[Task] In order to obtain a DC voltage with low ripple, the common practice is to use RC filter, LC filter and corresponding Π filter, but the use of capacitor or inductor filter has the following disadvantages: large volume, large leakage current, and large device discreteness , the filtering effect will not increase significantly with the increase of capacitance or inductance, and there will be a filtering bottleneck or the introduction of additional distribution parameters. Without increasing the capacitance, try to design an electronic filter based on a triode to filter the DC voltage (take +5V as an example).

[Concept] The starting point of the concept is the following capacitor filter circuit:

Now, our goal is to increase the filtering effect of capacitor C1 to βC1 times through a triode (β is the magnification of the triode). Obviously, only by connecting the C and E poles of the triode to the main circuit of the circuit can a large current flow to the load RL (take the NPN tube as an example, so the C pole must be connected to the +5V power supply terminal):

The above triode cannot be turned on, and the base must be positively biased to be turned on. The degree of conduction is related to the base current. The greater the base current, the deeper the degree of conduction until it is saturated. So, we thought of connecting a resistor between the +5V power supply and the base:

However, a new problem arises: the above circuit does not amplify the filtering effect by β times, but only cuts off the load current. If C1 is not thorough enough to filter +5V (exceeding the minimum ripple required by the design), then the remaining ripple will enter the base of the triode through R1, be amplified by β times and then sent to the load, which will become a ripple amplifier instead. meet the requirements of the design task. Therefore, we must add an additional capacitor C2 to "drain" the ripple at the base of Q1 to the ground (using capacitance characteristics: pass AC, block DC; pass high frequency, block low frequency):

The above circuit sends most of the DC component in +5V to the load RL through CE, and only a small DC component is sent to the base of Q1, and correspondingly "suppresses" the AC component, and the DC is magnified by β times, which is equivalent to Conversely, to suppress the AC component to 1/β, this is equivalent to connecting a βC1 capacitor (β is mostly between 50 and 2,3 hundred). Although it is equivalent to "amplifying" the capacitance, this is completely different from the effect of purely large capacitance filtering: large capacitance means large leakage current, large power-on shock, and electronic filters based on triodes do not exist These problems are due to the fact that the triode does not have such a large leakage current as the capacitor, and due to the existence of the junction capacitance, it has a certain buffering effect on the power-on surge current.

We can also optimize the above circuit by adding a small-capacity ceramic capacitor (usually not greater than 100nF) in front of the load to further filter out high-frequency signals:

[Calculation] For the convenience of calculation, assume that β=100, RL "eats" current 200mA, in order to make the voltage drop between CE of Q1 as small as possible, we can make the base current IB 1.5~2 times of the critical amplification current (for calculation convenience , I take 2 times):

IB=2IC/β≈2IE/β=2IL/β=2*200/100=4mA,

R1=(5-0.7)/IB=4.3/4≈1k (0.7 is the voltage drop of the triode BE).

Because the load current is 200mA, according to the filtering experience formula: 2000uF/A or 2uF/mA, we can get C1=2*200=400uF (take the 470uF electrolytic capacitor of the standard series).

Because the frequency of most DC ripples is about f=300kHz, in order to limit the capacitive reactance of C2 within 1Ω (close to the capacitive reactance grounding and enhance the filtering effect), that is, 1/(2πfC2)≤1, C2≥1/( 2*π*300k)≈53uF (take the 68uF of the standard series).

Check the triode selection manual, and see that the maximum collector current of the S9013 triode is Icm=400mA, which has double the margin compared with the load current (200mA) in this case, and the magnification is between 64 and 202 (close to 100 tube), and finally select the S9013 tube.

The circuit to determine the parameters is as follows:

[Description] The above circuit may cause a voltage loss to the +5V supplied to the load, so the front end of the load can increase the voltage regulation link, such as simple Zener diode voltage regulation, three-terminal voltage regulator voltage regulation, LDO voltage regulation, etc.