Multivibrator:
+v
|______________
| |
R1 1M e
| ________b Q2 PNP
| | c
| | |
| | |
| c |
|___b Q1 NPN |
| e |
| | |
| GND |
| |
C1 10uF |
| |
|______________|___
|
R2 100 Ohm
|
GND
*Spice Netlist
Q2 3 4 6 Q2N3905
Q1 4 5 0 Q2N3904
C1 3 5 10uF
V1 6 0 DC 4.5
R2 3 0 100
R1 5 6 1000000
Looking at Multisim Oscilloscope simulations, I can see how when the
circuit is first activated, the capacitor slowly charges up to .6 volt
and turns the NPN transistor on. This in turn turns on the PNP
transistor, charges the cap to -V and causes a voltage spike across
R2. The cap then slowly discharges and the cycle repeats.
What I _dont_ understand is how if you increase the resistance of R2
(to somewhere around 700 - 1000 ohms), a threshold is reached where
the circuit breaks down. Shouldnt a discharged cap been "seen" as a
short and, once Q2 turns on and clears a path to the voltage source,
be immediately charged no matter what the value of R2 is? Obviously
not, but why? If this could be explained in the most noobish (ie
minimal math
) terms possible I would greatly appreciate it.
I have questions about this version of an Astable Multivibrator:
+v
|______________
| |
R1 1M e
| ________b Q2 PNP
| | c
| | |
| | |
| c |
|___b Q1 NPN |
| e |
| | |
| GND |
| |
C1 10uF |
| |
|______________|___
|
R2 100 Ohm
|
GND
*Spice Netlist
Q2 3 4 6 Q2N3905
Q1 4 5 0 Q2N3904
C1 3 5 10uF
V1 6 0 DC 4.5
R2 3 0 100
R1 5 6 1000000
Looking at Multisim Oscilloscope simulations, I can see how when the
circuit is first activated, the capacitor slowly charges up to .6 volt
and turns the NPN transistor on. This in turn turns on the PNP
transistor, charges the cap to -V and causes a voltage spike across
R2. The cap then slowly discharges and the cycle repeats.
What I _dont_ understand is how if you increase the resistance of R2
(to somewhere around 700 - 1000 ohms), a threshold is reached where
the circuit breaks down. Shouldnt a discharged cap been "seen" as a
short and, once Q2 turns on and clears a path to the voltage source,
be immediately charged no matter what the value of R2 is? Obviously
not, but why? If this could be explained in the most noobish (ie
minimal math