Buck converter steps down voltage but does not step up current

crutschow

Joined Mar 14, 2008
27,020
The oscilloscope only outputs voltages and not currents
To measure current, place a 10mΩ resistor in series with whatever you want to measure the current and measure the voltage across that, or use a current-controlled-voltage-source in series with the line to be measured.

Alec_t

Joined Sep 17, 2013
12,009
∴Vin x Iin = Vout x Iout
I disagree (when the average values are being considered). Vout is the smoothed (average) output voltage, whereas Vin is the supply voltage which is chopped at the duty cycle. Hence the average input voltage is Vin*d and the average input power is Vin*d*Iin.

crutschow

Joined Mar 14, 2008
27,020
I disagree (when the average values are being considered). Vout is the smoothed (average) output voltage, whereas Vin is the supply voltage which is chopped at the duty cycle. Hence the average input voltage is Vin*d and the average input power is Vin*d*Iin.
Anyway you slice it, the average input current does not equal the average output current for either a step-up or a step-down converter.

Joined Nov 5, 2010
215
There is no such thing as "average input voltage". These are current-pump devices. The input voltage is the input voltage, as long as it can stand the current demands. The device draws a current at that voltage, chopped by the switch FET. The input voltage is not chopped or averaged or anything. It's always at the same level.

But there IS an average input current. The input current is very pulsatile, which leads me to the question, what does your ammeter read on the input? Is it the peak current? The average? And if average, what kind of algorithm does it use to average?

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Joined Nov 5, 2010
215
mugheesnawaz: One thing that really bothers me is your use of the battery as a load. If you have a battery of 12V and a resistance of 0.1 ohms at a point where the circuit is at a potential of 14.4V, you only have a voltage of 2.4V across that resistor. So you're going to get a maximum of 24A pulled through the load resistor. What was the problem of using a 10 ohm resistor in the first place? So you're set up for an output current of 1.34A and you can massage your circuit operation to get a lower input current. The battery as a load just confuses things, and the extremely high currents are going to confuse things.

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DickCappels

Joined Aug 21, 2008
7,303
Is Q1 really connected as a voltage follower? Imagine what that might do to the efficiency.

crutschow

Joined Mar 14, 2008
27,020
Is Q1 really connected as a voltage follower? Imagine what that might do to the efficiency.
It is connected as a source follower, but the gate driver is bootstrapped to the source so that the ON gate voltage is nearly equal to the source voltage plus the supply voltage, and thus is fully turned on.
The ON resistance is thus low and the efficiency good.

Alec_t

Joined Sep 17, 2013
12,009
Ignore my analysis. Senior moment (or two ). I was confusing the supply voltage with the input voltage to the inductor. An LTspice sim confirms that the input/output current ratio is roughly the same as the output/input voltage ratio.

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Joined Nov 5, 2010
215
Alec_t, try a higher value inductor and see what that does to the input vs output current. Like the original 2.8mH. It looks like you're seeing too much input current for the output, unless it averages to less. Capture a shorter timespan, like 2ms, so we can see the input current waveform better.

Alec_t

Joined Sep 17, 2013
12,009
Here you go ....

(Vout is 12.25V)

Joined Nov 5, 2010
215
Here you go ....
View attachment 91756
(Vout is 12.25V)
Sorry about the delay in response.

Nice. That's what I'd expect to see. The current draw of the input averages to about 0.7A, and the output is drawing 1.3A, about twice, reflecting the voltage cut in half.

By the way, Alec, in your analysis of input/output power, I think the input analysis is more about the current being chopped (this being a current-pumped device) than the voltage. That's why I've averaged the input current. The power calculation comes out the same, though.

Alec_t

Joined Sep 17, 2013
12,009
By the way, Alec, in your analysis of input/output power, I think the input analysis is more about the current being chopped (this being a current-pumped device) than the voltage.
Agreed. See my retraction in post #28.

mugheesnawaz

Joined Dec 10, 2014
10
@InspectorGadget if i use a 10 ohm resistor in proteus with output voltage of 12.5 volts then i get a fixed current of 1.25Amps no matter what.
All i wanted was to see the input waveform which doesnt seem possible in proteus. so thanks to alec, i switched to LTspice and i am verifying the results now

Joined Nov 5, 2010
215
@InspectorGadget if i use a 10 ohm resistor in proteus with output voltage of 12.5 volts then i get a fixed current of 1.25Amps no matter what.
All i wanted was to see the input waveform which doesnt seem possible in proteus. so thanks to alec, i switched to LTspice and i am verifying the results now
I was talking about the pulsing on the input. At the output, the current pumped through the system is integrated on the capacitor to produce a DC voltage with some ripple. That's probably why Proteus gave you a constant output current. With a significant load, like the 10 ohm resistor, the voltage and current will be DC with some ripple, both voltage ripple and current ripple.

mugheesnawaz

Joined Dec 10, 2014
10
I get it,proteus isnt best suited for power circuits where current waveforms are to be monitored.
I have simulated buck converter in LTSpice and is working perfectly.now comes the major problem which is driving the mosfet.i have used ir 2110 and am using its high side only.i have attached both LTSpice files.the results should be similar i.e bucking 24V to 12V with 50% duty cycle but it isnt the case with ir 2110.kindly guide me where i am wrong

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kubeek

Joined Sep 20, 2005
5,751
I get it,proteus isnt best suited for power circuits where current waveforms are to be monitored.
Well maybe it is ok and you just don´t know how to use it I would be suprised if it couldn´t do such simple thing as showing current over time.

mugheesnawaz

Joined Dec 10, 2014
10
yes it might be the case but i am sure i am missing something in design.the buck converter is working ok but when i drive the mosfet using ir2110,the results arent the same.

Alec_t

Joined Sep 17, 2013
12,009
The Vs pin of the IR2110 would need to be switched to ground periodically in order for the bootstrap arrangement to work. In conventional circuits using both high and low FETs, this is done automatically by the low FET.

analogdude

Joined Jul 14, 2015
14
yes it might be the case but i am sure i am missing something in design.the buck converter is working ok but when i drive the mosfet using ir2110,the results arent the same.
The charge pump should work even if Q2 were a diode. I am not sure why the circuit does not work. Can you provide more measured points?

Although it's common to hear that buck converters 'boost current', that's not really the case; what current goes into the inductor, comes out. A series circuit that is commutated by the mosfets.
The input and output capacitors average the inductor current over each part of the (commutated) cycle.
Inductor current can be more or less the the same as the output current. It's OK to allow LTSpice to calculate the average, but without input capacitors, the input current will be pulses equal to the inductor current.