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# KCC's Quizzes AQQ246 on a one-shot timer circuit

Here is a basic monostable circuit built with 2 BJTs: Questions:

1.  What is the timer delay of this monostable?
2.  How us the voltage waveform at the base of Q2 with the following Vin? 3. What if Vin is made of 2 pulses where the second one occurs before the timer ends? Good Luck!

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[edited by: StephenV at 3:23 PM (GMT -4) on 18 Oct 2023]
• 1. The delay time basically is given by the time that C1 needs to be charged by R2 to a voltage sufficient for Q2 to become conductive. With a typical silicon transistor, this is 0.7V, and the additional base current can be ignored, as it is roughly 1/100 of the collector current. With t=C*U/I; I (mean value)=(VCC-0.7V/2)/120k we get: I=39µA and t=10µF*0.7V/39µA=0.18s

2. The voltage jumps up to roughly guessed 1-2V due to the high current that the C2 and D1 causes (in principle only limited by the source resistance and diode and base resistance), where it becomes clipped and then it becomes held at roughly 0.7V by R4 and R3

3. The second pulse has no effect, because Q1 already is in conductive state.

• 1) 0.9 ms

2) Output: raise with the raising front of the signal, ends (descending front) 0.9 ms later.
Base of Q2:  drop at -4 volt (+/-1) with the ascending front of the input, then, charges (as the capacitor) to +1 volt (0.7 +/-) in 0.9 ms.

3) Output raise with the first ascending front of the signal, no change if a second ascending front occurs before 0.9ms.

I have to say that I used some help, using this circuit from circuitlab dot com
(
I used another NPN though, a BC337). Use the simulate button, and ask a time domain simulation without changing the parameters. May take a minute to complete the graph.

• I have to correct -my calculation was to quick and wrong and I used 10µ instead of 10nF. I admit I had to see vanderghast's reply to remark my error ;-)

With t=C*U/I; I (mean value)=(3/2 VCC-0.7V)/120k we get: I=57µA and roughly t=10nF*5V/57µA=0.88ms

2. The question was for Q2, not Q1 :-) Q2 Base jumps down from 0.7V by 5V, to roughly -4.3V and then becomes charged by R2 with slightly decreasing current, until it reaches the conduction region of about 0.7V again and the transistors toggle.

• OK correction of correction - C1 IS 10µF. With this, the time is a factor of 1000 larger: 0.88s!

• Haha, excellent Bernhard, your final solution is the correct one! Congratulations! The delay is defined by the time C1 is charged through R2 starting from -VCC+0.7until 0.7. By making a translation in such a way the charge starts from 0, we have the target full charge = 2VCC-0.7 and the trigger point at VCC. By negleting 0.7 versus 2VCC, we have approx t = 0.83 second. Thus slightly less than your value.

• Thanks @vandergast! But I have a a delay close to 1 sec; therefore I wonder if you have not made a typo (nF versus uF)

• Indeed, as Bernhard mentioned it.

• And here are the answers (this one is not evident because one has to remember the capacitor charging equation):

1. The timing is set by Q2 off time. Q2 is off from the moment Vin pulse makes Q1 ON, which put C1 left node to GND. Since this latter was precharged to +VCC, Q2 base will see immediately -VCC turning it immediately off. C1 will then be discharged (or charged) to VCC from -VCC+0.7 through R2. It can do so until it reaches 0.7 volt; which will make Q2 turn on again. By translating the vertical axe by Vcc-0.7, we have then the adapted C1 charge equation:

The time is therefore the time for C1 to charge from (-VCC+0.7) to +VCC

V(t) = (2*VCC-0.7) (1-Exp(-T/R2C1) for V(T) = Vcc

By resolving it for T :

Vcc = (2Vcc – 0.7) (1-Exp(-T/R2C1)

Giving T = R2C1 Ln[(2VCC-0.7)/Vcc]

With VCC large, 0.7 can be neglected versus 2VCC:

T = R2C1 Ln(2) ~ 0.7 R2C1 = 0.83 seconds

2. This is also the waveform at the base of Q2: Making a translation of the vertical axis in order to have 0 volt at time 0:

We can rewrite the equation as:

VCC= (2VCC-0.7) (1-Exp(-T/R2C1))

This gives : T = R2C1 * Ln[(2VCC-0.7)/VCC] = 0.7 * R2C1 ~0.744 sec

3. If a second pulse comes before the monostable re-stabilize, it will be ignored. Because the second pulse will try to get Q1 conducting. But this latter is still ON; it will not be more saturated than it is already!

Annyhow, congratulation Bernhard who seems to be the only one having found the solutions: