# Resistance of electronic circuits

I am currently trying to build a Regulated DC Power Supply for an electronic piano. I am trying to find the minimum capacitance that I need to able to reach the desired ripple voltage. To find my minimum capacitance I am using the following formula VDC = (1 - (1 / 2 x f x RL x C) x VPeak where:

f = Frequency, RL = Resistance of the load, C = Capacitance of Capacitor, VPeak = The Peak rectified voltage.

Before I can calculate the value of the capacitor I need to figure out the resistance of the load. With the current piano that I have I cant measure the resistance directly but I do have the maximum voltage and current ratings of the piano (12VDC @ 500mA). My question is:

Can I use Ohms Law to find the internal resistance of the piano? (12V / 0.5A = 24 Ohms)

• yes, you can. (comments must be must be at least 15 characters in length.) Jul 16, 2016 at 9:47
• Sure thing! Otherwise measure it because the name plate and reality can differ a lot. Jul 16, 2016 at 9:48

Yes you can - although "resistance" isn't quite the right word for a complex load such as an electronic piano.

$$V_{ripple} = \frac {1 }{ 2 f R_L C} V_{peak} = \frac {1 }{ 2 f C}\frac {V_{peak}}{R_L} = \frac {1 }{ 2 f C}I_{peak}$$

since I = V/R.

Since you have a steady current of 0.5 A you can plug that into the formula. You'll get the same result but without the intermediate calculation of the "resistance" of the piano.

Note that you should use the calculated value as a minimum. Doubling it won't do any harm other than raising the average voltage a little and a slight increase in power dissipated in the regulator.

• Thanks for accepting the answer. If you wish to un-accept for a day or two you might attract some other answers which could give you some other insights. The "accept" is a bit of a disincentive otherwise. Jul 16, 2016 at 10:24
• Good answer. I just wanted to add that that equation of the ripple voltage is the one for a full wave rectifier (half wave rectifier you don't have the 2 in the denominator). And since it is an approximation, it works better for small values of the ripple voltage, that is, Vr << Vpeak for the approximation to be valid.
– Big6
Jul 16, 2016 at 15:28

If you know the maximum current required from the supply, then you can write:

$$C = \frac {I\ \Delta t}{\Delta V}$$

Where:

• C is the required capacitance, in farads
• I is the required current out of the supply, in amperes
• $\Delta t$ is the period of the capacitor's charging waveform, in seconds, and
• $\Delta V$ is the allowable ripple into the regulator, in volts.

For example, let's say that your regulator is a 7812 and its dropout voltage is 2.5 volts with 500mA through it.

That means that the input voltage to the regulator must never drop below 14.5 volts.

If you're using a transformer with a 12 volt RMS output, that's about 17 volts, peak, and if you drop a volt across a full-wave bridge, that'll leave you with 16 volts, which is what the capcitor will charge to on the rectified AC peaks.

That's also what'll appear on the input of the regulator, and if that input must never fall below 14.5 volts, then the ripple must never exceed 1.5 volts.

Now, if your bulk supply is full-wave rectified AC, and if we know the frequency of the mains, we can now solve for the value of the capacitor.

For 50Hz mains,

$$C = \frac {I\ \Delta t}{\Delta V} = \frac {0.5A \times 0.01s}{1.5V}= 3333\text { microfarads.}$$

Most aluminum electrolytics' capacitance can run about 20% low, worst case, so to make sure that doesn't happen you should up the calculated capacitance by about 20% to make up for that. That comes out to about 4000uF and a voltage rating of 25 volts would be OK.

If you use a 7812 as described above, it should dissipate less than 750 milliwatts, so you may get away without using a heat sink, but you might want to use one just to be safe.

Check the data sheet.