Ok here's the problem. I have a power antenna in my car. The antenna goes up when a voltage is supplied to the voltage sensor in the antenna and goes down when the voltage is removed. I used this fact to install a switch allowing me to listen to CD's without deploying the antenna.
The problem is: Whenever I start the car when the radio is on, the voltage drop is enough so the sensor registers a loss of signal and begins to withdraw the antenna into the car only to redeploy three seconds later when the car is running and the voltage returns to normal. This is hell on the gears and the torque switch. My plan is to install a capacitor in the signal circuit so that the voltage sensor no longer registers a temporary loss of signal when the car is started but will still withdraw the antenna (after a bleed time ) when the car is off or the switch is turned off.
Two questions: Since I need about a 10 second bleed time how big a cap do I need? And I install the cap in parallel across the signal leads, correct?
\$Update\$
Got it done tested, soldered, boxed and installed it works great!!! Thanks to @Davetweed AND @RedGrittybrick for your help and insight.
Answer
In order to pick a capacitor for this application, you're going to need to know two things:
How much current does the antenna mechanism draw from the control signal when it is active? Call this I.
How low can the voltage drop on the control signal before the antenna starts to retract? Call this Vmin.
You have already stated that the hold time required is 10 seconds. Call this t.
We'll call the nominal voltage of the battery (when not cranking) Vsupply.
Therefore, the charge required on the capacitor is
$$\Delta Q = I\cdot t$$
The amount of voltage drop we can tolerate is
$$\Delta V = V_{supply} - V_{min}$$
Therefore, the minimum capacitance required is
$$C = \frac{\Delta Q}{\Delta V}$$
You should probably go with 2× this minimum value, in order to allow for variations in Vsupply and I under different operating conditions.
Finally, you're going to want to put a diode between the radio and the capacitor in order to keep the capacitor from discharging back through the radio.
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