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I have designed a current buck converter. The idea is that you the (voltage) buck converter and you take advantage of some symmetries of electronic elements, the capacitor is to a current source what the inductor is to a voltage source, and for all elements in series with a voltage source you need to put them in parallel to a current source. So I came up with a current buck converter which is a buck converter for current sources.

$ 1 3.125e-7 382.76258214399064 50 5 43 5e-11
i 192 208 192 112 0 0.01
r 336 112 448 112 0 10
l 448 112 448 208 0 0.001 0.002500045386706189 0
w 336 208 448 208 0
t 224 144 256 144 0 1 0.6323494705717794 0.670740192568514 100 default
w 224 144 224 272 0
R 128 272 64 272 0 2 100000 5 0 0 0.75
w 128 272 176 272 0
c 336 112 336 208 4 0.001 0.025003895623268373 0.001 0
d 272 112 336 112 2 default
w 272 112 256 112 0
w 256 112 256 128 0
w 256 160 256 208 0
w 256 208 336 208 0
w 192 208 256 208 0
w 192 112 256 112 0
g 256 208 256 240 0 0
r 176 272 224 272 0 1000
o 6 64 0 4099 5 0.0125 0 2 6 3

Project must be be opened with the Falstad simulator:https://falstad.com/circuit/

Here is a picture:

In the above circuit Io/Ii = 1-D where Io is the average current of the resistor Ii is the input current and D is the duty cycle of the NPN transistor.

I have been trying to do the same for a current boost converter. Here is my design:

and the code:

$ 1 3.125e-7 382.76258214399064 44 5 43 5e-11
i 192 208 192 112 0 0.01
r 336 112 448 112 0 0.01
l 448 112 448 208 0 10 -8.569084940961415e-8 0
w 336 208 448 208 0
w 224 224 224 272 0
R 128 272 64 272 0 2 100000 5 0 0 0.5
w 128 272 176 272 0
w 256 160 256 208 0
w 256 208 336 208 0
w 192 208 256 208 0
g 256 208 256 240 0 0
r 176 272 224 272 0 1000
t 256 80 256 112 1 1 -5.070559882847056 -5.035818518039943 100 default
w 224 224 112 224 0
w 112 224 112 32 0
w 112 32 256 32 0
w 256 32 256 80 0
c 224 112 224 160 4 0.01 0.07055988806897502 0.001 0
w 192 112 224 112 0
w 224 112 240 112 0
w 224 160 256 160 0
d 336 208 336 112 2 default
w 272 112 336 112 0
o 5 64 0 4099 5 0.0125 0 2 5 3
o 1 64 0 4099 0.0048828125 0.000390625 1 2 1 3

The relationship Io/Ii is supposed to be 1/D but every time I run it the output current oscillates between 15mA and 40ish mA and I don't understand why this is. The inductor should filter the AC part of the current so I really don't understand why this is happening. What is wrong with my design?

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  • \$\begingroup\$ Welcome! Where is your output/load? With an ideal current source input, the simulation will ramp up your input voltage to infinity if needed. \$\endgroup\$ Commented May 19 at 16:03
  • \$\begingroup\$ Hi.The output current is the current of the resistor(which in series with the inductor). \$\endgroup\$ Commented May 19 at 16:05
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    \$\begingroup\$ Ok. What's your goal? \$\endgroup\$ Commented May 19 at 16:06
  • \$\begingroup\$ And no it doesnt bcs it works with the same principle as the (voltage) dc-dc converters,due to the switching nature of the capacitor and the inductor. \$\endgroup\$ Commented May 19 at 16:06
  • \$\begingroup\$ I have explained it,I try to create a current boost converter(a circuit which boosts the dc current of a dc current source) but I get oscillations I dont understand why. \$\endgroup\$ Commented May 19 at 16:08

2 Answers 2

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Here is a working model of a "current boost" converter using CircuitLab. You can play with the model using the link below the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

When the model is set to 50% duty cycle, and the output resistance is sufficiently large, the output current is double the input current.

50% duty 1 ohm load

Increasing the load resistance from 1 ohm to 10 ohms does not affect the steady state output current.

schematic

simulate this circuit

50 % 10 ohms

However, when the output resistance is sufficiently small, the output current no longer depends only on the input current and the duty ratio, but also upon the output resistance. This is analogous to how a "voltage boost" circuit works in Discontinuous Conduction Mode (DCM) -- i.e. the output voltage no longer depending solely upon the input voltage and duty ratio.

schematic

simulate this circuit

50% and 10 mOhms

Diode D2 was added to the model to prevent C1 from having reversed polarity. That would not happen in the case where D1 is an ideal diode, but could occur if D1 were a real diode.

schematic

simulate this circuit

No D2

It may not be necessary in practice, but in creating my simulation, I did not want reverse polarity of C1 to be a confounding factor.

In the design of the "current boost" converter presented in the question, it may be the case that the transistor base is not properly biased with respect to the transistor emitter. The voltage at the emitter of the transistor varies quite a bit. It would make sense to have the biasing circuit referenced to the emitter of the transistor. I haven't (yet) played with the Falstad model, so I cannot say definitely that that is the problem in that particular simulation, but I suspect it would be a problem in general.

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  • \$\begingroup\$ Hi.Is falstad just bad then? \$\endgroup\$ Commented May 19 at 19:14
  • \$\begingroup\$ Why is the need for the extra diode(D2)? \$\endgroup\$ Commented May 19 at 19:17
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    \$\begingroup\$ "is falstad just bad then?" Falstad definitely has its limitations, but I believe that your problem may have been due to the transistor not being properly biased, and/or the load resistance being too low. \$\endgroup\$ Commented May 19 at 19:46
  • \$\begingroup\$ I have added a resistor in series with the current source to drive the transistor in saturation when the pulse of the base of the transistor is on. \$\endgroup\$ Commented May 19 at 19:47
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    \$\begingroup\$ @RootGroves Yes, the node connecting the transistor's emitter, the diode and the inductor has a fluctuating voltage. Proper biasing of the transistor is not trivial. One change you might consider is to use a PNP transistor (swap the emitter and collector) or a PMOS FET. However, you will still have a problem, because the voltage on the current source side of the transistor varies as well. But it will vary less quickly. If you want to explore further design, think about it for a little bit, then ask another question. This site frowns upon too much back and forth on the comments. \$\endgroup\$ Commented May 19 at 20:08
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Here you go, 20 mA out in your 10 mΩ resistor from your 10 mA current source. Feel free to substitute the ideal switch with your transistor.

enter image description here

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  • \$\begingroup\$ but why does it work well with the buck converter and not with the boost converter.The current buck converter works just fine in multisim. \$\endgroup\$ Commented May 19 at 16:44
  • \$\begingroup\$ You can either boost current and buck voltage, or vice versa. Conservation of energy law. \$\endgroup\$ Commented May 19 at 17:00
  • \$\begingroup\$ no it is not the same at all for a reason.1)In a buck voltage converter output current is arbitary(depends on the load) while in the boost current converter output current depends on the source ,which is a lot more practical \$\endgroup\$ Commented May 19 at 17:03
  • \$\begingroup\$ @RootGroves Conservation of energy always applies to all known systems. Please read up on it: wikipedia.org/wiki/Conservation_of_energy \$\endgroup\$ Commented May 19 at 18:43
  • \$\begingroup\$ Yes I agree and my point is that in a buck converter the current depends on the load itself while in a current boost converter current depends on the source. \$\endgroup\$ Commented May 19 at 18:44

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