Question

In the review of the theory, the following is a rule-of-thumb for selecting C and R values for Clamps: The RC time constant to fully charge the capacitor is 5RC (five times R times C). This value is equal to 1/4 of the period of the source (1/(4F)) (F=Frequency). To make this design work the fully charge event needs to be at least (minimum) 10 times this value. In short (equation): $5RC > 10 \cdot \frac{1}{4f} \implies 5 \cdot T_c > \frac{10}{4f}$ Where $T_c = RC$ Solving this for worst-case time constant: $T_c > \frac{1}{2f}$ (as least 10 to 60 times larger is preferred) Which leads to (assuming we know either R or C: $R > \frac{1}{2fc}$ or $C > \frac{1}{2fr}$ The best case is to set either R or C and determine the unknown component. Note: If C gets too large (much larger than 60 times), it will take too long to charge the capacitor and the clamp will not function. Note: if R gets too large (much larger than 60 times), the current will be essentially zero. Question: Assuming we set R to 10K (from 100K) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design) Assuming we set C to 1 uF (from 4.7 uF) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design)

          In the review of the theory, the following is a rule-of-thumb for selecting C and R values for Clamps:
The RC time constant to fully charge the capacitor is 5RC (five times R times C). This value is equal to 1/4 of the period of the source (1/(4F)) (F=Frequency).
To make this design work the fully charge event needs to be at least (minimum) 10 times this value. In short (equation):
$5RC > 10 \cdot \frac{1}{4f} \implies 5 \cdot T_c > \frac{10}{4f}$
Where $T_c = RC$
Solving this for worst-case time constant:
$T_c > \frac{1}{2f}$ (as least 10 to 60 times larger is preferred)
Which leads to (assuming we know either R or C:
$R > \frac{1}{2fc}$
or
$C > \frac{1}{2fr}$
The best case is to set either R or C and determine the unknown component.
Note: If C gets too large (much larger than 60 times), it will take too long to charge the capacitor and the clamp will not function.
Note: if R gets too large (much larger than 60 times), the current will be essentially zero.
Question:
Assuming we set R to 10K (from 100K) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design)
Assuming we set C to 1 uF (from 4.7 uF) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design)

In the review of the theory, the following is a rule-of-thumb for selecting C and R values for Clamps:
The RC time constant to fully charge the capacitor is 5RC (five times R times C). This value is equal to 1/4 of the period of the source (1/(4F)) (F=Frequency).
To make this design work the fully charge event needs to be at least (minimum) 10 times this value. In short (equation):
5RC > 10 ·(1)/(4f) 5 · Tc > (10)/(4f)
Where Tc = RC
Solving this for worst-case time constant:
Tc > (1)/(2f) (as least 10 to 60 times larger is preferred)
Which leads to (assuming we know either R or C:
R > (1)/(2fc)
or
C > (1)/(2fr)
The best case is to set either R or C and determine the unknown component.
Note: If C gets too large (much larger than 60 times), it will take too long to charge the capacitor and the clamp will not function.
Note: if R gets too large (much larger than 60 times), the current will be essentially zero.
Question:
Assuming we set R to 10K (from 100K) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design)
Assuming we set C to 1 uF (from 4.7 uF) in step 1, what value of capacitor is needed to produce the same output in our design? (assume we used a 56 times larger factor in our design)

Added by Mireia P.

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