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I'm wondering the possibility of a circuit where there is a varying input voltage to a DC buck converter, that would output a constant current with varying voltages?

Let me explain via this diagram:

A simple buck convert circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The output voltage \$\ V_o\$ is now less than \$\ V_p\$(or \$\ V_i\$), however, the output current is now higher than the input current as that is the functionality of the buck converter. What if I introduced \$\ V_1\$ in series with\$\ V_p\$ after some time \$\ t\$, is there a way to output the same current with the following: \$\ V_o + V_1\$? With \$\ V_1\$ as one of the input voltages(meaning can't have it in series with the output).

schematic

simulate this circuit

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    \$\begingroup\$ IIUC, you need buck converter with constant current output. That is easy, just take feedback voltage from a current sense resistor between load and ground. \$\endgroup\$ – venny Sep 16 '15 at 16:05
  • \$\begingroup\$ @venny Alright, but can the output voltage be \$\ V_o + V_1\$? Can \$\ V_1\$ specifically not change it's voltage while inputted to the converter? \$\endgroup\$ – Pupil Sep 16 '15 at 18:25
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Basically what you're asking is, do current controlled switching convertes exist. The answer is yes, as you can see here:

enter image description here

Yes, that looks complicated ! But it's not (for me at least). The MAX649 is a DCDC Buck converter controller IC, it does not have a build in switch, PMOS Q1 is the switch. If you would want a constant output VOLTAGE, you would need a voltage divider (2 resistors) and feed a divided value of the output voltage back to the FB (feedback) pin.

But in this circuit, has a constant output CURRENT, to charge a battery ! Instead of sensing the output voltage, now the output current is sensed by taking the voltage across R10, amplifying this voltage (the opamp does this, with a feedback network to set the gain) and where does ths feedback signal go ? Straight back in to the FB pin !

The simple buck converter in your schematic is too simple to be considered having a voltage or a current output. If you change the load then voltage and current change, it is not regulated in any way. Yes you can change the PWM of the switch signal but that PWM signal needs te be determined somehow. This is done with voltage or current feedback.

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  • \$\begingroup\$ The complexity(for me) really lies with the fact that V1 is varying, at times it's (V1) in series with (PS) at times it's 0. And I guess I can study & use voltage/current feedback to maintain a constant current output, with varying voltage input and output. Making the DC converter act similar to a constant current source! Thanks. \$\endgroup\$ – Pupil Sep 17 '15 at 9:58
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    \$\begingroup\$ It's not only the variations of V1 that make the output voltage and/or current vary. Also the value of the load, the Duty cycle of the PWM signal. Also, depending on the situation a small change can cause a large change in output. All in all it is unpredictable. Luckily feedback was invented to fix this :-) \$\endgroup\$ – Bimpelrekkie Sep 17 '15 at 10:02
  • \$\begingroup\$ Thank you, now I know with feedback, I can achieve my goals haha. \$\endgroup\$ – Pupil Sep 17 '15 at 17:04
  • \$\begingroup\$ FakeMoustache, I think my issue(confusing) coming back to this is the possiblity to have a varying input**(For both voltage and current) and a **constant output current and varying voltage? Is that kind/type of converter possible to design? \$\endgroup\$ – Pupil Sep 29 '15 at 11:32
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    \$\begingroup\$ Yes, the design from my answer does this exactly :-) Actually the input current is determined by the DCDC converter and the load, how much power is needed basically. As long as the input voltage is high enough for the DCDC to be able to make the output voltage, meaning the input voltage should be a few Volts higher than the voltage needed at the output. The voltage at the output depends on the load, the DCDC feeds the load a current and will do so as long as it can. If the load is 1 Mega ohm it obviously cannot feed it 1A as that would require 1 Mega Volt ! So within limits: YES. \$\endgroup\$ – Bimpelrekkie Sep 29 '15 at 11:46

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