In the Art of Electronics book, I don't understand the following phrase
.. because current is flowing down through R3, trying to pull it up.
Why does this current try to pull the cap up?
Collector of Q1 is saturated and 0V, and base of Q2 is -4.4V, so does this mean its flowing through C1?
In this case, why does it even have to change since no current would flow from -4.4V to 0V. ?
Also can electrolytic capacitors charge on either side?
Best Answer
Consider \$V_{in}\$ to be 0V and for that to have been the case for some considerable time. \$C_1 \$ will have 5V on its LHS (left hand side) because it has been charged by \$ R_2\$ and approximately 0.6V on its RHS (right hand side) because \$ Q_2 \$ is on and the voltage is limited by its base, \$R_3\$ is providing the base current.
We now take \$V_{in}\$ high to say 5V. This turns \$Q_1\$ on so the voltage on the LHS of \$C_1\$ falls to 0V, because the voltage across a capacitor can't change instantly the voltage on the RHS of \$C_1\$ falls to about -4.4V. Since the base of \$Q_2\$ is negative there is \$ 5-(-4.4) = 9.4\$ volts across \$R_3\$ so it must have \$940\mu\text{A}\$ flowing through it from top to bottom as drawn. \$Q_2\$ base-emitter junction is reverse biased (there is no base current) so the it must be flowing through \$C_1\$ into the collector of \$Q_1\$ and out of its emitter. This has the effect of discharging \$C_1\$ and starting to charge it in the other direction.
This turns \$Q_2\$ off and \$V_{out}\$ goes immediately to 5V.
\$R_3\$ now starts to charge the RHS of \$C_1\$ towards 5V, it will never get there because once it reaches about 0.6V \$Q_2\$ starts to turn on and \$V_{out}\$ begins to fall. The falling edge on \$V_{out}\$ is softer than the rising edge because transistors are current driven and because some of the current in \$R_3\$ is used to charge the capacitor as \$Q_2\$ starts to conduct.
We now take \$V_{in}\$ to 0V again and the LHS of \$C_1\$ charges via \$R_2\$ to 5V once it gets there we are ready to generate our next pulse.