# Switch voltage output with Zener diodes in series

I´m trying to produce a power supply that can give an output of both 24V and 12V. I want to switch between these two voltages with a signal of 5V, when the signal is low i want to get out 24V and then when it´s high i want an output of 12V.

I have been experimenting with two Zener diodes in series and try to bypass one of the diodes to get the lower voltage. I am not sure if this thery will work totally i practice. I got it to lower about 1V, maybe the transistor doesn´t saturate completely? Any ideas how to easy switch between the two voltages?

EDIT

Here is some extra information from the comments surrounding the project and an updated schematic:

So it´s suposed to be a power supply to a sensor that operates at 5V, the sensor sends out a 5V signal that should toggle a relay that operates at 24V. Since I want a low power consumption I want to lower the voltage to 12V when the relay doesn´t need to be on. Hope this clarifies a bit.

• What voltage is coming in to the bridge? That looks like some kind of capacitive dropper. Your 5V signal would have no ground in common with the transistor, so no go. The whole circuit is just icky. – JRE Nov 28 '17 at 12:50
• What's the input voltage to the rectifier? Seems like it's directly connected to the grid. Am I correct? What's the required output current? – Rohat Kılıç Nov 28 '17 at 12:51
• 230Vac input and output is about 40mA. The 5V signal is taken from the same circuit, I have a LDO that transform the 24/12 to the signal. So the ground would be the same, wouldn´t it? – TheOne Nov 28 '17 at 12:54
• You should state a crucial parameter: How much current will the output need to deliver. Also you do not mention your input voltage. Also note that the 5 V signal for switching has to be applied between X3-3 and the ground on the right (emitter of T4). That ground is not available on the X3 connector. – Bimpelrekkie Nov 28 '17 at 12:57
• You can make the whole circuit more energy efficient by removing the 12V/24V switching and dimensioning C1 (make C1 a smaller value) such that when the relay is off you get 24 V (or a bit more) across C2 and when the relay is on the voltage can drop to 12 V. That costs less components and actually does save power. I have seen this scheme being used in similar circuits. There is no need to switch between 12 V / 24 V if you just make the capacitive dropper such that the voltage drops when the relay is on. – Bimpelrekkie Nov 28 '17 at 13:33

Here's a circuit.

simulate this circuit – Schematic created using CircuitLab

When the signal is 5V the output will be 12VDC. Otherwise it will be 24VDC.

NOTES:

• Vin must be at least 30V and RZ should be calculated according to minimum input voltage, $V_{i-min}$: $RZ[k\Omega]=(V_{i-min}-24)/8$. Its power rating should be calculated from maximum input voltage, $V_{i-max}$.
• Q1 may need a heatsink. Because, if the input voltage is 30V and the output voltage is 12V, total power consumption of Q1 will be about 0.75W at a load current of 40mA. So a power transistor in TO-220 package (e.g. BD243) and a proper heatsink should be used.
• The voltage is not going to be 12V and 24V. Because of D1 and Q1, the voltage will be 1.2V above zener voltage. So it's going to be 13.2V and 25.2V – Chupacabras Aug 29 '18 at 21:03
• @Chupacabras nope. Please look again. Without D1 the output will be one Vbe below total zener voltage (11.4V or 23.4V). Adding D1 will will nullify the Vbe of Q1. – Rohat Kılıç Aug 30 '18 at 3:58
• I looked again. You are right :) I was wrong about the polarity of Vbe. – Chupacabras Aug 30 '18 at 4:55

I don't see anything wrong with your circuit conceptually.

Not sure why you picked 39K for the base resistor of T3- perhaps you have a big bag of them.. anyway, the peak current through the transistor with 240VAC/60Hz applied to the circuit will be a bit less than $C\cdot dv/dt$ or about 51mA if I did the derivative correctly (do the calculation yourself, using whatever numbers make sense in your application). So your base current should be in the 2-3mA range- implying more like 1.5K.

Or replace the BJT with a small logic-level MOSFET.

While you're at it, calculate the base resistor for the relay driver transistor and make sure you're giving it enough drive for whatever the coil current is.

• Yeah got lots of 39K that i need to get rid of, I think that this might solve the problem that i´m having! Thanks for the advice will try it later when I have the time! – TheOne Nov 29 '17 at 6:53

It does not make sens, as energy saving is concerned, because what you gain by limiting the voltage down to 12V will be lost through the zener. Normaly if IC1 is a 5V voltage regulator, it should deal with this issue intelligently and you wouldn't need to worry about it.