Please could someone tell me how to connect this TVS diode?

[DarrenB]

I am trying to copy someone else's idea and drop the voltage to my ebike display by using a TVS diode (P6KE24CA). The displays max voltage is 60V and I want to connect a 20S li-ion battery to the system which has a fully charged voltage of 84V. I know that I need to add the TVS diode in series with the displays power cable but I am unsure which way around to put the diode or even if I have selected the right diode. I think the P6KE24CA is a bi directional TVS, is that ok to drop 24V? The person who did this originally used the same part but with a 18V drop. The display only uses a few mA when powered on. I would appreciate it if someone could explain how this works and why a bi directional TVS diode is used over a uni directional (I don't know what either are)?

https://www.mouser.co.uk/datasheet/2/240/Littelfuse_TVS_Diode_P6KE_Datasheet_pdf-741008.pdf

 

[Diver300]

TVS diodes are designed to conduct only when the voltage across them exceeds a certain voltage. They are designed to absorb the energy of voltage spikes.

Unidirectional TVS diodes will conduct like a diode in one direction, so they will start to conduct at about 0.6 V, for any voltage rating. A unidirectional TVS can be useful to limit positive voltages to whatever is required, while keeping negative voltages very low.

One downside of unidirectional TVS diodes are that some circuits, such as ac circuits, have similar positive and negative voltages in normal operation so limiting the negative voltages to a low value would stop the circuit working.

Another downside of unidirectional TVS diodes is that they can be fitted the wrong way round. Some dc circuits, such as relay coils, will work either way round, and so they may be connected either way round. Fitting a unidirectional TVS to a dc relay would stop it working if the relay were connected one way round and the relay manufacturer won't know which polarity it will be used with.

In either ac applications, or in dc applications where the polarity is unknown, bidirectional TVS diodes are often used. The bidirectional TVS is basically two unidirectional TVS diodes in series in opposite directions. The bidirectional TVS diode will start conducting at much the same voltage in either direction.

For your application a bidirectional TVS diode would have the advantage of working either way round, so as long as you put it in series with the power supply to the display, it should be fine and you won't have to work out which way round to fit it.

The TVS diode will get hot when running. The one that you have quoted has Thermal Resistance Junction to Ambient of 75 °C/W, so that means it will be 75 °C hotter than the air around it when it is dissipating 1 W of electrical power. A rise of 75 °C is probably acceptable. You will get 1 W of power with 24 V across the TVS diode when there is 42 mA flowing.

I would regard that as the maximum. If the display takes more than that, you need a different design. If the display takes less than that, the TVS diode will run cooler. You need to consider the minimum and maximum voltage that the display needs, and the minimum and maximum battery voltages. The current that the display takes may vary as the voltage varies.

 

[DarrenB]

TVS diodes are designed to conduct only when the voltage across them exceeds a certain voltage. They are designed to absorb the energy of voltage spikes.

Unidirectional TVS diodes will conduct like a diode in one direction, so they will start to conduct at about 0.6 V, for any voltage rating. A unidirectional TVS can be useful to limit positive voltages to whatever is required, while keeping negative voltages very low.

One downside of unidirectional TVS diodes are that some circuits, such as ac circuits, have similar positive and negative voltages in normal operation so limiting the negative voltages to a low value would stop the circuit working.

Another downside of unidirectional TVS diodes is that they can be fitted the wrong way round. Some dc circuits, such as relay coils, will work either way round, and so they may be connected either way round. Fitting a unidirectional TVS to a dc relay would stop it working if the relay were connected one way round and the relay manufacturer won't know which polarity it will be used with.

In either ac applications, or in dc applications where the polarity is unknown, bidirectional TVS diodes are often used. The bidirectional TVS is basically two unidirectional TVS diodes in series in opposite directions. The bidirectional TVS diode will start conducting at much the same voltage in either direction.

For your application a bidirectional TVS diode would have the advantage of working either way round, so as long as you put it in series with the power supply to the display, it should be fine and you won't have to work out which way round to fit it.

The TVS diode will get hot when running. The one that you have quoted has Thermal Resistance Junction to Ambient of 75 °C/W, so that means it will be 75 °C hotter than the air around it when it is dissipating 1 W of electrical power. A rise of 75 °C is probably acceptable. You will get 1 W of power with 24 V across the TVS diode when there is 42 mA flowing.

I would regard that as the maximum. If the display takes more than that, you need a different design. If the display takes less than that, the TVS diode will run cooler. You need to consider the minimum and maximum voltage that the display needs, and the minimum and maximum battery voltages. The current that the display takes may vary as the voltage varies.

Thank you for the detailed yet understandable explanation! Here are the displays specifications https://manual.eggrider.com/display/specifications/#current-consumption , they say that the max is 20mA. I am assuming that that includes when the motor controller is being reprogrammed and bluetooth active.
This diode was the last missing piece to the puzzle as I am trying to convert the whole motor system over to 72V from 48V. I have 100V capacitors and 100V FETs for the power side and now I will get the TVS diode to install in series with the display. When everything is finished I am hoping to achieve 4.5kW peak output whilst keeping everything more or less stock. I recently found an absolutely beautiful adjustable Huawei R4875G1 rectifier that has been modified with a bluetooth CAN interface and tweaked guts that will allow me to charge the battery at up to 4kW. This thing is dense at 29x12x4cm! It will be paired with a type 2 EV connector for the ultimate ebike charging.
https://www.aliexpress.com/item/1005005471090863.html?spm=a2g0o.cart.0.0.43be38daH6ydd4&mp=1

Here is the bike. It's full carbon minus the wheel set and seat post.
View attachment Ultra X4 - 1.jpeg
Thanks once again Diver300!

 

[Diver300]

Anything giving electrical assistance over 15 mph is considered a motorbike in the UK, so would need VED, a license and insurance if used in public.

When I had a petrol motorbike, it was 7.3 kW, and it weighed about 100 kg

 

[DarrenB]

Anything giving electrical assistance over 15 mph is considered a motorbike in the UK, so would need VED, a license and insurance if used in public.

When I had a petrol motorbike, it was 7.3 kW, and it weighed about 100 kg

It's possible to limit the speed and power to appropriate legal levels if it were to be used on public streets.

 

[Nigel Goodwin]

It's possible to limit the speed and power to appropriate legal levels if it were to be used on public streets.

As far as I'm aware it's only legal to have electric assistance in the UK, so you have to peddle and the electric motor can provide varying degrees of assistance. A solely electrically powered bike isn't street legal.

 

[DarrenB]

As far as I'm aware it's only legal to have electric assistance in the UK, so you have to peddle and the electric motor can provide varying degrees of assistance. A solely electrically powered bike isn't street legal.

Yes you are right. If I were to do that I can switch off the throttle plus limit the power to 250W and 15mph.

 

[DarrenB]

TVS diodes are designed to conduct only when the voltage across them exceeds a certain voltage. They are designed to absorb the energy of voltage spikes.

Unidirectional TVS diodes will conduct like a diode in one direction, so they will start to conduct at about 0.6 V, for any voltage rating. A unidirectional TVS can be useful to limit positive voltages to whatever is required, while keeping negative voltages very low.

One downside of unidirectional TVS diodes are that some circuits, such as ac circuits, have similar positive and negative voltages in normal operation so limiting the negative voltages to a low value would stop the circuit working.

Another downside of unidirectional TVS diodes is that they can be fitted the wrong way round. Some dc circuits, such as relay coils, will work either way round, and so they may be connected either way round. Fitting a unidirectional TVS to a dc relay would stop it working if the relay were connected one way round and the relay manufacturer won't know which polarity it will be used with.

In either ac applications, or in dc applications where the polarity is unknown, bidirectional TVS diodes are often used. The bidirectional TVS is basically two unidirectional TVS diodes in series in opposite directions. The bidirectional TVS diode will start conducting at much the same voltage in either direction.

For your application a bidirectional TVS diode would have the advantage of working either way round, so as long as you put it in series with the power supply to the display, it should be fine and you won't have to work out which way round to fit it.

The TVS diode will get hot when running. The one that you have quoted has Thermal Resistance Junction to Ambient of 75 °C/W, so that means it will be 75 °C hotter than the air around it when it is dissipating 1 W of electrical power. A rise of 75 °C is probably acceptable. You will get 1 W of power with 24 V across the TVS diode when there is 42 mA flowing.

I would regard that as the maximum. If the display takes more than that, you need a different design. If the display takes less than that, the TVS diode will run cooler. You need to consider the minimum and maximum voltage that the display needs, and the minimum and maximum battery voltages. The current that the display takes may vary as the voltage varies.

I forgot to ask you if I can use the diodes in parallel and achieve the same result? For example if I use 2x P6KE24CA in parallel it will still drop 24V without any negative effects?

 

[Nigel Goodwin]

I forgot to ask you if I can use the diodes in parallel and achieve the same result? For example if I use 2x P6KE24CA in parallel it will still drop 24V without any negative effects?

Two in parallel should double the current capability, while leaving the voltage drop the same. However, as your current requirement is very low, why use two?.

My main worry is the low current, as it's a very crude way of dropping the voltage, and the low current means it might not be very accurate.

 

[DarrenB]

Two in parallel should double the current capability, while leaving the voltage drop the same. However, as your current requirement is very low, why use two?.

My main worry is the low current, as it's a very crude way of dropping the voltage, and the low current means it might not be very accurate.

I want the added mechanical strength and redundancy as it will live on a keychain. I plan to have swappable plug in keys so I can access the lower portion of the battery when the voltage decreases past a certain point. The controller has a hard coded cut off of 38V so to use all the battery I will need to swap out the 24V drop TVS at around 10% remaining capacity. I need to account for internal resistance as well. It will be like having a fuel reserve.

 

[Diver300]

It really doesn’t seem like a good idea to reduce the voltage in an inaccurate way and then have the controller act on the result.

You would be better to use two resistors as a potential divider with an emitter follower.

That would reduce the voltage as a proportion of the supply.

 

[DarrenB]

It really doesn’t seem like a good idea to reduce the voltage in an inaccurate way and then have the controller act on the result.

You would be better to use two resistors as a potential divider with an emitter follower.

That would reduce the voltage as a proportion of the supply.

I don't know how to do that? The voltage from the battery is 84V-55V and the voltage on the display is 60V-38V, what do I need to do to do it the way you described?

 

[Diver300]

An NPN transistor with the collector connected to the battery +ve.
Connect the controller +ve to the emitter. Connect the battery -ve and the controller -ve together.
Connect a 22 kOhm resistor between the base of the transistor and -ve
Connect a 10 kOhm resistor between battery +ve and the base of the transistor.

I suggest you draw out that circuit.

The two resistors will divide the voltage to give 22/32 or 68.75% of the battery voltage. The transistor will allow more current to flow to the controller than you could get from the resistors, while supplying about the same voltage as the resistors give, less about 0.6 V.

The transistor needs a voltage rating of 100 V or more and a power rating of 1 W or more. If you use a TO220 package transistor, the tab is connected to the collector so will be at battery voltage and needs to be insulated from -ve.

Check that you are getting around 2/3rds of the battery voltage before connecting the controller.

 

[DarrenB]

An NPN transistor with the collector connected to the battery +ve.
Connect the controller +ve to the emitter. Connect the battery -ve and the controller -ve together.
Connect a 22 kOhm resistor between the base of the transistor and -ve
Connect a 10 kOhm resistor between battery +ve and the base of the transistor.

I suggest you draw out that circuit.

The two resistors will divide the voltage to give 22/32 or 68.75% of the battery voltage. The transistor will allow more current to flow to the controller than you could get from the resistors, while supplying about the same voltage as the resistors give, less about 0.6 V.

The transistor needs a voltage rating of 100 V or more and a power rating of 1 W or more. If you use a TO220 package transistor, the tab is connected to the collector so will be at battery voltage and needs to be insulated from -ve.

Check that you are getting around 2/3rds of the battery voltage before connecting the controller.

Usually the battery is connected to the controller and the display is powered directly from this voltage. Everything was designed to run off of the battery voltage which was 60V max. The power FET side will now run 84V but I still need to keep the voltage for the display and logic side to 60V. Do you get what I mean?

 

[Diver300]

On my post about the emitter follower, you should substitute “display” for “controller”. I was talking about a way of reducing the voltage to a low power load, which is the display in your case, even if it has some control functions.

 

[DarrenB]

On my post about the emitter follower, you should substitute “display” for “controller”. I was talking about a way of reducing the voltage to a low power load, which is the display in your case, even if it has some control functions.

Ok I will give it a try. If it works it will be great.

 

[DarrenB]

On my post about the emitter follower, you should substitute “display” for “controller”. I was talking about a way of reducing the voltage to a low power load, which is the display in your case, even if it has some control functions.

Diver300 I tried what you said and it works, thanks. I bought a ST TIP3055 TO-247 because I want to connect a small relay which will add another 0.5W, will that be ok? Also I have had to change the plan slightly and I will now be using 19S rather than 20S. Is it possible to change the values of the resistors to get a different ratio? Ideally 25% because 60V is fully charged and the lowest cut off I can select is 38V.

 

[Diver300]

Diver300 I tried what you said and it works, thanks. I bought a ST TIP3055 TO-247 because I want to connect a small relay which will add another 0.5W, will that be ok? Also I have had to change the plan slightly and I will now be using 19S rather than 20S. Is it possible to change the values of the resistors to get a different ratio? Ideally 25% because 60V is fully charged and the lowest cut off I can select is 38V.

Can you draw out the circuit? It's not clear where the relay goes.

You can make the voltage divider with any ratio you want, but it will only give you one ratio, so you would need to match one voltage. Can you explain what the fully charged voltage and the cut off voltage will be for the battery pack, and what the maximum supply is to the display, and what is the range of cut-off voltages you can set?

 

[DarrenB]

Can you draw out the circuit? It's not clear where the relay goes.

You can make the voltage divider with any ratio you want, but it will only give you one ratio, so you would need to match one voltage. Can you explain what the fully charged voltage and the cut off voltage will be for the battery pack, and what the maximum supply is to the display, and what is the range of cut-off voltages you can set?

I need to match 79.8V to 60V, this will be more than good enough as I can still use 95% the battery. Would you be able to explain in really simple way how to achieve a 25% drop? Also is it ok to add the NPN in series with the brown cable where the TVS would have gone? So base to 22k to black wire, base to 10k to collector, collector and emitter in series with the brown wire with the collector being connected on the battery side and the emitter on the display side. Obviously ignoring the value of the resistors at the moment.

The relay is to switch a LED light from a 24V step down which is connected to battery +-. The relay will be powered by the logic side of the controller as it can be switched via the display, it a bit strange but I think the display is in series with the controllers logic side which means it all runs through the display? I think this is why the person who first did this mod added a TVS in series. I have taken a few pictures for you if it helps. Do you know what those SMD parts are without any markings?

 

[Diver300]

I'm still not clear where the relay is connected.

Here is the schematic for the voltage reducer. The output voltage is Vbatt * R2 / (R1 + R2) - 0.6 V.

The current taken by the resistors will be Vbatt / (R1 + R2). With the ratios that you are wanting, R2 will dissipate more power than R1. R2 will dissipate R2 * (Vbatt / (R1 + R2)^2

The output voltage will reduce slightly depending on the current taken by the display. If G is the gain of the transistor, and the display takes a current of I, the voltage will drop by:- I / G * R1 * R2 / ( R1 + R2 )

You can try various values for R1 and R2. To reduce the voltage by 25%, R2 will be three times the value of R1.

If R1 and R2 are too low, they will take too much current and get too hot. If R1 and R2 are too large, the voltage will reduce too much from the current taken by the display.