# Charge time is t1 = 0.693?(R1 + R2)?C1, but how is 0.693 obtained?

Latest Updated:01/01/2007

## Question:

The data sheets of the uPC1555 and uPD5555 describe astable multivibrator examples, stating that "when the output voltage is high, the charge time is t

_{1}= 0.693·(R_{1}+ R_{2})·C_{1}", but how is the 0.693 coefficient obtained?## Answer:

Take the case of the astable multivibrator of the uPC1555 as an example.

The expression is calculated under the assumed ideal conditions that the output pin signal is repeatedly inverted by repeatedly charging and discharging of capacitor C

When the trigger pin voltage drops to 1/3Vcc or lower, the internal latch inverts, turning the discharge pin off, and the charge voltage of capacitor C

Regarding the application circuit example shown in Fig. d of the data sheet, the charge and discharge times are obtained as follows.

<Charge Time>

The charge time being the time it takes for t

Designating the pin voltage of capacitor C

Vc=Vs{1-exp(-t/CR)} -----(1)

First, substituting t = t

1/3Vcc=Vcc{1-exp(-t

1-exp(-t

exp(-t

t

Therefore, t

Next, substituting t = t

2/3Vcc=Vcc{1-exp(-t

1-exp(-t

exp(-t

t

Therefore, t

Finally, since t

t

The coefficient can therefore be calculated as 0.693.

<Discharge time>

During discharge, the discharge current flows from C

t

Therefore, oscillation period T being the sum of charge time t

T= t

* Caution: For more information on this expression, see the "Time constant" FAQ in the Background Knowledge section.

The expression is calculated under the assumed ideal conditions that the output pin signal is repeatedly inverted by repeatedly charging and discharging of capacitor C

_{1}between a trigger pin input voltage of 1/3Vcc and a threshold pin input voltage of 2/3Vcc.When the trigger pin voltage drops to 1/3Vcc or lower, the internal latch inverts, turning the discharge pin off, and the charge voltage of capacitor C

_{1}rises. When this voltage reaches 2/3Vcc, the internal latch pin is inverted by 2/3Vcc applied to the threshold pin, turning the discharge pin on, and the charge voltage of capacitor C_{1}drops. When it reaches 1/3Vcc, the internal latch is inverted, turning the discharge pin off, and the charge voltage of capacitor C_{1}rises.Regarding the application circuit example shown in Fig. d of the data sheet, the charge and discharge times are obtained as follows.

<Charge Time>

The charge time being the time it takes for t

_{3}to change to t_{4}in Figure 3 below, t_{3}and t_{4}are obtained and then ( t_{4}- t_{3}) is calculated as the charge time.Designating the pin voltage of capacitor C

_{1}(in other words, the voltage of pin 6) as Vc and the charge supply voltage as Vs, based on the time constant formula*, the following expression is obtained:Vc=Vs{1-exp(-t/CR)} -----(1)

First, substituting t = t

_{3}, Vc = 1/3Vcc, Vs = Vcc, R = R_{1}+ R_{2}, and C = C_{1}in expression (1) above, time t_{3}, which is the time it takes for the difference in potential between the two pins of the capacitor to change from 0 V to 1/3Vcc, is calculated as follows:1/3Vcc=Vcc{1-exp(-t

_{3}/((R_{1}+R_{2})·C_{1}))}1-exp(-t

_{3}/((R_{1}+R_{2})·C_{1}))=1/3exp(-t

_{3}/((R_{1}+R_{2})·C_{1}))=2/3t

_{3}/(R_{1}+R_{2})·C_{1}=ln(3/2)=0.405Therefore, t

_{3}= 0.405·(R_{1}+ R_{2})·C_{1}.Next, substituting t = t

_{4}, Vc = 2/3Vcc, Vs = Vcc, R = R_{1}+ R_{2}, and C = C_{1}, in expression (1) above, time t_{4}, which is the time it takes for the difference in potential between the two pins of the capacitor to change from 0 V to 2/3Vcc, is calculated as follows:2/3Vcc=Vcc{1-exp(-t

_{4}/((R_{1}+R_{2})·C_{1}))}1-exp(-t

_{4}/((R_{1}+R_{2})·C_{1}))=2/3exp(-t

_{4}/((R_{1}+R_{2})·C_{1}))=1/3t

_{4}/(R_{1}+R_{2})·C_{1}=ln3=1.098Therefore, t

_{4}= 1.098·(R_{1}+ R_{2})·C_{1}.Finally, since t

_{1}, which is the time it takes for the charge of the capacitor pin voltage (= 1 - exp (-t1/(( R_{1}+ R_{2}) C_{1}))) Vcc^{*}) to change from 1/3Vcc to 2/3Vcc, is determined as t_{1}= t_{4}- t_{3}, the following expression is obtained:t

_{1}=1.098·(R_{1}+R_{2})·C_{1}-0.405·(R_{1}+R_{2})·C_{1}=0.693·(R_{1}+R_{2})·C_{1}-----(2)The coefficient can therefore be calculated as 0.693.

<Discharge time>

During discharge, the discharge current flows from C

_{1}through R_{2}, so that, obtaining discharge time t_{2}in a similar manner with R = R_{2}, the following expression results:t

_{2}=0.693· C_{1}· R_{2}-----(3)Therefore, oscillation period T being the sum of charge time t

_{1}and discharge time t_{2}, the following expression is obtained from expressions (2) and (3):T= t

_{1}+ t_{2}=0.693· C_{1}·(R_{1}+2 R_{2}) -----(4)* Caution: For more information on this expression, see the "Time constant" FAQ in the Background Knowledge section.

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