Inductorless DC-DC converters for space application

[p1ayaone1]

I am designing a small satellite which uses a magnetometer and some coils (which interact with Earth's magnetic field) to provide attitude control.

The strength of the geomagnetic field is 30 uT to 60 uT (according to Wolfram Alpha), and the magnetometer has a resolution of 0.015 uT.

The satellite observes the CubeSat specification, which means it is basically a 10-cm cube. It is really small.

Now I need to power this thing. What I really need is voltage regulators with the highest possible efficiency. Also, because of the magnetometer, I don't want to have inductors in the power system, because they would be "always on".

The desired output voltage is 3.3V with 150 mA absolute minimum, but I would much rather be closer to 200 mA. (It should be able to go lower than 150 mA, but that figure is the minimum maximum. ) I will probably also put a 5 to 8 V bus, but my main concern is the 3V3 bus because that's where all my electronics are powered.

The input voltage is two Li-Ion cells. These could be in series (5.5 Vmin to 8.2 Vmax) or in parallel (2.75 Vmin to 4.1 Vmax). I'm ambivalent wrt series/parallel connection, as long as the regulator is sufficiently efficient.

Linear Regulator
OK, fine, it would *work* but this is an efficiency-critical application, and I don't want to lose all that extra voltage as heat.

DC-DC Converter
Buck/boost, Sepic all use inductors. No good.

Charge Pump
There are a tiny few that support variable input voltage, but the dominant functionality is voltage inversion and voltage doubling. Also, current ratings are quite low. I suspect the solution lies in this class of DC-DC converters.

Anybody got any neat ideas?

 

[gnurf]

Use a DC-DC converter with a shielded toroid-wound inductor. With your low current setup I'm guessing you'll be fine. What reasons do you have to believe otherwise?

 

[skeptic2]

How about a pulse width modulated switch between the higher voltage batteries to the 3.3 V supply. The supply voltage would be compared with a reference and when the voltage drops below a preset threshold, the comparator would turn on the switch which would allow the supply capacitors to charge. When the capacitors charge up to the upper voltage threshold the comparator would turn off the switch. The biggest disadvantage would be ripple on the 3.3 V supply. That could be mitigated by setting the upper and lower supply voltage thresholds close together or reducing the hysteresis of the comparator. Additional filtering could be done with resistors and capacitors at some loss of efficiency. I'm picturing the switch operating in the range of 10s of kilohertz to make it easier to filter.

How do you get rid of the heat from the electronics?

 

[p1ayaone1]

gnurf, I know that toroid-wound inductors will have EMI (albeit quite low) unless the toroid is exactly 360 degrees. And shielding only goes so far. The question would them become: what is the B at a particular distance from the inductor at a particular current? Any idea how to estimate that? In any case, the value should be less than 0.015 uT.

skeptic2, the idea you proposed is similar to the charge pump, and I've seen some switching in the MHz range. The problem is that without the comparator, the charge pump gives only integer multiples of the input voltage. I'll search for an integrated solution.

The question of heat is interesting as well. Since we are designing for space, a fan is obviously useless. Hot components will get heatsinks, but I really have no idea how effective those will be with no atmosphere.

The way I understand it, is that space (rather, the few air molecules up there at 800 km) is extremely hot. But, since the molecules are so far apart from one another, the perceived sensation is one of cold. Thermodynamics in space should perhaps be reserved for another thread...

 

[berkeman]

I am designing a small satellite which uses a magnetometer and some coils (which interact with Earth's magnetic field) to provide attitude control.

The strength of the geomagnetic field is 30 uT to 60 uT (according to Wolfram Alpha), and the magnetometer has a resolution of 0.015 uT.

The satellite observes the CubeSat specification, which means it is basically a 10-cm cube. It is really small.

Now I need to power this thing. What I really need is voltage regulators with the highest possible efficiency. Also, because of the magnetometer, I don't want to have inductors in the power system, because they would be "always on".

The desired output voltage is 3.3V with 150 mA absolute minimum, but I would much rather be closer to 200 mA. (It should be able to go lower than 150 mA, but that figure is the minimum maximum. ) I will probably also put a 5 to 8 V bus, but my main concern is the 3V3 bus because that's where all my electronics are powered.

The input voltage is two Li-Ion cells. These could be in series (5.5 Vmin to 8.2 Vmax) or in parallel (2.75 Vmin to 4.1 Vmax). I'm ambivalent wrt series/parallel connection, as long as the regulator is sufficiently efficient.

Linear Regulator
OK, fine, it would *work* but this is an efficiency-critical application, and I don't want to lose all that extra voltage as heat.

DC-DC Converter
Buck/boost, Sepic all use inductors. No good.

Charge Pump
There are a tiny few that support variable input voltage, but the dominant functionality is voltage inversion and voltage doubling. Also, current ratings are quite low. I suspect the solution lies in this class of DC-DC converters.

Anybody got any neat ideas?

Hard problem. Can you just use the parallel connected batteries and not bother with any voltage regulation? What is the input voltage range of your 3V3 circuitry?

 

[p1ayaone1]

berkeman, there is a good deal of 3V3 circuitry (an MCU, a camera, an SD card, temperature sensors, the magnetometer itself) so I would prefer not to run those directly from the battery, but your point is well-taken.

Additionally, the Tx, Rx, and attitude coils will all run on a higher-voltage bus (at least 5V, perhaps up to 12V) to that would require boost circuitry (read: inductors) as well.

 

[p1ayaone1]

A hackish solution would be two of these in parallel, since the current rating to too low.

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=LTC1751EMS8-3.3%23PBF-ND

I'm sure a more elegant solution is out there!

 

[berkeman]

berkeman, there is a good deal of 3V3 circuitry (an MCU, a camera, an SD card, temperature sensors, the magnetometer itself) so I would prefer not to run those directly from the battery, but your point is well-taken.

Additionally, the Tx, Rx, and attitude coils will all run on a higher-voltage bus (at least 5V, perhaps up to 12V) to that would require boost circuitry (read: inductors) as well.

Fair enough. Then I would put the batteries in series, and use a buck topology DC-DC, with a toroidal inductor, with a mu-metal shield around it.

BTW, when you mentioned full-circumferential winding of the toroid, that is for immunity to external B-field interference, not for minimizing external leakage B-field from the toroid itself (AFAIK).

Also, remember that with mu-metal shields, they need to be anealed after they are formed, so you can't just bend up a shield out of a sheet of mu-metal and have it work. I've had good luck with this company:

http://www.magnetic-shield.com/

They do offer kits for prototyping shields, and can help you with annealing prototype shields.

 

[berkeman]

A hackish solution would be two of these in parallel, since the current rating to too low.

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=LTC1751EMS8-3.3%23PBF-ND

I'm sure a more elegant solution is out there!

But those are doublers, right?

 

[berkeman]

BTW, I don't think that a switched cap solution for lowering the source voltage will be very efficient. Maybe no more efficient than just using a linear regulator.

Remember the classic question about where the energy goes when you connect a charged capacitor to an uncharged capacitor through some resistance, and then take the limit as that resistance goes to zero... It's a classic problem that shows that you lose energy when you transfer charge to lower the voltage with capacitors...

 

[p1ayaone1]

nope they are proper regulators. The only problem is the output current is too small!

As for mu-metal, would you wrap the inductors in that then have it treated, or could I build a "box" for the whole circuit, except the magnetometer and attitude coils, and then have it treated?

(Note that the MCU algorithm will prevent the magnetometer from measuring while the coils are being fired.)

 

[p1ayaone1]

and you're right about efficiency. Page 4 of the datasheet has the efficiency of the 3V3 regulator vs. output current for different input voltages. It is 50% efficient when Vin = Vout, and the efficiency is greater when Vin < Vout (that ESR is conducting for less time)

 

[berkeman]

nope they are proper regulators. The only problem is the output current is too small!

I'll have to look into them. Are you familiar with the classic capacitor energy question that I mentioned? Take two caps of equal value C, charge one to Vi, connect them in parallel to make a new lower voltage Vf. What is the energy content of the one charged cap C at Vi? What is Vf? What is the total final energy of the two caps at Vf? Where did the other energy go?

As for mu-metal, would you wrap the inductors in that then have it treated, or could I build a "box" for the whole circuit, except the magnetometer and attitude coils, and then have it treated?

I'm not sure of the best shape shield to use to shield a toroid -- maybe contact Magnetic Shield Corp (or browse their website) to find out what other folks use. Probably something like two 5-sided boxes that interleave along the edges when put together...?

(Note that the MCU algorithm will prevent the magnetometer from measuring while the coils are being fired.)

What does that mean? The DC-DC will be running somewhere in the 50kHz to 150kHz range. How could you interleave magnetometer measurements with that? Do you briefly shut off the DC-DC to make the measurement quickly, while capacitors hold the 3V3 and other rails to minimum droop? If so, do you even need a shield at all?

 

[p1ayaone1]

First point:
I can't say I know which equation you're talking about... I know the energy in a cap is the product of the charge stored and the voltage over 2. (1 coulomb at 1 volt inside a cap has 0.5 joules of energy)

I understand that all caps have an ESR which dissipates power during conduction, causing energy loss in the system. Ceramic caps can be used to mitigate this loss.

Second point:
The question is whether to focus on shilding the inductors themselves, or the whole circuit? Overall weight is also a consideration.

Third point:
That comment was pre-emptive and a bit unclear on my part. I was not referring to the DC-DC coils (what I've been calling inductors), but the attitude coils (which are admittedly inductors themselves, but they will be designed and hand-made to provide the highest possible radiated B)

I was half-expecting someone to say: "well, if you're so concerned about EMI, how can you use coils to change the satellite's attitude?" What I mean is that the coils (one on each axis) will be used to correct the position of the satellite by inducing a magnetic field of some strength, but those fields will interfere with the magnetometer (also on three axes). So I will take a reading of Bx, By, and Bz, and fire the attitude coils (say Lx, Ly, and Lz after the reading is complete.

I want *lots of* B from the attitude coils because I don't need to run them all the time, but I want as *little* B as possible from the DC-DC converter(s) becuase they need to run always, including them the magnetometer is running.

 

[p1ayaone1]

the B from the attitude coils will interact with the Earth's field, and generate a tiny little torque.

 

[berkeman]

I would just shield the toroidal inductor on the buck DC-DC.

On the capacitor switching question, it goes like this:

Start with one cap C charged to Vi.

Connect it in parallel with another cap (initiall discharged) of the same value C. This reduces the voltage from Vi to Vf = Vi/2, because the total charge stored has to be conserved, but you have twice the capacitance in the final configuration. And since Q = CV, doubling the capacitance leaves you with half the voltage.

But the initial energy stored was Ei = 1/2 C*Vi^2, and the final energy is half that (because the capacitance doubled, but the voltage term gets squared in the energy calculation).

You've lost half of your energy by switching the charge from one cap onto the parallel combination of two caps. Where did the energy go? If you model the setup as having a resistor between the two caps when they are connected, you can show that the energy is dissipated in the resistor as the 2nd cap is charged up to Vf. But you can also show that the energy is lost, even as you take the value of the resistor to be vanishingly small.

So I'm not sure you will be able to make an efficient version of a buck DC-DC using only switched caps. I could be wrong -- maybe there is some trick I'm missing...

 

[p1ayaone1]

Wow. half the energy is lost!

Ei = (C*Vi^2)/2

Doubling the capacitance, and noting that Vf = Vi/2

Ef = (2C*Vf^2)/2
Ef = (2C*(Vi/2)^2)/2
Ef = C*(Vi/2)^2
Ef = (C*Vi^2)/4 = Ei/2

-QED-

So back to that part a few posts back, how do you suppose they get very nearly 90% efficiency for the 5V regulator (LTC1751-5) when Vin = 2.7V and Vout = 5V?
(It's on page 5 of the datenblatt)

 

[berkeman]

Boosting with the caps is different, I think. You're pushing the charge up onto a storage cap, where it is held with the diode as the pump cap comes back down to fill with charge. I guess I should go through the math to figure out what's different, in case it shows some topology trick where the buck down in voltage could be made more efficient too.

 

[p1ayaone1]

I guess I should go through the math...

As you wish, but it's not something I really need. I'm more interested in terminal characteristics than IC implementation...

 

[RocketSci5KN]

http://focus.ti.com/docs/prod/folders/print/tps60501.html

 

[berkeman]

http://focus.ti.com/docs/prod/folders/print/tps60501.html

Fascinating! Thanks RocketSci. So with different topologies of flying caps, you can make pretty high efficiency for some voltage down-conversion ratios. Cool.

And that looks just like what the OP is asking for.

 

[p1ayaone1]

Physics Forums is the greatest!

Application circuit is simple and small, and meets all design requirements. Will hit up TI for some free samples and see if I can get this circuit working.

Thanks all!

 

[gnurf]

Additionally, the Tx, Rx, and attitude coils will all run on a higher-voltage bus (at least 5V, perhaps up to 12V) [...]

The TI step-down device has an input voltage of 1.8V to 6.5V, so, to avoid exceeding this limit, the two liion cells must be connected in parallel (with a voltage of < 4.2V). What then with the transceiver and attitude coils?

 

[p1ayaone1]

The TI step-down device has an input voltage of 1.8V to 6.5V, so, to avoid exceeding this limit, the two liion cells must be connected in parallel (with a voltage of < 4.2V). What then with the transceiver and attitude coils?

Good question. I will use the MAX1680 or something similar (gotta love free samples from TI) to provide a regulated 5V supply. Doubling that further is not out of the question, we'll have to see how the Tx/Rx perform.

http://datasheets.maxim-ic.com/en/ds/MAX1680-MAX1681.pdf

For the record, this satellite will not be launched. Because of this, the Tx/Rx are intentionally underpowered to save precious $$$. The rest of the system will hopefully be declared spaceworthy once it passes environmental, electrical, and functional testing.

 

[Carl Pugh]

? Use a super capacitor and a switching power supply.
Turn the magnetometer off when the super capacitor is charging.

 

[p1ayaone1]

? Use a super capacitor and a switching power supply.
Turn the magnetometer off when the super capacitor is charging.

I don't really know anything about supercapacitors... What is the advantage over Li-Ion?

 

[skeptic2]

On the capacitor switching question, it goes like this:

Start with one cap C charged to Vi.

Connect it in parallel with another cap (initiall discharged) of the same value C. This reduces the voltage from Vi to Vf = Vi/2, because the total charge stored has to be conserved, but you have twice the capacitance in the final configuration. And since Q = CV, doubling the capacitance leaves you with half the voltage.

But the initial energy stored was Ei = 1/2 C*Vi^2, and the final energy is half that (because the capacitance doubled, but the voltage term gets squared in the energy calculation).

You've lost half of your energy by switching the charge from one cap onto the parallel combination of two caps. Where did the energy go? If you model the setup as having a resistor between the two caps when they are connected, you can show that the energy is dissipated in the resistor as the 2nd cap is charged up to Vf. But you can also show that the energy is lost, even as you take the value of the resistor to be vanishingly small.
So I'm not sure you will be able to make an efficient version of a buck DC-DC using only switched caps. I could be wrong -- maybe there is some trick I'm missing...

Yes, Berkeman, I was familiar with this problem and what you say is true, if the two caps are the same size. But what if the source cap is twice the size of the load cap?

Given:
C1 = 2 F
C2 = 1 F
Vi = 1 V
Qi = 0.5*C2*Vi^2 = 0.5 Q
Vf = C1*Vi/(C1+C2) = 0.667 V
Qf = 0.5*(C1+C2)*Vf^2 = 0.667 Q
Eff = 66.67%
Clearly the 50% efficiency is only true if both capacitors are the same size.

You didn’t say how large your battery is so I’m guessing somewhere around 2000 mAh. That is equivalent in energy to 7.2 kF. If C2 is 100 uF then the ratio between the two capacitors is 72 MF. This time we have:
Vf = 72MF*Vi*(72 MF + 100uF) = ~1.0 V
Qf > 0.9999999 Q
Eff > 0.9999999%

Granted, we haven’t taken into account the power consumed by the comparator and switch but if care is taken choosing your components, a switched capacitor has the potential to be more efficient than the 80 – 90 % of the two referenced ICs.

 

[sophiecentaur]

If the quantity you are measuring with your magnetometers is very slowly varying then why should a small amount of emi, at many tens of kHz, be a problem? Why would your measuring algorithm not be able to filter that out? Using an 'intelligent' DC converter, you would need to have no inherent energy losses. A crude voltage doubler followed by a regulator seems a bit low tech.

@berkeman - this is a well known 'question for the student'. The only way to connect the two capacitors without energy loss is to involve an inductor during the switching process and, with a switch, grab a time when the resonance gives you no magnetic energy again. 1. Connect a 'suitable' inductor in parallel across C1; volts will start to fall as current flows into L.
2. When the volts have dropped to 1/√2 (I think it would be), change over from C1 to C2.
3. Wait until the resonance in the L C2 circuit has brought the volts on C2 back up to 1/√2.
4. Disconnect L.
This gives you two capacitors, both with the same polarity of voltage and with half the original energy in each.

 

[Carl Pugh]

playaone1,
Supercapacitors can be be used at any voltage below their rated voltage.
Li-Ion batteries have a limited useful voltage range.

Google supercapacitor and go to the wikipedia website for an explanation on how supercapacitors work. wikipedia say, supercapacitors typically have 1000 of times higher capacitance than electrolytic capacitors.

? Could the magnetometer be off for long periods of time?
If the magnetometer was off, this would save enough power that a series regulator could be used.

 

[skeptic2]

Sophiecentaur, there is something I don't understand about your example. Perhaps you can walk me through it and explain the part where I'm getting lost.

Assume a 1 F capacitor, C1, charged to 1 V containing 1Q.
1. Connect a 'suitable' inductor in parallel across C1; volts will start to fall as current flows into L.
2. When the volts have dropped to 1/√2 (I think it would be), change over from C1 to C2.
4. Wait until the resonance in the L C2 circuit has brought the volts on C2 back up to 1/√2.
5. Disconnect L.
This gives you two capacitors, both with the same polarity of voltage and with half the original energy in each.

It seems that in order for each capacitor to have half the original energy, each capacitor must be charged to √2/2. I think this is what your example does. That means that each capacitor must contain C*V coulombs or 0.7 Q for a total of 1.4 Q.

Where did the extra 0.4 Q come from?

 

[sophiecentaur]

It comes from the fact that current flowed around the circuit, through the inductor. In this case, the 'charge conservation' law doesn't apply - it doesn't need to because it's not a total charge involved but an 'imbalance' of charge. The only respect in which the conservation of charge applied is in the fact that the whole circuit starts off neutral and ends neutral.
(I must say, you had me there for a moment )
What is conserved, however, is Energy - because (ignoring resistance) none is dissipated.

 

[sophiecentaur]

And why is everyone so inordinately chuffed about super capacitors? They have their uses and are eminently suited to many applications but their one huge disadvantage is surely that their voltage is so variable. Not so much of a problem is the circuits they feed have been designed to cope with it but it makes a lot of designs a lot more complex. It seems to me that batteries are doing pretty well at the moment - with increasing capacities and fast charging times - for most applications.

 

[skeptic2]

I'm sorry but your explanation is a little vague. I wonder if you could either go into more detail or point me to a reference. I have searched for and looked at references but haven't found anything that addresses this point.

It seems to me that it must be the charge that is conserved as it is in the traditional two capacitor problem. That problem is a paradox only when zero resistance is assumed. When resistance is added, the lost energy is clearly dissipated in the resistance. What happens in your example when a small amount of resistance is added? On the other hand, charge represents a quantity of electrons, something that to me would be a great deal harder to create or lose.

 

[sophiecentaur]

VAGUE ?!
The cheek of it. (I thought it was more than adequate and not too bad as I was making it up as I went along)
Nevertheless, it seems right enough - helped by the fact that the answer produced the conservation of energy - quite a good case for the prosecution M'lud.

In the 'two capacitor problem' there is no 'alternative' current path to the wires joining the two capacitors. What is being shared is the imbalanced charge as no charge can flow anywhere else. I would agree that, as no 'discharging' can take place then the imbalanced charge must not change: a kind of charge conservation, if you like but not in the 'particle Physics' sense of the phrase.
Connecting an inductor across the terminals of C1 allows the charge on C1 to be any value from +CV to -CV and back up to +CV, depending on which part of the oscillation cycle you disconnect the L. You would agree that is what happens in an LC circuit? Where is your requirement for 'conservation of charge' during that process? Energy is not lost, of course, because you have a current flowing in the inductor so the Electrical Potential Energy is transferred to Magnetic Energy.
I repeat - no electrons are ever lost or gained by the circuit; they are just re-arranged about it according to the permitted rules. There may be more or fewer excess electrons on one plate but that number is totally balanced by a change in the number of electrons on the other plate.

 

[mheslep]

A problem with using any kind of a switching circuit as a solution here is that for a given inductance the B field is proportional to the current I, and current pulses/spikes intrinsic to a switcher are many multiples of the steady state current through a linear converter. And, it is not just the switching magnetics that have inductance: capacitor-diode-resistor leads, circuit traces, all of it does; therefore everything that conducts switching currents may be radiating. The switching currents may also couple into everything else connected to the battery power supply - an undesirable effect that can be attenuated many orders of magnitude by prudent design but perhaps not sufficiently in this case.

Therefore I'd be disinclined to use a switcher unless the application measurements can be made relatively quickly while the switcher is actively disabled/blanked as berkeman suggested earlier:

Do you briefly shut off the DC-DC to make the measurement quickly, while capacitors hold the 3V3 and other rails to minimum droop? If so, do you even need a shield at all?

I've seen disable / blanking circuits like that used before; you can generally oversize your output capacitor to holdup for whatever time you require. I suspect it will cost you little to design in blanking, and if later during integration you find your magnetometer is not impacted by your converter running continuously, you can just forget about syncing up the blanking circuit.