Previous blogs in this series:
In the last blog, I wrote about how to use the 555 as an one bit latch - the simplest of use cases. Incidentally the resulting circuit is also called bi-stable mode of 555. In this blog, we will build further on it. For ease of reference here is the circuit we built last time.
In the first blog, we learnt that Pin 3 will go high if SET (Pin 2) is pulled low (SET button is pressed) and CLR (Pin 6) is low (CLR button is not pushed). But, what about the t=0 condition in the circuit above? t=0 condition is the initial state of Pin 3, the moment power is turned on.
In the above circuit, assuming both the buttons are open (not pushed) - Pin 6 will be pulled down via R2 and Pin 2 will be pulled up via R1. Now let's refer to the diagram below which shows how the inputs at Pin 6 and 2 influence the output.
All good so far! So, can we be assured that the output will be always low on power-up. :) Unfortunately, not. Relying on the above logic to get a deterministic low output on startup is not a great idea. The most common cause is stray capacitance in the circuitry which feeds to the trigger pin or the stray capacitance of any switches connected to pin 2.
Secondly, what about the case where we want the output to be high on power-up?
The above use cases are called power-up set/reset use cases and we need extra circuitry to cater for them in a deterministic way.
Low output state at startup
There are multiple strategies to force a low output state at startup. The two most common ones are to
Shown below is a simple strategy to pull Pin 6 to high at power up. At the start C2 is effectively shorted, forcing Pin 6 to be high. As time passes, C2 charges up and in the process Pin 6 is pulled down to ground in due course. As Pin 6 starts up high, the output is effectively 0 on power up.
High output state at startup
There are cases where you might want 555 to start up on a high output state at power up. For example, I have a garden irrigation system which operates via a monostable pulse of 555. In this case, I want the 555 to operate my solenoid valve as soon as I power up the circuit and subsequently on a trigger. To achieve this, we simply build on the strategies discussed above - in this case, we keep Pin 2 and 6 pulled down at startup.
I have intentionally left other pins in the diagrams as the strategies discussed above are independent of the configuration of other pins.
Stay tuned for the next instalment in this series, where I will start discussing about the monostable and astable operation modes of the 555 chip.
In the last blog, I wrote about how to use the 555 as an one bit latch - the simplest of use cases. Incidentally the resulting circuit is also called bi-stable mode of 555. In this blog, we will build further on it. For ease of reference here is the circuit we built last time.
In the first blog, we learnt that Pin 3 will go high if SET (Pin 2) is pulled low (SET button is pressed) and CLR (Pin 6) is low (CLR button is not pushed). But, what about the t=0 condition in the circuit above? t=0 condition is the initial state of Pin 3, the moment power is turned on.
In the above circuit, assuming both the buttons are open (not pushed) - Pin 6 will be pulled down via R2 and Pin 2 will be pulled up via R1. Now let's refer to the diagram below which shows how the inputs at Pin 6 and 2 influence the output.
With Pin 6 at 0 - CMP1 will output 0 (low), while with Pin 2 at 1 (high) - CMP2 will output 0. So that initial input into the SR flip-flop is 0-0 for S-R respectively. This implies that the output will remain unchanged. Since right before power up, the output was low it will start with a low state when the power is turned on.
Secondly, what about the case where we want the output to be high on power-up?
The above use cases are called power-up set/reset use cases and we need extra circuitry to cater for them in a deterministic way.
Low output state at startup
There are multiple strategies to force a low output state at startup. The two most common ones are to
- Solution A ) Force the reset pin (Pin 4) to be low at startup, transitioning to high in some finite but small period of time.
- Solution B ) Force the threshold pin (Pin 6) to be high at startup, transitioning to low in some finite but small period of time.
Below are two common strategies to force Pin 4 to a grounded state at power up. In the first strategy, the collector of T1 is grounded as C1 is charging up through R2. In the second strategy, (which works just as good), C2 is effectively shorted at startup and slowly builds up voltage as it charges through R3 - effectively pulling Pin 4 to ground.
Shown below is a simple strategy to pull Pin 6 to high at power up. At the start C2 is effectively shorted, forcing Pin 6 to be high. As time passes, C2 charges up and in the process Pin 6 is pulled down to ground in due course. As Pin 6 starts up high, the output is effectively 0 on power up.
High output state at startup
I have intentionally left other pins in the diagrams as the strategies discussed above are independent of the configuration of other pins.
Stay tuned for the next instalment in this series, where I will start discussing about the monostable and astable operation modes of the 555 chip.
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