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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:

enter image description here

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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.

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  • \$\begingroup\$ Hi Bitrex, Thanks for your answer, and it is perfect. In my furmula I missed the (2/3) factor that's why it was coming wrong. Now the consolidated formula is (2/3)*(V/(V-VBE))*R*C, assuming high value of beta Iref=Ic \$\endgroup\$ – Debojit Kundu Apr 30 '13 at 15:12
  • \$\begingroup\$ Should I use this circuit to create very accurate time delay ramp circuit like 10 ms. Actually I want to use this ramp generator for ac dimming purpose, and want to trigger on every zero crossing (half cycle). \$\endgroup\$ – Debojit Kundu Apr 30 '13 at 15:15
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In a current mirror the current in the collector that is connected to the capacitor is constant until the voltage on the cap nearly reaches the supply rails then something has to give-out BUT your circuit uses a 555 that discharges the cap before it gets to this point so you can reasonably say the collector current is constant.

That collector current is determined by the 100k resistor current - it's the same - the current mirror does what it says on the tin and mirrors (or replicates) the current through the 100k. Work-out what that will be on a 12 V PSU given also that there is about 0.6V lost in the base-emitter region.

That current divided by capacitance tells you the rate at which the voltage on the cap is rising. The 555-theshold value will allow you to calculate how long it takes to reach this voltage after it has been discharged.

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