22/04/202110/11/2021.

I recently rediscovered voltage multipliers: doubler, tripler, quadrupler… These are circuits which, as their name indicates, are capable of supplying a higher voltage at the output than at the input, provided that the latter is supplied with alternating current.

You will tell me that the transformer is here for this. Of course, but this component is always rather cumbersome, too cumbersome in our applications, even at relatively high frequency.

But you will see that with very small components, diodes and capacitors, we can achieve the same result.

- Compared to a transformer, a multiplier has a limited current output. But as I intend to apply this circuit to coach lighting, a few milliamperes will suffice.
- How do I get the alternating current? Two possibilities:
- from a DC source, with an oscillator;
- but we have an alternating source: the DCC signal!

- What is the point of raising the voltage?

Here we are talking about powering LEDs for coach lighting. To sum up, it allows to decrease the current (which is already good in itself), but also to use smaller capacitors for the anti-flickering. So less problem to find a discrete place for them. Incidentally, it also simplifies the design of the circuit.

Let’s consider the case of a coach requiring 12 LEDs. The numerical values chosen will be as follows:

- Rectified DCC supply voltage: V
_{CC}= 15 V; - Current in an LED: I
_{F}= 0.5 mA. This may not seem like much, but the brightness obtained is perfectly sufficient for me; - Voltage drop in an LED: V
_{F}= 2.5 V; - Hold time in case of cut-off: 1 s.
- Acceptable brightness drop at the end of the hold time: that corresponding to half the normal current in the LEDs, i.e. 0.25 mA. Value determined by testing.

We will consider three cases:

Each LED has its own limiting resistor (it is not advisable to use a single resistor for a group of LEDs in parallel). This is the classic arrangement.

There are twelve branches; the current drawn by the whole is 12 times that of one LED, so 12 × 0.5 = 6 mA.

Each resistor value is: V_{CC} – V_{F}) / I_{F}
= (15 - 2.5) / 0.5 10^{–3} = 25 10^{3} = 25 kΩ.

This is the solution I used most often. The total current consumption is only 4 times that of an LED, i.e. 2 mA.

Each resistor value is (V_{CC} – 3 V_{F}) / I_{F}
= (15 – 7.5) / 0.5 10^{–3} = 15 10^{3} = 15 kΩ.

Attention: the current is lower, but the voltage limit below which the capacitor cannot discharge is higher: 7.5 V instead of 2.5 V. Won’t this compensate for that? This needs to be checked.

Furthermore, this solution is not always practical, because the number of LEDs is not necessarily a multiple of three. If, for example, a branch has only two LEDs, we must:

- either calculate a different resistor, in which case the brightness will not be the same if the supply voltage varies;
- or replace the missing LED with a component with approximately the same voltage drop. A Zener diode does the trick.

The problem of balancing the branches does not arise, since there is only one branch!

The current consumed by the assembly is that of one LED, here 0.5 mA. The (single) resistor is:

(4 V_{CC} – 12 V_{F}) / I_{F}

= (60 – 30) / 0.5 10^{–3}

= 60 10^{3} = 60 kΩ.

Before going on, I would like to answer an objection that is often made: in this circuit, if one LED is defective, the whole circuit will not work. This is true, but I answer this:

- an LED very rarely fails, except during assembly (overheating, bad soldering). If it works at the beginning, you are safe for a long time;
- Industrial LED lamps, for the home, for car headlights or for traffic lights, are equipped with LEDs in series (not just one series in general, I admit). However, the requirement for proper functioning is much higher than in our model trains.

Knowing this capacitance is important to see in which case the anti-flicker capacitors will be the smallest for the same efficiency.

Let’s consider the following circuit. It is assumed that the
capacitor is charged to V_{CC} voltage. At time t = 0, the
switch S is closed.

Theoretical calculation of the voltage V as a function of time is
of little interest here. Let’s just say that the discharge of a
capacitor into a resistor follows a so-called *exponential decay*
curve, which I’ll boldly replace with a straight line (in red) that
is nothing more than its tangent at the origin. This straight line
intersects the voltage limit line V_{LED} at a point which
corresponds, on the time axis, to the *time constant* which is
equal to the R × C product.

We can see that by replacing the real curve by the tangent, I am very pessimistic: in reality, the voltage drops much less quickly. But this greatly simplifies the calculation. When the voltage at the capacitor terminals reaches that of the LEDs, they have long since been switched off. However, a very small current goes on flowing through them, so that the capacitor continues to discharge very slowly, as shown by the line with a very small slope.

I made an Excel spreadsheet to calculate the capacitance to be expected, as well as other interesting quantities, such as P, power consumed by the whole circuit, according to the parameters indicated above. Here are the results.

Number | V_{CC} |
Resistor | I_{tot} |
P | C | |
---|---|---|---|---|---|---|

of LED | V | Qty | kΩ | mA | mW | µF |

12 × 1 | 15 | 12 | 25 | 6 | 90 | 960 |

4 × 3 | 15 | 4 | 15 | 2 | 30 | 533 |

1 × 12 | 60 | 1 | 60 | 0.5 | 30 | 33 |

We can see that, by putting LEDs in series:

- we decrease the total current and power consumption;
- the capacitance needed to attenuate voltage variations is reduced;
- this reduction is much greater for a voltage multiplied by 4.

The last solution is a winner on all counts: simplicity of the diagram, capacitor capacitance and power consumption! Of course, this must be tempered a little, as the quadrupler will require some additional components.