Power Supply circiut for Propeller with 3 levels.

DavidM

Hi,

Can someone verify my schematic for a power supply circuit for the propeller.

I need to run off a 12 volt SLA battery, and provide 3 power levels as follows..

12 volts up to 6-9 amps ( from the 12 volt battery)
5 volts at 1 amp ( from the LM2595T-5.0)
3.3 volts at 500 mA ( from the LM2937ET-3.3)

I also want to run separate ground lines as indicated as well.
I believe this will not require heat-sinks?
All my parts need to be through-hole.

The 12 volts is required to run power mosfets or relays.


Please excuse my drawing as I haven't learnt how to use eagle PCB CAD yet ( and besides, not all of the parts I need are in the libraries)
So I did this with a simple drawing application on my MAC


regards

David M

QuattroRS4

DavidM,
TChapman has designed a nice power supply unit - details of it can be found in the Sandbox forum

it may be worth a read ..

http://forums.parallax.com/showthread.php?p=641261

Regards,
Quattro

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'Necessity is the mother of invention'

Harley

DavidM,

D1 and D2 are polarized parts; they should be marked so as to ID the proper orientation.

B1 hopefully cannot be connected backwards. There isn't any protection for that possibility. Probably lose the first cap, and regulator and any circuitry connected to the 12v line.

Just precautions to consider. Else, you may not even see the 'puff of smoke' of the lost parts.

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Harley Shanko
h.a.s. designn

DavidM

Thanks Harley,

yes I need to state the polarity for the diode ( D2) and led (D1), thanks for pointing that out.

1) Do I need a protection diode between the battery and the switch? If so which one should I use? I do want to protect from an incorrectly connected battery.

2) Why would I want to "Lose" the first cap ( C1) this is required by the regulator according to the PDF for the LM2595?

regards

Dave M

Harley

DavidM said...

1) Do I need a protection diode between the battery and the switch? If so which one should I use? I do want to protect from an incorrectly connected battery.

2) Why would I want to "Lose" the first cap ( C1) this is required by the regulator according to the PDF for the LM2595?

response....
1. Not knowing what your 'heavy load' components and circuitry might be on the 12v line, hard to say what diode to use to protect against reverse battery connections. Maybe separate the 12v right after the switch and protect at least the regulator from 12v with a smaller diode. Since I don't know what all the 12v uses, maybe it doesn't matter if the 12v is reversed to the load? To carry Amps might require a hefty sized diode; I've not dealt with this heavy a current draw since college days in electrical lab dealing with dc motors.

2. I meant, if the 12v were reversed it would back bias the polarized cap and destroy (= lose) it most likely.

Difficult to give much more advice on further protection without knowing about the whole design

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Harley Shanko
h.a.s. designn

Mike Green

By saying "lose", Harley was saying that hooking the input supply backwards would probably destroy the input capacitor (C1) since they're usually polarized.

To protect the power supply from a reverse battery connection, all you need is a diode in series with the positive lead from the battery with the anode connected to the battery and the cathode connected to your power supply. The diode probably should be rated at 50V or higher and about twice the current that your supply will draw. As usual for any power supply running from a storage battery, you should also have a fuse in the circuit also rated somewhat more than the maximum expected to be drawn. This is mostly to protect everything from some kind of part failure or other short circuit that would otherwise result in a fire or damage to a variety of components "in the way".

DavidM

Thanks Mike & harley

Ok I understand the "lose" now.
I will revise my schematic with an added diode and fuse, and will post it shortly.


Dave M

Lawson

While a diode to protect from hooking up the battery up backwards is a nice idea it does prevent regenerative braking. (not so bad) Also, if any of the parts can back feed power to the 12v line (like regenerative brakes) the 12v line can raise in voltage a lot! (very bad) This is because the reverse polarity diode is preventing the battery from absorbing excess power and keeping the 12v line at 12v. A better reverse polarity protection method uses a relay and takes advantage of an electronic speed controller or some other device that shorts reverse voltages. see the attached picture. (I oridgonally found this on the http://www.4qdtec.com site but can't find it again!)

Hope this helps,
Marty

DavidM

Hi,

I'm note sure about the relay?


but here is my updated Schematic.

Dave M

Mike Green

The easiest way to fix the issue with the diode and regenerative braking is to connect any motors to the other side of the diode. The diode is intended to protect the switching regulator and the 220uF electrolytic capacitor, not the motors and their associated circuitry. These can have their own reverse polaritly protection if needed. On the Shottky diode ... you need to show polarity (see the regulator datasheet). You can also have separate fuses for the logic and the motors. The schematic looks good to me otherwise.

DavidM

Thanks Mike and others!

Ok, I have change the position of the diode, I still dont know which one to use?

I will have other protection in the 12 volt circuit that runs my relays/power mosfet switches.

Thanks

Dave M

Mike Green

All kinds of power diodes will do. The 1N4001 should work. It is rated at 50V and can handle an average current of 1A.

DavidM

Thanks Mike,

Is there any problem with using the SCHOTTKY DIODE 1N5820 as it is rated at 3 amps 40V, I actually have 1.5 amps running for this part of the circuit, it also keeps down the number of different pars required,

Mike Green

The Schottky diode should be fine. They have lower voltage drops than standard diodes, but have slightly higher leakage. Not an issue in a power circuit.

DavidM

Thanks Mike,

Is leakage "reverse" current?

Dave M

DavidM

By The way, Will these two regulators need to be placed "FLAT" on the board, or can they stand vertically?

regards

Dave M

inservi

Hello,
About the D2 if the + is the anode and the - is the cathode then i think that she is not in the right sens. I think you must connect the anode to the ground and the cathode to the OUT.
I have an other question about the separate ground lines. As i understand your schematic, there is only one ground. You will probably use 3 cables but it is the same ground for all ?

dro.

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in medio virtus

DavidM

Hi Inservi, ( and others )

You are right, I have update my schematic accordingly.

My Outstanding Questions Are..

1) What kind of capacitor should C3 be? ( i.e Electrolytic/ceramic?) What voltage?
2) Do I need to use heat-sinks for the regulators , according to the specs it says I don't
3) Can the regulator ( which are through-hole) be left standing vertically.
4) How do I add a BATTERY VOLTAGE LEVEL circuit to this, ( ideally I2C ) so the Propeller can monitor the battery level, I would probably like a 8 pin DIP ADC to I2C if one exists?

Thanks

Dave M

Leon

C3 is a 10 uF, you will only find electrolytics or tantalums in that value. I'd use an Al electrolytic. 10V would be OK.

You won't need heat sinks.

It will help with heat dissipation if it is vertical.

The simple sigma-delta ADC with a couple of resistors and capacitors might be good enough. Otherwise I'd use an MCP3201, they are cheap and have an SPI interface.

Leon

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Amateur radio callsign: G1HSM
Suzuki SV1000S motorcycle

DavidM

Thanks Leon,

I like your answer!

I will choose the electrolytic as you suggested. I am glad that the heat-sinks aren't required, vertical = heat dissipates!

Also I will look into the specs for the MCP3201, but it does sound like the device I need.

Thanks

Dave m

DavidM

Hi,

I have decided to not use the ADC chip, overkill/over complicated compared to RC Time method!

I have now updated my schematic to include RCTIME circuit to monitor the battery level back to the propeller via 1 x I/O pin, this will take up less space and cheaper ( and also easier to code)

I am unsure of the values, but I have read some BS2 docs as well as the RCTIME docs for the Propeller and have come up with the following circuit V2.4

The components for the RCTIME circuit are in red text.

Q1) Will this work and have I done this correctly?


Thanks

Dave M