How the transistor works in this 555 circuit?

Question

555 Is not able to provide long timing due to its leakage current.

The following circuit is very useful to build a long time timer such as 1 hour or more.

The accuracy of the time is not important for me. I don't care if the 1 hour become 50 min or 70 min.

I know how monostable operation works but I don't understand how the transistor works and What the principle of operation is. How can it prevent the leakage current of 555 and provide long timer? and how is the capacitor charged and discharged?

Thank you,

Answer 1

When the button is pressed the capacitor is charged instantaneously to 8.74 V. It will then turn on the transistor and a current will flow through RT, the value of this current is (VC-0.6)/RT where VC is the voltage in the capacitor.

When the button is released the transistor will be kept ON by the capacitor and a small fraction of the previous current \$\approx \frac 1 \beta\$ is provided by the capacitor itself, slowly discharging it.

The final effect is that the capacitor is discharging through an equivalent \$ \beta \cdot RT\$ resistor instead of the \$RT\$ resistor you would have without the transistor. Because \$\beta\$ in transistors is around 50-150, you can have very long discharge times.

The leakage current of the pins in 555 (trigger and threshold) puts a limit on the maximum effective resistor you can use as RT (because the leakage can be thought as a parallel resistor internal to the 555). Your configurations allows to multiply this value by \$\beta\$.

Answer 2

The capacitor is charged when you press the switch.

The capacitor is discharged through the B-E junction of the transistor and RT. The current gain of the transistor means that the current flowing out of the capacitor is a tiny fraction (typically 1/100) of the current flowing through RT. It also means that any bias current required by the 555 inputs is also reduced by a similar factor.

In the "normal" monostable mode of the 555, the capacitor starts out at 0V and charges to 2/3 Vcc, taking 1.1×R×C seconds to do so. In this configuration, the capacitor starts out at Vcc and discharges to 1/3 Vcc plus the V_BE drop of the transistor (about 0.65 V), taking somewhat less time to accomplish this.

Note that the sense of the output is reversed: Pin 3 goes low when you push the button, and then goes high again when the time period expires.

This configuration of the transistor is called "emitter follower", and it is used specifically when you want to "buffer" a voltage. The input impedance of this configuration is much higher than the output impedance, and the voltage gain is very nearly unity (albeit with an offset, as noted above).