# Buck(-boost) with constant voltage?

Discussion in 'The Projects Forum' started by JMD, Apr 13, 2010.

1. ### JMD Thread Starter Member

Dec 9, 2009
96
0
I got a problem, that i hope any of you can assist me with.

I got a supply of around 24V (varies a bit) and about 60mA (also varies a bit). I need a constant output (when the output is enabled) of 12V and 800mA (the load is a resistor @ 15 Ohm).

The load needs to be connected for 10ms, with as high duty cycle as possible.

So.. i figured i would use a buck-boost converter. But.. is that the best choice? Im open to ideas!

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Think wattage. You have a supply of 24V an 60ma, that is 1.44W. Taking it down with perfect efficiency you would get 12V at 120ma (also 1.44W). Since nothing is perfect, 80% would be excellent, and you could get 12V at 96ma.

TANSTAFL.

You might be able to charge a cap, and get very short bursts of current. Is this what you're after?

3. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
I'm wondering where you are going to get all of the extra power from?

Let's say that you could build an 80% efficient DC-DC converter, which is reasonable.

The theoretical 80% efficiency of a converter means you'll require 9.6/.8=12 Watts

You say your supply is 24v at 60mA, which is 1.44 Watts.

This means that you will need 1.44W-12W = 10.56 Watts over what your available power supply is.
[eta]
Gosh, when you stop to take out the trash, it really slows down your posting.

4. ### JMD Thread Starter Member

Dec 9, 2009
96
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Bill --> What i need, is to dump 12V across a 15 Ohm resistor (9,6W), for 10 ms - and repeat that as fast as possible. Say, if i got a supply of 0,96W - that would mean i would have to "charge" for 100 ms, and then drive the load for 10 ms.

What i need, is to be able to repeat those 10ms with the least possible delay (the delay is the time needed to store enough energy to drive the load at 9,6W for 10ms).

SgtWookie --> The key-word here is "energy storage". The best case, would be where i had a supply of 9,6W - that would mean i could run the 10ms cycle 100 times pr. second.

The case here, is that i got around 1,44W - meaning that in an ideal world, i would have to store that in a total of 66,67ms, in order to be able to drive the load for 10ms. Meaning that i would be able to repeat the cycle 13 times pr. second.

Does it make more sense now? If not, let me know

Dec 9, 2009
96
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Someone?

6. ### ifixit Distinguished Member

Nov 20, 2008
639
110
Does the 12V need to be regulated to maintain 12V during the 800mA discharge? If so, how accurate?

7. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Why not just use a lead-acid battery for the energy storage, with some rather large caps on the output?

8. ### ifixit Distinguished Member

Nov 20, 2008
639
110
To get 9.6W from 1.44W...
1. Charge time with a constant current is: T = (C x ΔV)/I.
2. Charge a 1500uF cap from the 24V with a constant current at 60mA.
3. Allow 2V for constant current device headroom.
4. Initial charge takes 1500u x 22V / 60mA = 0.55 Sec.
5. Power a 1A, 3-terminal regulator from this cap and adjust for a 12V output.
6. Allow 4V for the regulator headroom.
7. With a 800mA load on the regulator output, the cap will discharge at 533V/sec. or from 22V to 16V in 11.3mS.
8. Discharging for 10mS would take the cap V from 22V to 16.667V.
9. Re-charging from 16.667V to 22V at 60mA takes 134mS.
10. You could also just charge the cap with a resistor to 24V, but will take a little longer.
Play around with this example and come up with a circuit that works for you.

Have Fun,
Ifixit

9. ### Bernard AAC Fanatic!

Aug 7, 2008
4,240
414
A little more information on power source might be helpfull, ie, solar cells? I would use a buck converter to drop V down to about 15 V, & charge Sgt Wookie's 10 Ahr Pb-H2SO4 battery; would skip big cap on output. Might use sample and hold during on time to monitor battery V & adjust duty cycle as necessary to keep V above 12.

Last edited: Apr 18, 2010
10. ### JMD Thread Starter Member

Dec 9, 2009
96
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That is preferred, yes.

SLA's or any other batteries is not an option. We are to imitate the power grid - power source is a generator with 3-phases. (very tiny one).

True, a LM7812 could be used - i just have to compare the power loss with a buck-converter Ill give it a shot.

The power source is a 3-phase generator (actually a hobby motor). Batteries are not an option.

11. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
69
Charge a large capacitor direct from the generator.

Produce the 12V with a National 'Simple Switcher' IC to get decent efficiency and operation as the input voltage droops.

I think? that some of the simple switcher ICs have a logic enable control. You could use that with a couple of comparators monitoring the input capacitor charge to switch the regulator on & off.

12. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Ive thought about the caps too - actually dual layer super caps. Didnt think of the ICs tho.. got any model numbers?

13. ### Bernard AAC Fanatic!

Aug 7, 2008
4,240
414
LM2575T- adjustable or 12V, pin 5 ON-OFF.

14. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Thanks! Ill give it a look !