Hi,

I am scratch building a model of the Dryden Shuttle MDD which includes a lot of LED's as I want to depict it fully lit at night.

I have two towers with 54 LED's, and a horizontal gantry with some others which I have yet to work out. I will be wiring the LED's for a tower as 4 parallel strings of 12 with a 220 ohm resistor for aach string powered by a 45v DC source, the extra 6 LED's, I have yet to think about (they will be red rather than white).

There are other LED's scattered else where such as on telegraph poles etc. Each tower or other set of LED's will have it's own power source which gives me 6 or possibly 8 separate circuits, some may have different source volts but most will be 45v DC.

To turn this lot on and off I would like to use DC SSR's, 3v switch with a max load of 60v, powered by a 2032 coin cell. Is it sensible to wire the SSR's in series ie in- to in+ of the next SSR while connecting the load to different circuits or is there a better way of going about this?

Thanks and regards,

Rich

 

I think ...
The +/- inputs to a Crydom DC SSR is voltage controlled.
Given the above, the SSR inputs should be connected in parallel, not series.

Consider a Constant Current Source for the LED's, which eliminates the power wasting resistor.

 

Start by making a diagram (schematic) of your proposed setup.

Post it here and we'll help you improve and deploy it.

 

Potentially it depends on the specific SSR model choice, but I think wiring the SSR control signals in parallel is most the most likely solution, especially with such a low control voltage.

Speaking of the low control voltage, take a look at the discharge curve of a 2032 battery here:
http://data.energizer.com/pdfs/cr2032.pdf

Notice that the graph shows current draw at only 0.19mA, far less than multiple SSR control inputs will probably draw. Even at the low current in the graph, the voltage is below 3V, and gradually drops over time. If your SSR inputs actually need 3V, they may not work with a coin cell control voltage. I'd suggest taking a close look at your SSR datasheet to see what the minimum voltage that's guaranteed to turn it on is, and also how much current it will draw.

I suspect you'll want a better power source for control voltage, but I could be wrong. Only the datasheets and a little math can tell you for sure.

 

Panasonic makes a Low Power SSR with 0.2ma input current (min) at 1.5 volts (min)
A lithium 2032 Coin Cell can provide 210mah
1,050 hours of on-tIme = 210 / 0.2

Also, a MOSFET as the "power switch", might use even less power from the Coin Cell.

 

Start by making a diagram (schematic) of your proposed setup.

Post it here and we'll help you improve and deploy it.

I'll get onto that as soon as I can. I have to have Chemo every two weeks so please bear with me on this.

Rich

 

Hi,

Thanks for being patient, please find attached my (admittedly amateurish) schematic of what i hope to achieve. I've also been looking at the prices of some of the suggested SSR's online, and unless I'm looking in the wrong place the cost (> 25 GBP each??) is too high for me to justify using them in this project so I have found Takamisawa's JV3S-KT which seems to fit the bill.

The PDF contains the following image:

In reply to ebeowulf17, for the JV3S-KT, the coil nominal power is .2w and the operate power is .113w.

Thanks very much for all your help.

Regards

Rich

 

Looks ok, but your 2032 battery switch is wired wrong, it should be in series with one of the terminals, your circuit has it connected to a relay coil.

 

Looks ok, but your 2032 battery switch is wired wrong, it should be in series with one of the terminals, your circuit has it connected to a relay coil.

Hi Dave,

I'm not sure if I understand what you're saying, or if I have what the diagram intends to show wrong.

The relays are in parallel on the coil? circuit and the battery switch connects directly to the positive terminal of the 2032.

I have a feeling that I've drawn it wrong though. Any chance of a quick diagram to show me how to depict it properly?

Thanks and regards

Rich

 

Hi,

Thanks for being patient, please find attached my (admittedly amateurish) schematic of what i hope to achieve. I've also been looking at the prices of some of the suggested SSR's online, and unless I'm looking in the wrong place the cost (> 25 GBP each??) is too high for me to justify using them in this project so I have found Takamisawa's JV3S-KT which seems to fit the bill.

The PDF contains the following image:

In reply to ebeowulf17, for the JV3S-KT, the coil nominal power is .2w and the operate power is .113w.

Thanks very much for all your help.

Regards

Rich

That battery and relay combo may not be a good match. Let's do some math:
0.113W / 3V = 0.038A (38mA) per relay
0.038A * 4 relays = 0.152A (152mA) total
235mAh battery capacity / 152mA = 1.55hrs

So, your battery contains enough energy to run those relays for just over an hour and a half. In reality, battery behavior is much more complicated than just running the numbers on total amp hours as I've just done. I suspect that in realty the battery will completely fail to operate those relays.

The battery is spec'd for two discharge conditions: continuous drain at 0.19mA, and pulsed drain at 6.8mA - note that they expect you to run it only 24 seconds per day at that current. Your current draw would be more than 20 times higher. If the battery was capable of delivering that much current at once, they'd provide specs at higher values.

Looking at it another way, the battery probably has internal resistance of 15 ohms or higher. Your four relays in parallel present a resistance of roughly 20 ohms. Add that together, and you've got 35 ohms total. They circuit delivers just over half the expected voltage, at just over half the expected current, resulting in around one-fourth the expected power to the relay coils. Not good!

Anyway, sorry I'm beating this into the ground. Sometimes I just enjoy research and math.

How do you feel about soldering? Instead of buying special SSRs, you can buy through hole MOSFETs and do a little soldering. We can help you choose the right ones. Depending on how fancy you want to make it, the wire connectors and housings/mounting will cost far more than the MOSFETs - they're dirt cheap!

 

That battery and relay combo may not be a good match. Let's do some math:
0.113W / 3V = 0.038A (38mA) per relay
0.038A * 4 relays = 0.152A (152mA) total
235mAh battery capacity / 152mA = 1.55hrs

So, your battery contains enough energy to run those relays for just over an hour and a half. In reality, battery behavior is much more complicated than just running the numbers on total amp hours as I've just done. I suspect that in realty the battery will completely fail to operate those relays.

The battery is spec'd for two discharge conditions: continuous drain at 0.19mA, and pulsed drain at 6.8mA - note that they expect you to run it only 24 seconds per day at that current. Your current draw would be more than 20 times higher. If the battery was capable of delivering that much current at once, they'd provide specs at higher values.

Looking at it another way, the battery probably has internal resistance of 15 ohms or higher. Your four relays in parallel present a resistance of roughly 20 ohms. Add that together, and you've got 35 ohms total. They circuit delivers just over half the expected voltage, at just over half the expected current, resulting in around one-fourth the expected power to the relay coils. Not good!

Anyway, sorry I'm beating this into the ground. Sometimes I just enjoy research and math.

How do you feel about soldering? Instead of buying special SSRs, you can buy through hole MOSFETs and do a little soldering. We can help you choose the right ones. Depending on how fancy you want to make it, the wire connectors and housings/mounting will cost far more than the MOSFETs - they're dirt cheap!

Hi Ebeowulf17

I wouldn't consider it being beaten into the ground - just being thorough!

I have no qualms about soldering, in f

That battery and relay combo may not be a good match. Let's do some math:
0.113W / 3V = 0.038A (38mA) per relay
0.038A * 4 relays = 0.152A (152mA) total
235mAh battery capacity / 152mA = 1.55hrs

So, your battery contains enough energy to run those relays for just over an hour and a half. In reality, battery behavior is much more complicated than just running the numbers on total amp hours as I've just done. I suspect that in realty the battery will completely fail to operate those relays.

The battery is spec'd for two discharge conditions: continuous drain at 0.19mA, and pulsed drain at 6.8mA - note that they expect you to run it only 24 seconds per day at that current. Your current draw would be more than 20 times higher. If the battery was capable of delivering that much current at once, they'd provide specs at higher values.

Looking at it another way, the battery probably has internal resistance of 15 ohms or higher. Your four relays in parallel present a resistance of roughly 20 ohms. Add that together, and you've got 35 ohms total. They circuit delivers just over half the expected voltage, at just over half the expected current, resulting in around one-fourth the expected power to the relay coils. Not good!

Anyway, sorry I'm beating this into the ground. Sometimes I just enjoy research and math.

How do you feel about soldering? Instead of buying special SSRs, you can buy through hole MOSFETs and do a little soldering. We can help you choose the right ones. Depending on how fancy you want to make it, the wire connectors and housings/mounting will cost far more than the MOSFETs - they're dirt cheap!

Hi Edowulf17

Thanks for the comprehensive reply. You're not beating it into the ground, just being thorough.

I appreciate your help and as for the soldering I even bought a new iron before I started this thread because I thought a hundred or so LED's needed a bit of soldering...

I have little experience in designing circuits but did do a lot of bread board stuff "when I was a lad"....
.
I would appreciate your help with this.

If you need any 6502 or Z80 assembler, Cobol or other programs written let me know!!

 

Hi Ebeowulf17

I wouldn't consider it being beaten into the ground - just being thorough!

I have no qualms about soldering, in f


Hi Edowulf17

Thanks for the comprehensive reply. You're not beating it into the ground, just being thorough.

I appreciate your help and as for the soldering I even bought a new iron before I started this thread because I thought a hundred or so LED's needed a bit of soldering...

I have little experience in designing circuits but did do a lot of bread board stuff "when I was a lad"....
.
I would appreciate your help with this.

If you need any 6502 or Z80 assembler, Cobol or other programs written let me know!!

Cool! No promises on timing, but I'll try to pick out a few suitable MOSFETs and draw an example schematic of how to use them. If not tonight, probably Monday (tomorrow will be busy.) Cheers!

 

Ok, here's an example of how I would do this with one MOSFET and one resistor per lighting zone/power supply. Note that the position of the MOSFET in the circuit matters - it must be between the LED cathodes and the negative side of the battery. Also note that this system only works if all of the circuits share a common ground for the negative sides of the power supplies.


All of the MOSFETs I looked at list the same 100nA leakage current specification as a maximum, although I would guess the actual leakage value is usually much lower. Most of the current consumption in the 3V side of this circuit is from the pull-down resistors I've added to the MOSFET gates. Lower consumption could be achieved with weaker pull-downs, which would probably be fine. I was playing it safe with this design out of ignorance. As drawn with 100k resistors, these MOSFET switch circuits would draw ~0.03mA each, or a total of ~0.12mA for all four of them. This is below the continuous current spec'd in the battery datasheet, so it should work well with the coin cell and have a reasonable run time.

Here's a DigiKey search I did as a starting point for choosing MOSFETs:
https://www.digikey.com/short/phdq9c

Here are a few that I picked out that looked promising. I may have gone overboard by limiting myself to units that can switch 100V, but I wasn't sure how high your power supplies would go, and I always like to leave a large margin on specs whenever possible.
https://www.digikey.com/product-detail/en/on-semiconductor/FQP13N10L/FQP13N10L-ND/1053764
https://www.digikey.com/product-detail/en/vishay-siliconix/IRLD110PBF/IRLD110PBF-ND/812489
https://www.digikey.com/product-detail/en/diodes-incorporated/ZVN4210A/ZVN4210A-ND/152484

Other folks here may have better advice on MOSFET selection - I'm relatively new to the hobby and don't have any standard favorite parts, so I'm always choosing blindly based on random search results and reading datasheets. A lot of the regulars here already know numerous components by heart and may be able to point directly to better candidates.

Still, I think the ones above would get the job done, and they're only about $1 each through DigiKey, much better than previously discussed SSRs and relays, and with much lower current consumption too!

 

Ok, here's an example of how I would do this with one MOSFET and one resistor per lighting zone/power supply. Note that the position of the MOSFET in the circuit matters - it must be between the LED cathodes and the negative side of the battery. Also note that this system only works if all of the circuits share a common ground for the negative sides of the power supplies.
View attachment 180612

All of the MOSFETs I looked at list the same 100nA leakage current specification as a maximum, although I would guess the actual leakage value is usually much lower. Most of the current consumption in the 3V side of this circuit is from the pull-down resistors I've added to the MOSFET gates. Lower consumption could be achieved with weaker pull-downs, which would probably be fine. I was playing it safe with this design out of ignorance. As drawn with 100k resistors, these MOSFET switch circuits would draw ~0.03mA each, or a total of ~0.12mA for all four of them. This is below the continuous current spec'd in the battery datasheet, so it should work well with the coin cell and have a reasonable run time.

Here's a DigiKey search I did as a starting point for choosing MOSFETs:
https://www.digikey.com/short/phdq9c

Here are a few that I picked out that looked promising. I may have gone overboard by limiting myself to units that can switch 100V, but I wasn't sure how high your power supplies would go, and I always like to leave a large margin on specs whenever possible.
https://www.digikey.com/product-detail/en/on-semiconductor/FQP13N10L/FQP13N10L-ND/1053764
https://www.digikey.com/product-detail/en/vishay-siliconix/IRLD110PBF/IRLD110PBF-ND/812489
https://www.digikey.com/product-detail/en/diodes-incorporated/ZVN4210A/ZVN4210A-ND/152484

Other folks here may have better advice on MOSFET selection - I'm relatively new to the hobby and don't have any standard favorite parts, so I'm always choosing blindly based on random search results and reading datasheets. A lot of the regulars here already know numerous components by heart and may be able to point directly to better candidates.

Still, I think the ones above would get the job done, and they're only about $1 each through DigiKey, much better than previously discussed SSRs and relays, and with much lower current consumption too!

Ok, here's an example of how I would do this with one MOSFET and one resistor per lighting zone/power supply. Note that the position of the MOSFET in the circuit matters - it must be between the LED cathodes and the negative side of the battery. Also note that this system only works if all of the circuits share a common ground for the negative sides of the power supplies.
View attachment 180612

All of the MOSFETs I looked at list the same 100nA leakage current specification as a maximum, although I would guess the actual leakage value is usually much lower. Most of the current consumption in the 3V side of this circuit is from the pull-down resistors I've added to the MOSFET gates. Lower consumption could be achieved with weaker pull-downs, which would probably be fine. I was playing it safe with this design out of ignorance. As drawn with 100k resistors, these MOSFET switch circuits would draw ~0.03mA each, or a total of ~0.12mA for all four of them. This is below the continuous current spec'd in the battery datasheet, so it should work well with the coin cell and have a reasonable run time.

Here's a DigiKey search I did as a starting point for choosing MOSFETs:
https://www.digikey.com/short/phdq9c

Here are a few that I picked out that looked promising. I may have gone overboard by limiting myself to units that can switch 100V, but I wasn't sure how high your power supplies would go, and I always like to leave a large margin on specs whenever possible.
https://www.digikey.com/product-detail/en/on-semiconductor/FQP13N10L/FQP13N10L-ND/1053764
https://www.digikey.com/product-detail/en/vishay-siliconix/IRLD110PBF/IRLD110PBF-ND/812489
https://www.digikey.com/product-detail/en/diodes-incorporated/ZVN4210A/ZVN4210A-ND/152484

Other folks here may have better advice on MOSFET selection - I'm relatively new to the hobby and don't have any standard favorite parts, so I'm always choosing blindly based on random search results and reading datasheets. A lot of the regulars here already know numerous components by heart and may be able to point directly to better candidates.

Still, I think the ones above would get the job done, and they're only about $1 each through DigiKey, much better than previously discussed SSRs and relays, and with much lower current consumption too!

Wow! Monday comes fast! I shall print this off to read first.

Thanks for your help and suggestions. I'll try not to ask too many questions! Have a good relaxing Sunday and don't be too busy.

I'll post a pic of the model here when it's done.

Thanks and regards

Rich

 

Wow! Monday comes fast! I shall print this off to read first.

Thanks for your help and suggestions. I'll try not to ask too many questions! Have a good relaxing Sunday and don't be too busy.

I'll post a pic of the model here when it's done.

Thanks and regards

Rich

Ask as many questions as you like. Nothing wrong with that.

Good luck with the project, and I look forward to seeing pictures! It sounds like a cool project. That's quite a beast of a rig that you're recreating!

 

You have five red LEDs in series with a 9VDC supply.

The red LEDs have a Vf of 2.2V. 4 times 2.2 is 8.8. There is 0.2V remaining. That may not be enough to stably control the current with a resistor.

 

You have five red LEDs in series with a 9VDC supply.

The red LEDs have a Vf of 2.2V. 4 times 2.2 is 8.8. There is 0.2V remaining. That may not be enough to stably control the current with a resistor.

And as the battery dies, the red LEDs won’t light at all. You’d be better off recalculating your current resistor for two LEDs and with your red lights in a series|parallel circuit. Two sets of two LEDs and a resistor, which are then wired in parallel.