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.

enter image description here

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,

  • \$\begingroup\$ Please provide a link which shows why you think this circuit will work - I'm pretty sure it won't. \$\endgroup\$ Jun 19, 2016 at 21:24
  • \$\begingroup\$ @WhatRoughBeast No, it works properly and I'm very happy that's why I would like to know how it works. I've tested it four times. \$\endgroup\$ Jun 19, 2016 at 21:52
  • \$\begingroup\$ @WhatRoughBeast The resistor value is 1Mohm. When the capaictor is 470uf, The time was 20 min. When it is 940uf, The time was 40 min. When it is 1410 uf , The time was about 1 hour. When it is 1880 uf The time was 1 hour and 20 minutes. I connect capacitors of 470 uf in series to increase the capacitance. \$\endgroup\$ Jun 19, 2016 at 21:55
  • \$\begingroup\$ @WhatRoughBeast Here is the link of the circuit: homediyelectronics.com/basic/longtimedelays \$\endgroup\$ Jun 19, 2016 at 21:56
  • 1
    \$\begingroup\$ @MichaelGeorge. Note that in spite of the benefits gained by the transistor, at some large value (or poor quality), the leakage of the capacitor will exceed the needs of the 555, and time delays will begin to shrink. If the capacitor is brand new, seek a low leakage type, and prime it (its electrolyte) by using it for a days worth of cycles before considering it stable. \$\endgroup\$
    – user105652
    Jun 19, 2016 at 22:21

2 Answers 2


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\$.


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 VBE 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).


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