22/04/202125/04/2021.

## Diagram of the quadrupler circuit

I don’t intend to explain the operation of this circuit. If you are interested, you can consult the website Electronique et informatique, or Wikipedia already quoted.

In this diagram, capacitors C3 and C4 participate both in the operation and in the anti-flickering. Caution: their operating voltage must be greater than 30 V (i.e. twice the input voltage assumed to be 15 V peak). Capacitors C1 and C2 are small size ceramics. ### For comparison: diagram of the “classic” circuit

The rectifier composed of diodes D1 to D4 can of course be an integrated device, which is suggested by the framing in mixed line.

The value of R1 is low, usually around 100 Ω. ## First tests

The first tests revealed a problem: on power-up, when the capacitors are discharged, the diodes are subjected to large current spikes, of the order of 3 A, which diminish over time. But these are far in excess of the maximum values permitted by the small diodes I use, and I fear that they will also interfere with the DCC command station itself.

Here is the current flow through a diode after adding a 10 Ω resistor in the supply circuit, to attenuate these spikes: Readings taken on a PicoScope 2204A oscilloscope.
Voltage-to-current conversion factor: 1 A / V; time base: 50 µs / div.

Measured values: IMAX = 0.47 A; T = 15 µs.

The current peaks still reach almost 0.5 A during 15 µs, knowing that the 1N4148 diodes have a non-repetitive surge current (IFSM) of just 0.5 A during 1 µs! This is clearly still too much.

With a 100 Ω resistor, things are better, for a voltage drop that remains negligible compared to the resistor to be put in series with the LEDs, which is 45 kΩ in the previous example. Voltage-to-current conversion factor: 1 A / V; time base: 50 µs / div.

Measured values: IMAX = 0.11 A; T = 40 µs.

### Evolution of the current in a diode

This is the same reading as above, but with a much longer time base: 100 ms instead of 50 µs. The voltage-current conversion factor is still 1 A / V; the duration of the reading is 0.5 s. We can see that the current peaks start at 110 mA and end at 10 mA. The average current is much smaller, of the order of 1 mA.

## New diagram

Finally, here is the diagram including resistor R1: ### Output characteristic

It’s interesting to know the output characteristic I = f (V) of the quadrupler considered as a DC power supply (in steady state).

For this purpose, I made a test set-up on a breadboard, and I measured these quantities for different loads, including the short circuit. In theory, the ICC values (short-circuit current) and V0 (open-circuit voltage) allow to draw the curve, if the characteristic is linear.

The small drawing shows the equivalent diagram of the short-circuited power supply, with V0 being its EMF (electromotive force), and r being its internal resistance. Values found: Voltage (V) Current (mA) 61.7 57.8 55.8 54.6 48.9 0 0 1.23 2.11 2.55 5.26 28.2

The internal resistance r is approximately 2.2 kΩ.

The above diagram has one small drawback: it requires two electrolytic capacitors. In general, in the coaches to be lit, there are two toilets, so there is enough space to accommodate them. But this is not always the case. In this case, the diagram below can be used.

Here there are four small ceramic capacitors plus a single electrolytic capacitor, but with a higher operating voltage (63 V instead of 35 V), and therefore a larger footprint for the same capacitance — or a smaller capacitance for the same footprint. However, as we have seen, this arrangement requires only small capacitance values. ## Association with a DCC decoder

Personally, I usually control the lighting in my coaches with simple latching reed switch, which give full satisfaction.

However, I wondered if it was possible to combine the quadrupler with a DCC decoder. At first sight, it’s not obvious: the voltages are too different. The solution exists, but it becomes a bit complicated: the decoder output that will control the lighting has to be isolated from the 60 V part by means of an optocoupler (or optoisolator), as follows: This must have an off-state output voltage (VCE0) greater than 60 V. Note that its location in the LED circuit is irrelevant.

Operation: when the decoder output is inactive, the optocoupler LED is off, the output transistor is blocked, the lighting is off. If the decoder output is activated, the transistor saturates (provided that the resistors are well calculated! ), it becomes conductive and the lighting turns on.

However, it will be necessary to provide an additional energy reserve circuit on the decoder side, which will add to the overall footprint. In the event of a power loss, the decoder outputs will be deactivated. The LED circuit will also be switched off, although it has a power reserve.

So I don’t think this is a very interesting solution. Personally, I only use a decoder for a complete train (TEE for example), and in this case, there is generally no difficulty in finding space for large enough capacitors.

## Conclusion

In the models to which I have applied the quadrupler diagram, the effect is spectacular: with one or two 100 µF capacitors, and sometimes 15 LEDs, I obtain holding times of about 5 seconds.

Moreover, I have a standard schematic in which I can only change the number of LEDs and the value of the series resistor to adapt it to any particular case.

I only regret not having thought of this before.

SMD Optocoupler
FODM124R2 On Semiconductor
Collector-Emitter Voltage 80 V
CTR 100-1200% @ 1 mA
at TME.eu

### Calculation of R1

The data are: the current in the LEDs, here called IC (collector current), = 0.5 mA; the current transfer ratio (CTR), = 100% to 1200%; the decoder output voltage, VCC = 15 V.

The transfer ratio is the ratio between the output current and the input current: CTR = IC/IF. Therefore, IF = IC/CTR.

The difficulty lies in the great imprecision of the CTR value. It must work in the worst case, i.e. when the CTR is the lowest — who can do more…

So, we take CTR = 100% = 1; therefore, IF = IC/1 = IC = 0.5 mA. We see that the decoder is not likely to be overloaded! In the input circuit, the optocoupler’s LED drops about 1.5 V; this leaves VR1 = 15 − 1.5 = 13.5 V across resistor R1. Applying Ohm’s law:

R1 = VR1 / IF = 13.5 / 0.5 10−3 = 27 103 = 27 kΩ.