# Electronic – Calculating the value of bypass capacitors for an amplifier

amplifierdecoupling-capacitor

I want to know if there is a rough any ready way of calculating the capacitance required to decouple the power supply used in an amplifier. (Rough and ready because I'll be adding 30% to the value anyway!)

Questions:
1) Does this depend on the type of amplifier used (in my case, common emitter stage going into an AB pushpull)?

2) Does this depend on the signal to be amplified, (in my case just a simple sinewave).

3) How is this linked to the power rating of the amplifier (in my case only a few watts.)

4) Does the voltage effect it? (apart from the voltage rating of the cap being high enough for the supply! (in my case 25v))

5) Does the frequency effect this, (in my case 50kHz to 5MHz)

In short, what steps/calculations should I make to calculate the values of the bypass caps needed to decouple the supply and prevent distortion due to supply ripple voltage?

I think that the general rule of thumb for decoupling caps is "the bigger, the better!".

The best way to figure out whether your decoupling is good enough is to build the circuit and measure the ripple. The second best way is to simulate the circuit in Spice and measure the ripple.

However, if you do want to have a rough estimation of the order of magnitude, you need to take into account the following parameters:

• Output impedance of your power supply - \$R\$
• Average switching current and its duration - \$I_A\$ and \$T\$
• The maximal allowed ripple in the supply's voltage - \$V_{r_{max}}\$

It is clear that the ripple of the supply's voltage is due to the voltage drop on its internal output impedance, and that this voltage drop is equal to voltage drop on the cap:

$$V_{r}(t)=I_{R}(t)R=\frac{1}{C}\Delta Q(t)$$

Note that \$I_{R}\$ in the above equation is not the total current drawn by the load, but the fraction of this current which is drawn from the power supply (the other part is drawn from the capacitor).

If you do the algebra and substitutions, you'll get to the following equation:

$$V_r(t)=\frac{1}{C}\int_{0}^{t}I_C(t')dt'$$

Where \$I_C(t')\$ is the current drawn from the capacitor.

In order to find the maximal voltage drop you need to find the maximum of the above function. This requires the differentiation with respect to \$t\$ and finding the value of \$t\$ for which the derivative is equal to zero. Due to the fact that the current drawn from the capacitor depends on the voltage on the capacitor and the current drawn by the load, the above differentiation is not simple and requires an exact characterization of the switching current profile.

However, you do not want an exact solutions, but just estimations, therefore we can make several assumptions which will simplify the problem:

• The current drawn from the power supply when the ripple is at maximum is \$\frac{V_{r_{max}}}{R}\$. We can assume linear ripple, which means that the average current drawn from the supply is \$I_{R_{A}}=\frac{V_{r_{max}}}{2R}\$
• The average current drawn from the capacitor is then \$I_{C_{A}}=I_A-I_{R_{A}}=I_A-\frac{V_{r_{max}}}{2R}\$.
• The total voltage drop on the capacitor due to the above average current which flows during time period of \$T\$ (switching time) is \$\Delta V_C=\frac{1}{C}I_{C_A}T=\frac{T}{C}*(I_A-\frac{V_{r_{max}}}{2R})\$

Accepting all the above assumptions and requiring \$\Delta V_C=V_{r_{max}}\$ leads to the following capacitance value:

$$C=T\left ( \frac{I_A}{v_{r_{max}}}-\frac{1}{2R} \right )$$

Disclaimer:

I've derived the above equation just now. It may be completely wrong. However, I see that the dependency of the required capacitance on the parameters of the problem is intuitively correct:

• The higher the switching current the bigger the capacitance you need
• The lower the desired ripple the bigger the capacitance you need
• The higher the internal output resistance of the supply the bigger the capacitance you need
• The switching time dependence is a bit tricky: it has no affect on the first term in parentheses due to the averaging of the current. Therefore, the shorter the switching time, the bigger capacitance you need.

It will be wise to test this model, and, as you said, anyway take the capacitor which is bigger than predicted by this equation.

I'll be glad to get a feedback on this model.