# Simplest way to reduce voltage from ~48v to ~36v

(I'm very much a beginner with electronics, so I apologise in advance)

I’m trying to use a 48v battery on my 36v ebike controller. The controller itself can deal with the higher voltage, but it has a hardcoded limit of 44v, which it reads from a wire coming from the display.

So my aim is therefore to lower this voltage somehow. The parameters are:

• Input is 39-55v, output needs to be between 29-44v

• Current on the wire is 80-120ma

• It would be great it if this could be done proportionally, or with a constant voltage drop, instead of just regulating to a set output
voltage

• Single component solution would be best, since it would be easier to wateproof for riding in rain

Solutions I have researched so far:

Tvs diode:

• Guy here used one for this purpose (52v battery but otherwise same situation) and he said it worked

• But I’ve read that they’re not supposed to be used continuously like this, only for spikes in voltage - does that mean it could fail if
used like this?

• Also have no idea how to choose one for specific voltage drop - have read many explanations and data sheets but don’t understand
difference between clamping, breakdown and working voltage. The guy
in the linked post seemed to choose based on breakdown voltage =
desired voltage drop, but this doesn't seem to be what breakdown
voltage should mean?

Zener diode:

• To calculate the series resistor, I would need to know the resistance of the load, correct? I’m not sure how to find this - do I measure resistance across the voltage sense wire and ground while the display

Step down converter:

• I have one of these (appears to use a LT3800) This has a constant output and is large, so not ideal. But I would also worry that it would draw too much current? I've tried to measure the current it draws with no output, but I doubt that is particularly useful information. I tried to figure it out from the datasheet, but I don't really know what half the symbols mean

Resistor:

• Every mention of using a resistor for this purpose (that I've seen) has said not to do it, but not said why (although I assume this is because fluctuating current would change the resistance too much?)
• Also wasn't sure how to select the resistor value for a desired voltage drop. Is the V in V=IR meant to be the drop, or the voltage of the circuit?)

Voltage divider

• From what I’ve read, this doesn't work once you apply a load?

Other ideas:

My main reasons for asking instead of just trying these things are:

• Not wanting to exceed the mA level that I have witnessed on the line, because I don’t want to break anything
• Not knowing whether my lack of understanding might lead to something breaking
• Need something reliable, i.e. that won't fail halfway up a hill

Sorry if I've missed anything obvious - I feel like I could have researched some of these things more, but I did try. And in fairness they are bloody confusing if you have no real prior knowledge!

• I doubt a typ 500W e-bike uses only 80-120ma..More like 4kW full start and 11A draw*36V=4kW surge then reduce to 500W full speed Oct 7, 2019 at 14:34
• This is is just some small voltage sense wire, not the main power to the motor. The measurements given in the question and noted by @DwayneReid are correct
– Tpw
Oct 7, 2019 at 15:05
• I'm concerned about why that hardcoded 44V limit is there. Are you sure it can safely charge your battery without blowing it up? Oct 7, 2019 at 15:45

You are able to use a fixed voltage drop of about 11 Vdc at about 120 mA. This is fairly easy.

simulate this circuit – Schematic created using CircuitLab

The transistor is a Darlington device in a TO-220 package and has a reasonable gain of greater than 1000. The Vbe drop is about 1.2V. Choose the appropriate Zener diode for the desired voltage drop.

The total power dissipation is about (11V) * (0.12A) or about 1.3 Watts. You will need s small heatsink to keep the temperature rise reasonable. A small heatsink from an old computer motherboard or CPU is a good choice. Note that the transistor tab is connected directly to the incoming supply voltage. Don't let it touch Ground.

Note that there is NO current limit and no other protection. It's up to you to keep bad things from happening. Do NOT allow the output to short to Ground.

1) First thought was to simply use a 5W Zener diode. But those can be hard to come by these days. Next easiest is to use a smaller Zener diode coupled with a buffer transistor - the Zener handles only a few milliwatts and the bulk of the heat comes from the transistor - it's easy to get increased surface area (small heatsink) and thus keep the transistor cool.

2) I'm simply mentioning that there isn't any current limit in this solution. If there isn't any chance that you will "Oops" and touch the output to ground, then don't worry about it.

We can also substitute a LM317HV regulator for the transistor - this does have significant over-current and thermal protection. But the transistor is less expensive and may be easier to get. Mention in your comments if you want to explore using a LM317HV.

Do note that any of that transistor family will work: TIP120, TIP121, TIP122. The transistor is dropping only about 11 Vdc.

• Thanks for your answer. Did you include the warning about the lack of a current limit there, simply because of the potential for the output to touch ground? Or is there some way this circuit could force more current down the wire (I'm pretty sure that's not possible, but I'm more sure that I don't know much about electronics..) If it's the former, I could just put a fuse before this circuit - that would work fine right?
– Tpw
Oct 7, 2019 at 15:00
• Additionally, is there a reason to choose this more complex solution over Simon B's answer?
– Tpw
Oct 7, 2019 at 15:09
• @Tpw I can't speak for Dwayne's design choices but I imagine it's largely because this solution puts most of the power dissipation is in that transistor, which is better able to handle the power due to being designed to mount on a heatsink (zener diodes are rarely in packages appropriate for high power dissipation) Oct 7, 2019 at 15:44

Probably the simplest solution would be a zener diode in series with the load, chosen to drop just the amount of voltage you want dropped (not the voltage you want to get). No resistor is required. In your case, a zener of around 11V would do. The zener will need an appropriate wattage rating, as it will be dropping up to about 1.3W.

simulate this circuit – Schematic created using CircuitLab

• if you chose this solution, you must have large regions of copper wiring or copper foil, on the 2 leads of the Zener, so the heat is removed. Oct 7, 2019 at 13:03
• Is there a reason this doesn't need a series resistor like the example circuits using zener diodes to regulate power supplies do? Is it because the input is already regulated? Or because I just need a constant drop, not constant output?
– Tpw
Oct 7, 2019 at 15:07
• @analogsystemsrf - could I get away with cutting the leads short? Will the diode itself be able to deal with the heat?
– Tpw
Oct 7, 2019 at 15:08
• @Tpw cutting the leads short would make it worse, the leads act like heatsinks on their own since they're directly bonded to the silicon. You really ought to get one designed to be mounted to a heatsink if you can, maybe in a TO-220 package for example. Oct 7, 2019 at 15:42
• Is the power rating of the diode not staying how much power it can dissipate without a heat sink though? I can’t see any mentions of passive cooling being required in the data sheets. Would bonding to the frame using thermal paste work though?
– Tpw
Oct 7, 2019 at 19:20

We can design the heat-removal from the Zener. Shall we do so?

We'll assume the leads of the Zener are copper, and are 1mm square. Yes, they likely are round, but I'll let you insert a square-to-round correction factor. Copper, in the default thickness of PCB foil, which at 1 ounce/squareFoot is 1.4 mils or 35 microns, has thermal resistance of 70 degree Centigrade per watt per square of foil.

This assumes the heat enters one of the 4 edges, flows laterally thru the foil, and exits the opposite edge. Thus if we place 30 squares end-to-end, the Rthermal will be 30*70 = 2,100 degrees Centigrade per watt. We want to avoid that much temperature rise.

Lets design for 20 degree C rise. And because we don't know how the Zener silicon die is attached to the 2 leads, we'll design this heat removal to be used on EACH lead. If lucky, you'll end up with 20/2 = 10 degree C and the Zener should be very reliable.

How to do this design of heat removal?

We are going to think about a 1mm^3 of copper.

We've assumed the leads are 1,000 micron by 1,000 micron. Which is about 30 layers of PCB foil. With a square being 1mm by 1mm. The thermal resistance of each 1mm piece of the leads is 70/35 or 2 (TWO) degree Centigrade per watt. [yes, I rounded 30 up to 35. Its my math, and we should not carry along more precision, such as 2.2317 degreesC, than we deserve.]

The leads are our best way to remove heat, not air cooling, and not tiny pieces of foil soldered to short leads. Again, the leads are the best way to remove heat, but ultimately we have to dump heat to AIR or to the PCB (air) or to metal regions of the chassis to move heat to the outside of the case.

Remember the leads are 2 degree per watt, whereas the foil is 70 degreeC per watt.

simulate this circuit – Schematic created using CircuitLab

0 1 It could also be a simple voltage divider with 2 resistors and 1 or 2 power transistors in parallel 3n2055 with a few small resistors at the