capacitorchargingcurrentremote control
I want to use a 555 in monostable mode to make a ramp signal generator. The capacitor would be charged using a current mirror circuit.
Given a input voltage, what would be the charging time of capacitor?
My understanding is: \$ \dfrac{V}{V-V_{EB}} \cdot R \cdot C \$
where \$ V_{EB} \$ is 0.545V and constant.
I am using a circuit from here:
Fuse blowing during capacitor charging can be prevented by using a "slow blow" fuse which tolerates moest overcurrent peaks of say 5x rated value for a second or so but which blows much more rapidly under heavier overloads. Getting the right balance between allowable overload and good protection can be an interesting exercise.
Slow Blow Fuses:
A slow blow fuse will take seconds to minutes to "blow" at 2 X its rated current but will typically blow in 10's of mS at 10x rated current and in under 1 mS at 100 x rated current. Exact current time profile will vary with brand and type. This time delay effect is usually but not always achieved by giving the fuse "thermal inertia" and by controlling the cooling available to the fusible element.
The graph below shows the current versus time profile for Littefuse HF461 fuses.
This is NOT the cheapest slow-blow fuse available. They can be purchased for rather less that the $1.50/10 or so per fuse that these cost, but these have a good data sheet to show you what you can expect. (Or hope to expect).
Below it can be seen that the 0.5A version will blow in about
20 seconds at 1A (2 X load)
0.6 s at 2A (4X overload)
50 ms at 5A (10X overload)
Under 10 mS at 10A
Under 1 mS at 30A
In the case in question a 2A fuse with a 6A load should blow in 1 to 2 seconds.
These results will depend to some extent on temperature, mounting, enclosure and which way the wind is blowing (in some cases literally so) but will serve as a guide which should be good enough..
Slow Blow fuses.jpg
Start your Digikey search here
Click on SlowBlow and then set filters
Here is a $US0.20/10 offering
bel 5ST / 5 STP series
Also provide a good data table:
Note the very long periods when Iin is not much > Irated.
With a 800 Volts DC supply, and a 200 kΩ resistor in series with the capacitor, the maximum current the charging will draw is 4 mA.
How to calculate this:
I = V / R, V = 800 Volts and R = 200 kΩAs this is lower than the 20 mA supply limit, the supply-limited maximum current will not come into play at all.
EDIT: Note that the time constant of the resistor and capacitor is 47 seconds, which means that it will take that long to reach 63% of its final voltage. It will take 3× that long (2:21) to reach 95% and 5× that long (3:55) to reach 99%. You can reduce those numbers by a factor of 5 by reducing the resistor to 40KΩ, which would just touch the source limit of 20mA at the beginning of the charge cycle. If you simply charge the capacitor with a constant current of 20mA, it will take 9.4 seconds.
Best Answer
From the equation here we have
\$ I_{ref} = I_c(1 + \frac{2}{\beta})\$,so
\$I_c = \frac{I_ref}{1 + \frac{2}{\beta}}\$
where \$I_{ref} \approx \frac{12 - 0.6}{100k} = 0.114 mA\$.
From \$I = C\frac{dv}{dt} \$ we get \$ V(t) = \frac{I}{C}t\$.
Assuming the transistors in the current mirror have a high \$\beta\$ then \$ I_c \approx I_{ref}\$. The 555 timer upper comparator will fire when \$V_c = \frac{2}{3}V_{cc} = 8v.\$ Plugging these numbers in to the equation above along with the capacitor value and solving for t we get:
\$ 8 = \frac{1.14*10^{-5}}{470*10^{-6}}t \implies t \approx 330\$ seconds.
Even assuming that the capacitor is "ideal", high transistor beta, and that the 555 timer inputs draw no current, the actual charging time will probably be greater than this, due to the Early effect of the right hand transistor in the current mirror.