Bsrhrki

Using a regulated current source to light them, wire the LEDs in series and short out the segment which you want to be dark.

^{simulate this circuit – Schematic created using CircuitLab}

You can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.

Here's a simple way to build one using a LM2596S module:

1. Remove the potentiometer and both large capacitors


2. Connect one of the salvaged capacitors between +in and +out (positive to +in), and fit a 1uF ceramic capacitor where the output capacitor was.


3. Connect a 100 ohm resistor from the -output to the centre potentiometer terminal.

Modified in this way, it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out.

^{simulate this circuit}

or a XL6009 buck-boost module can be modified. this time just remove the potentiometer and add a 100 ohm resistor, connec 3-30V to the nirmal input terminals and connect the LED string to the output and resistor.

^{simulate this circuit}

If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.

One way to achieve your progressive LED lightup as you turn the rotary switch is to use a current sink on the common of the switch and then wire the LEDs across the selector switch terminals as shown below. The constant current sink shown is a low cost way to get a 20mA sink for the LEDs so that there is no brightness variation as the number of lit LEDs changes. This scheme does require a high enough supply voltage that overcomes the forward voltage drop of the series string of up to nine LEDs.

Oldfart and Mattman944 give very similar answers involving complex diode networks. If brightness variation is acceptable a simple diode ladder is sufficient. Red LED's typically have 2V voltage drop and diodes typically have 0.6V voltage drop, so the combined effect of the diode voltage drops in a ladder is can be significant.

With a 9V battery and the switch in position 9, the current limiting resistor for LED 9 will see 9-2=7V and the current limiting resistor for LED 1 will see 9-2-(0.6*8)=2.2V, which will lead to a more than threefold difference in current through the LEDs if the current limiting resistors are of the same value.

If you insist on equal brightness it would be necessary to include all the diodes recommended by Oldfart and Mattman944, but with only a few extra diodes you can mitigate the variation in brightness to hopefully imperceptible levels. By adding three more diodes at the left as in the drawing above, we ensure that with the switch in position 9, LED 5 sees the same voltage as LED 8. The actual voltages across the current limiting resistors are as below. Note that an additional diode between LED's 5 and 2 (not considered in the table below) would improve the circuit further.

LED Voltage across current limiting resistor
9 7
8 7-0.6 =6.4
7 7-0.6*2 =5.8
6 7-0.6*3 =5.2
5 7-0.6 =6.4
4 7-0.6*2 =5.8
3 7-0.6*3 =5.2
2 7-0.6*4 =4.6
1 7-0.6*5 =4

Another way of balancing the brightnesses is to install diodes in the lines to some LED's to deliberately increase the voltage drop. In the drawing above an additional diode is inserted into the line from switch contact 1 to LED 1, so that LED 1 sees the same voltage regardless of whether the switch is in position 1 or 2. The current limiting resistor for LED 1 can then be a smaller value than the others in order to balance the brightness of this LED with the others.

These are just ideas - for this type of project the best balance of even brightness versus complexity may be best found by experimentation.

Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.

Sunyskyguy has a clever solution if you have a high voltage available.

You can use a buffer per LED like this.

In this diagram, R1 through R3 are pullup resistors. Closing any of the
switches will cause the buffer directly connected to it to go to 0,
which turns low all the buffers below it. 4050 has 6 buffers. You will
need 2 of them for 9 LEDs.

This solution does only needs a voltage to power the 4050 (3V to 20V for
CD4050B). You can chain up as many 4050s as you like.

If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

(Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)

^{simulate this circuit – Schematic created using CircuitLab}

It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.

I wouldn't suggest this unless you're eager to climb the learning curve for FPGAs (including buying a programming pod and dealing with a SMT part with lots of pins), but you could use a Lattice LCMXO2 series with internal flash and oscillator. Circuit would look like this (plus some power supply connections, a programming connector and bypass caps):

^{simulate this circuit – Schematic created using CircuitLab}

The programming software (Lattice Diamond) supports VHDL and Verilog.

If you're feeling lucky you could set the outputs to minimum current drive and omit the resistors.

An alternative approach is to use a LM3914 to drive the LEDs, with an external 10-resistor ladder powered from the reference voltage. Then the rotary switch simply selects a voltage from the ladder which will light the required number of LEDs.

This is just an outline; for example, the topmost resistor of the ladder would be selected to set the step voltages within the tolerance (which in my experience is pretty tight) of the LM3914 comparators.

In addition, the whole thing will run off 3.3V supply

Similar to the micro controller method, another way is to use a OP amp ICs. The positive inputs are all connected together and they connect to a potentiometer that produces varying voltage, instead of a switch. The negative connections connect to a series of resistors to give each one a different Voltage. As the knob is turned, the lights turn on one by one.

This type of circuit is used in power inverters that have those 10 segment LED strips to tell you how many Amps the inverter is putting out. I believe they have all the OP amps in one IC.

I know it's not an exact answer to the question since it doesn't use a switch, but it likely accomplishes what you want.

Edit 2: It's still possible to use a normal switch that connects only one contact at a time. Connect all the negative OP amp inputs to a low Voltage like 1V. Then connect each switch output to each op amp Positive input. Put a large resistor like 100k on the switch input and connect that to the positive power supply. It needs to be a big resistor to not let enough current through to make the LED above turn on noticeably, since the positive inputs will be connected to an LED anode from another OP amp. Now when you turn the switch, one LED will come on at a time. To make all the LEDs next to it come on too, just connect the output of each OP amp to the positive input of the one under it. The forward voltage drop of the LEDs will much too high compared to the 1V reference Voltage to take enough Voltage away from the positive input of the OP amp under it, so the LED won't prevent the OP amp from turning on, but other non LED loads might. This assumes that the OP amps are the current source only type. Current source and sink op Amps can't be used since it will prevent the positive input of the other op amp from going high. Many OP amps are current sink only, so in that case the LEDs would have to be arranged with the cathodes connecting to the OP amp inputs, and the rest of the circuit switched around. Don't forget to use pull up or pull down resistors for the OP amp inputs that are connected to the switch. The same resistor value that was used to connect the switch to the positive Voltage supply should be fine. I hope that's not too confusing.

Edit 3: It looks like someone else posted a similar but simpler solution using buffer ICs instead of OP amps.