I'm no expert, but how about rigging the first pack with a motor to drop food pellets to encourage a chicken to run on a treadmill connected to a generator hooked up to the second pack. You could, alternately, use the first pack to shock the chicken into running, but that may raise ethical questions.

 

current controlled buck converter.

 

20khz Mosfet PWM chopper or similar

Yes it's basically the way to transfer without the heat loss of a linear circuit.
Here is a diagram found on the net which gives an overall idea of what you need.
The value of the coil is important, but I don't have the knowledge to calculate it.

For feedback I suggest a hall effect sensor which, unlike a shunt resistor, will isolate the output you are measuring.

 

Yes it's basically the way to transfer without the heat loss of a linear circuit.
Here is a diagram found on the net which gives an overall idea of what you need.
The value of the coil is important, but I don't have the knowledge to calculate it.
View attachment 97773
For feedback I suggest a hall effect sensor which, unlike a shunt resistor, will isolate the output you are measuring.

A spare iGBT module with heat sink and fan might be quite useful here. It would be nice to charge up a bunch of LTO packs at a "normal" rate, then dump their contents into the car as fast as heat dissipation will permit.

 

I have a 3kw Zivan NG3 EV charger kicking about. Hmm.

That will probably run with a 200-300V or so DC input and it has the right output voltage for the Lithium target pack I'm working with.

I could bypass the NG3 input noise filtering and AC bridge rectifier to feed my buddy pack DC directly into the NG3 filter caps. Might need a precharge resistor for the inrush current.

If I'm clever with a changeover setup I could use the same NG3 to charge the Buddy pack from the mains when back at base.

 

Nice little buck converter calculator tool.

https://www.ti.com/tool/BUCK-CONVCALC

I have loads of inductors and big filter caps laying around.
Need a schottky diode and a power device.

Just throwing some values into it gives this.
Does it look sensible?

The Transfer Function of a Buck Converter:
VOUT =	168​	V
180​	V
IOUT =	50​	A
Output power of the converter:	POUT =	8400​	W
IOUTMIN =	5​	A, assumed to be 10% of IOUT
fSW =	25​	kHz
1.68​	V, assumed to be 1% of VOUT
0.25​	V
RDSon =	0.1​	W
Voltage drop across RDSon:	VRDSon =	5​	V
Conduction losses of switch:	PCOND =	241.158​	W
Duty Cycle:	D =	0.961
Switching Period:	T =	40​	ms
On-time of the switch:	tON =	38.457​	ms
The minimum inductor value is calculated assuming the minimum output current is equal to 10% of the nominal current. The inductor is sized such that the converter will remain in the continuous current mode through this range.
Minimum inductor value:	L =	26.92​	mH
Inductor stored energy:	E =	40717​	mJ
The drain current waveform is a ramp on a step. The value of the current at the center of the ramp is equal to the output DC current. The peak inductor current is equal to the output current added to half the peak to peak ripple.
Peak-to-peak ripple current:	IppRIPPLE =	10​	A
Peak switch current:	IPEAK =	55​	A
RMS current:	IRMS =	49.108​	A
A Schottky rectifier is chosen because of its low forward voltage, VF, and its excellent reverse recovery characteristics. Replacing this diode with a FET and using synchronous rectification will give even more efficiency benefits. This rectifier must meet the following criteria:
DC blocking voltage:	VR =	180​	V
Average rectified output current:	IAVE =	1.929​	A
The switch must be selected to meet the above current requirements. The major Drain to Source voltage stress occurs at switch turn-off when the Source could possibly ring up to 5V below ground.
Minimum rated Drain to Source voltage:	VDS =	185.25​	V
The output capacitor is chosen such that it provides significant filtering of the switching ripple. The selected capacitor must be large enough so that its impedance is much smaller than the load at the switching frequency, allowing most of the ripple current to flow through the capacitor, not the load. The ripple current flowing through the output capacitor is equal to the inductor current waveform with the dc component removed. The output capacitor's ESR must also be taken into account because this parasitic resistance, which is out of phase with its capacitance, will cause additional voltage ripple. Be sure to select capacitors based upon their maximum ripple current and ESR ratings at the temperature and frequency of the application.
Output capacitor RMS ripple current:	IRMScap =	2.887​	A
Minimum output capacitance:	COUT =	29.76​	mF
Chances are, a bank of capacitors will be required to handle the output ripple current. This capacitance will have an ESR associated with it:
Total capacitance of output bank used:	COUTbank =	10000​	mF
Maximum ESR required:	ESRMAX =	0.168​	W
Actual ESR of output capacitor bank used:	ESR =	0.015​	W
Peak-to-peak voltage ripple due to output capacitance:	VPPcap =	0.005​	V
Peak-to-peak voltage ripple due to output ESR:	VPPESR =	0.150​	V
Resultant total peak-to-peak output voltage ripple:	VPPtotal =	0.150​	V
The same logic is applied when selecting the input capacitor. This capacitor, or bank of capacitors, will experience very high ripple current; the same current that is at the switch drain. An acceptable level of input voltage ripple which would still maintain regulation is assumed to be 5%.
Input capacitor RMS ripple current:	IRMS =	49.108​	A
Acceptable input voltage ripple:	VrippleIN =	9.000​	V, assumed to be 5% of VIN
Minimum input capacitance:	CIN =	30.556​	mF
Total capacitance of input bank used:	CINbank =	10000​	mF
Maximum ESR required:	ESRMAX =	0.1636​	W
Actual ESR of input capacitor bank used:	ESR =	0.1​	W
Peak-to-peak voltage ripple due to input capacitance:	VPPcap =	0.028​	V
Peak-to-peak voltage ripple due to input ESR:	VPPESR =	5.500​	V
Resultant total peak-to-peak input voltage ripple:	VPPtotal =	5.500​	V

 

Hiho Pete,

somehow your topics always draw my attention hehe,

i 'designed' a 1kW buck that used a speaker-coil 'Visaton 1,5mH' as in the attached photo...
There are versions with several different inductances, mine is L=1,5mH and R=0,17ohm, the smallest value that can handle about 9Amps tops before it starts to melt.

It works in my buck-converter for a couple of years now and converts a 250VDC source to the IMA pack DC voltage, with amperage limit 9A.

Set 3 of those parallel to get a 25Amps buck?

 

Hiho Pete,
somehow your topics always draw my attention hehe,

Can you post a pic of your converter and schematic please?

 

I don't have any experience with magnetics. I suppose it is time I taught myself. I could use some good resources for this.

 

behold of... --the beast--

 

warning 1: my 'design' does not imply fitness for a particular purpose
warning 2: high voltage dc could kill

i also want to attach the LTspice Schematic '.asc' file but website says 'not an allowed extension'

 

Just rename you asc extension as a pdf to upload it.

 

ok, it became a zip file, so rename the pdf to zip

I use a free tool, LTspice IV from Analog Devices, to work with these schematics.

 

@Sondair very interesting. What resource did you use to determine the magnetic requirements for the inductor?

 

To basic parameters are critical here: Induction and resistance value.

Induction, L, in Henry

Basically i simulated a buck converter circuit with the LTspice schematic tool which is also a simulator for electronic circuits, in order to find out a coarse min and max for the inductor value. buck in picture runs at 20kHz switching frequency. Apart from that, it was trial and error. I remember playing with low-voltage side of mains transformers for a short time, but a trafo has an iron core that is lossy and heats itself up from within fast and melts soon. So, I needed a core material with low loss: ferrite. I ended up using a speaker coil normally used for filtering audio current to a speaker I think. Could easily be, there's a better solution for this out there.

Resistance, R, in ohm

due to resistance a coil-component dissipates energy that heats itself up from within when a current flows through it. In a power converter current is high, so resistance must be as low as possible to keep the coil's core temperature below a certain limit: say 100C. Too much current and coil will melt itself from within, blowing converter up. Coil must be inside force ventilated casing and ucontroller must monitor temp inside casing and reduce current when temp goes too high. In my prototype 1kW buck converter in the picture, coil is glued to casing, over time i figured out coil could dissipate a max of 12Watts (i started lower and increased it slowly over time). With P = I*V, V = I*R and P = I^2*R we can calculate current that gives 12Watts dissipation: Imax = sqrt(P/R) Imax = sqrt(12/0,17) = 8,3 Amps. Up till now i didn't blow up a single coil (-yet-) (Before coil blows, fets blow first, started with one, but needs 3 in parallel for 1kW throughput, still coil is not at it max here and can handle more, could be, it can handle 1500 or even 2000W)

Eventually turned out i had to use a ferrite coil with windings that have the lowest resistance, in order words, that have the lowest number of windings, or lowest induction value, L, that can still provide sufficient induction. This boils, in turn, down to the minimum switching frequency you can use. The lower L gets, the higher your switching frequency must get. Losses are lower in the coil, but higher in the mosfets. A balance must be found there. I'm using 20kHz in the buck in the picture, what already works for a couple or years now.

 

For those following this, I replaced the regular silicon diode 'FESF16JT' in the buck circuit with a Silicon Carbide Schottky version 'WNSC6D20650CW6Q'.

The new diode is a recent development in high voltage Schottky diodes, that have virtually zero reverse recovery time. With that low reverse recovery the buck circuit emits much less EM noise that interfered with other equipment like the FM radio.

I run the buck on 1.5kW now without interference on the FM radio.