So i want to run a highpower IR spotlight and raspbery pi based webcam 100-150 ft from my house. I plan on low V ac then using a full bridge rectifer but havent actually built a circuit like this b4 and have no idea what size the cap should be. Also i would like input on lightning isolation of the 24v ac line from the house. Finally i would appreciate any other input!

my plan is as follows

a safe estimate for the total power consumption is 1amp @ 12v for the IR light and 500mA @ 5.5v for the raspi and camera.

this leaves me with low voltage ac to the camera. my envisioned set up is

• 120 to 24v 3a transformer (i belive this gets around the legal/permit issues with running 120/mains)
• 150 ft of 18g wire either over head or under ground.
• full wave bridge rectifier rated for 3a http://www.digikey.com/product-search/en?pv721=76&FV=fff40015,fff8052c&k=bridge+rectifier&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25
• smoothing cap (no idea how to size this)
• 2 dc-dc buck converters to go from what ever i get after line loss To the 12v and 5.5v for the light and raspi/camera. something like this http://www.amazon.com/LM2596-Conver...&sr=1-10&keywords=step-down+voltage+converter

things im unsure on:

1. lightning arresting
2. cap size
3. calculating the load the dc-dc converters will put on the demand side ... aka if the demand is drawing 5.5v @ 500mA what is the demand on the 24v source side.

Thanks!

 

If you use a 24Vac tranny, full-wave rectify, filter, you will wind up with an unregulated voltage north of 35V. I'd be more inclined to look for a transformer that outputs 12Vac. That will give you an unregulated average voltage of about 16V, which is plenty to feed to the buck regs.

Figure that those switchers are 80% efficient, so your power requirements are 12W for the IR light plus 2.7W for the Pi, for a total of ~15W.
15W/80% = 19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25A. This sets the required filter cap; see below.

Putting the Full-Wave-Bridge (FWB ) and filter at the camera end of the feed, and feeding AC is problematic, because with a capacitor-input filter, the current in the feed wires will be very big pulses of short duration (just a fact of life with a power supply), causing large I^2R losses. I would put the FBW and filter at the transformer end, and feed filtered, but unregulated DC down the feed wire.

To calculate the size of the filter cap, q=C*ΔV, but q =I*t, so substituting and rearranging gives C=I*t/ΔV. Figure that the filter will charge to a peak near 1.4*12V = 16.8V, and it is ok to let the filter sag 3V for 20ms until the next current pulse, so C=1.25*0.02/3 = 0.008F = 8000uF.

 

I tried the 12-0-12 approach. The 1.5 ohms in the 18 AWG wire at 3 amps queers the deal.
I think this is going to need at least a 15 volt winding.
This is how I fell on my face with the wrong voltage:
When I went to add in the wire losses, the answer was FUBAR

 

Here, see if this is better.

 

If you use a 24Vac tranny, full-wave rectify, filter, you will wind up with an unregulated voltage north of 35V. I'd be more inclined to look for a transformer that outputs 12Vac. That will give you an unregulated average voltage of about 16V, which is plenty to feed to the buck regs.

Figure that those switchers are 80% efficient, so your power requirements are 12W for the IR light plus 2.7W for the Pi, for a total of ~15W.
15W/80% = 19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25A. This sets the required filter cap; see below.

Putting the Full-Wave-Bridge (FWB ) and filter at the camera end of the feed, and feeding AC is problematic, because with a capacitor-input filter, the current in the feed wires will be very big pulses of short duration (just a fact of life with a power supply), causing large I^2R losses. I would put the FBW and filter at the transformer end, and feed filtered, but unregulated DC down the feed wire.

To calculate the size of the filter cap, q=C*ΔV, but q =I*t, so substituting and rearranging gives C=I*t/ΔV. Figure that the filter will charge to a peak near 1.4*12V = 16.8V, and it is ok to let the filter sag 3V for 20ms until the next current pulse, so C=1.25*0.02/3 = 0.008F = 8000uF.

thanks! a few follow up questions as i do have to do it with 24v for a different project (pi sprinkler)

Here, see if this is better.

might as well go with the 24v as i need to built another rectifier circuit for a lawn sprinkler system anyway, i found a DC-DC converter thats rated for 48v max which should handle the smoothed power from the rectifier. http://www.amazon.com/gp/product/B00CZ6422Y/ref=ox_sc_act_title_1?ie=UTF8&psc=1&smid=A67FFQ1POP4BF which should make it work even better given the 18G wire correct?

 

I agree with #12. A 15V secondary would be better.

 

If you rectify and filter 24 VAC, that's going to put about 50 watts on the regulator chip. You will need a fan to keep that from burning up.

 

If you rectify and filter 24 VAC, that's going to put about 50 watts on the regulator chip. You will need a fan to keep that from burning up.

He is using switchers, so they don't care...

 

Oh. Never mind.

 

If you use a 24Vac tranny, full-wave rectify, filter, you will wind up with an unregulated voltage north of 35V. I'd be more inclined to look for a transformer that outputs 12Vac. That will give you an unregulated average voltage of about 16V, which is plenty to feed to the buck regs.

Figure that those switchers are 80% efficient, so your power requirements are 12W for the IR light plus 2.7W for the Pi, for a total of ~15W.
15W/80% = 19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25A. This sets the required filter cap; see below.

Putting the Full-Wave-Bridge (FWB ) and filter at the camera end of the feed, and feeding AC is problematic, because with a capacitor-input filter, the current in the feed wires will be very big pulses of short duration (just a fact of life with a power supply), causing large I^2R losses. I would put the FBW and filter at the transformer end, and feed filtered, but unregulated DC down the feed wire.

To calculate the size of the filter cap, q=C*ΔV, but q =I*t, so substituting and rearranging gives C=I*t/ΔV. Figure that the filter will charge to a peak near 1.4*12V = 16.8V, and it is ok to let the filter sag 3V for 20ms until the next current pulse, so C=1.25*0.02/3 = 0.008F = 8000uF.

Thanks! but i do have a few follow up questions as i do desire to do a 24v version for another project (pi lawn sprinklers..comercial irigation valves are 24v ac and i only want one transformer in the box) also i academically want to understand the equation better. (I found a higher rated buck switch as well)

1) "19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25" this confuses me... if i do the I=w/v shouldnt it be a 1.58A requirement? or is there something im missing about VA?

2) is the t (time in sec) = .02 you used a rounded 0.016? 1sec/60hz?
3) where did the 1.4*12v come from? i get that the cap increases the voltage but not necessarily why or how to calculate by how much
4) when buying the cap, is it correctly the voltage rating on the cap just has to be a bit higher than the max v in the system and the UF number calculated is all that matters?

 

Thanks! but i do have a few follow up questions as i do desire to do a 24v version for another project (pi lawn sprinklers..comercial irigation valves are 24v ac and i only want one transformer in the box) also i academically want to understand the equation better. (I found a higher rated buck switch as well)

1) "19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25" this confuses me... if i do the I=w/v shouldnt it be a 1.58A requirement? or is there something im missing about VA?

2) is the t (time in sec) = .02 you used a rounded 0.016? 1sec/60hz?
3) where did the 1.4*12v come from? i get that the cap increases the voltage but not necessarily why or how to calculate by how much
4) when buying the cap, is it correctly the voltage rating on the cap just has to be a bit higher than the max v in the system and the UF number calculated is all that matters?

never mind on number 3, i did some googling and found the info on rms vs peak V

1) "19W, so your transformer will have to have to be rated at >20VA. 20VA/16V = 1.25" this confuses me... if i do the I=w/v shouldnt it be a 1.58A requirement? or is there something im missing about VA?
2) is the t (time in sec) = .02 you used a rounded 0.016? 1sec/60hz?
3)--------
4) when buying the cap, is it correctly the voltage rating on the cap just has to be a bit higher than the max v in the system and the UF number calculated is all that matters?

 

The energy cost to the transformer is related to peak voltage so you can't get the labeled current when you rectify. The voltage rating of the capacitor must be at least the actual maximum voltage and up to twice that much.

I can't tell you about the 0.02. MikeML is a, "real" engineer and can prove it with numbers. I was merely taught by real engineers, built the circuit, measured it, and verified the performance to within 1% of the predicted value. Strange as it seems, MikeML and I agree from time to time.
Probably accidental.

 

The energy cost to the transformer is related to peak voltage so you can't get the labeled current when you rectify. The voltage rating of the capacitor must be at least the actual maximum voltage and up to twice that much.

I can't tell you about the 0.02. MikeML is a, "real" engineer and can prove it with numbers. I was merely taught by real engineers, built the circuit, measured it, and verified the performance to within 1% of the predicted value. Strange as it seems, MikeML and I agree from time to time.
Probably accidental.

sounds good! and thanks again!

 

Here is what I found using a 24Vrms transformer. I'm trying to model the transformer, the filter, the wire resistance, and the SMPS load.



R1 was inserted just so I can plot the current though it to show why it is a bad idea to put the full-wave-bridge and filter at the distal end of the loop. Note the 8A peaks in I(R1) that would otherwise have to flow down the loop through R2. By placing D1-4 and C1 at the proximal end of the loop, the current that flows is the almost-steady I(R2).

The #18 awg wire used for the loop has 6.5Ω per 1000 ft, so the loop resistance is lumped into R2 = 2*(150*6.5/1000).

Based on the final output voltage/current, after adding power lost in SMPS inefficiencies, the power delivered to the far end of the wire is 20W. To get 20W, the current is I=P/V(smps), so I use a behavioral current sink B1 to simulate what the SMPSs do..., namely, dynamically adjust their input current such that their total input power is 20W at whatever the voltage is at V(load), or put another way, the product of V(smps)*I(B1) = 20 (Gotta love the behavioral current source).

Note the voltage drop (difference between V(filt) and V(smps) due to the wire resistance R2 and the loop current I(R2).

With a 1000uF filter capacitor C1 at the transformer end, there is ~5.5V of ripple at V(filt). The voltage at the load end V(smps) sags to a minimum of 25.4V, which is more than enough head room for the SMPSs.

To get the 20W to the distal end, the rms line current is about 703mA, and the rms voltage at V(smps) is 28.6V, which is real close to 20W.

The power delivered by L1 (transformer secondary) is 22.1W; hence the recommendation that it be rated for ~25VA. The lost power is in the FWB and the line loss R2.