# Ramp current in inductor

When DC voltage is applied across an inductor, would the current through it be ramp? According to voltage current relationship in an inductor, the current should be ramp if voltage across inductor is constant.

1. Is this practically possible only when there is no resistance connected in series with inductor? 1. Why would the current through inductor be ramp in this circuit of dc-dc buck converter?

Edit

The large value of capacitance reduces the change in output voltage and so the voltage across inductor remains constant. This would mean that load current and hence output voltage is not really constant. Why is this acceptable?

What would be the nature of voltage across the capacitor(output voltage?

When DC voltage is applied across an inductor, would the current through it be ramp?

Yes

Is this practically possible only when there is no resistance connected in series with inductor?

Yes, or low enough resistance to not affect the result much. Remember that all real inductors have parasitic series resistance, so being able to neglect a small resistance matters.

Why would the current through inductor be ramp in this circuit of dc-dc buck converter?

The capacitor C should be large enough to prevent $V_o$ changing very much.

Of course the output voltage is not perfectly constant. There is always some voltage ripple in this circuit. So the inductor current is not a perfect ramp. But if it's designed reasonably, it will be close enough to a ramp to do useful analysis with that approximation.

• Thanks. In the buck converter circuit above, there is a resistor connected to inductor. Why would the current through inductor be ramp in this case? – Aditya Patil Oct 6 '15 at 16:30
• The resistor is not in series with the inductor. In series means that all the current through one part also goes through the other. In this case, the capacitor can supply current to the load when the inductor current is below the load current, and the capacitor gets recharged when the inductor current is greater than the load current. – The Photon Oct 6 '15 at 16:32
• :) understood. Could you please answer the edited question – Aditya Patil Oct 6 '15 at 16:41
• @AdityaPatil, See my edit – The Photon Oct 6 '15 at 16:49
• Even if the output voltage is assumed to be constant, the voltage across inductor happens to be Vpulse - Vout. This is not a constant. Then why would the current be ramp? – Aditya Patil Oct 9 '15 at 3:33

Yes, from the relationship $V = L \frac{di}{dt}$ we can see that for a constant voltage, we get a constant rate of change of current - i.e. a linear current ramp. If we include a series resistor, we get a decaying exponential curve caused by the resistor dropping increasing amounts of voltage as the current increases - $i(t) = \frac{V}{R}( 1-e^{-\frac{Rt}{L}})$. For small resistance values this doesn't deviate much from an ideal ramp - the deviation increases with increasing R (and over longer time periods).

For the buck converter, C resists the change in output voltage $V_O$ such that the voltage across the inductor, $V_L = V_O - V_I$ remains relatively constant, and hence you get a simple current ramp. Varying current flows in the inductor and output capacitor, with the load seeing (ideally) fixed output current defined by an (ideally) fixed output voltage across the capacitor. The Buck converter Wiki page shows many of these waveforms along with a more general overview of operation.

According to voltage current relationship in an inductor, the current should be ramp if voltage across inductor is constant. Is this practically possible only when there is no resistance connected in series with inductor?

1. Yes, increased inductor current would cause larger voltage drop across the resistance, which would drive the voltage across the inductance lower.

Why would the current through inductor be ramp in this circuit of dc-dc buck converter?

1. Typically, the analysis of switched mode power supplies is done in a piece-wise linear fashion. This means that the capacitor voltage is considered constant for the duration of a switching period, which would then cause the inductor current to look like a ramp. In reality, the switching frequencies are in hundred's of kHz (up to MHz with GaN devices) and the output voltage does not really change much during this period. Since inductors have quite small resistance (to minimize losses) and the inductor current typically changes ~20% max during one switching period, the voltage across the inductor can truly be considered constant.

The resistor in the picture symbolizes the dc/dc converter load and since it is in parallel with the capacitor, the voltage across the load (=resistor) is considered constant as well. The cycle-to-cycle ripple in the output voltage is ~ very small, typically <1%. This is the reason why the simplified first-order (inductor only) analysis is feasible.

The large value of capacitance reduces the change in output voltage and so the voltage across inductor remains constant. This would mean that load current and hence output voltage is not really constant. Why is this acceptable?

What would be the nature of voltage across the capacitor(output voltage?

See the simulation below. The capacitance is made (too) small to emphasize the ripple.  • Careful- the capacitor voltage is constant, which is also the output voltage. So the load current is constant as well. The inductor current, which resembles a triangle, is absorbed by the output capacitor and has very little effect on the capacitor voltage. – SunnyBoyNY Oct 6 '15 at 16:44
• Even if the output voltage is assumed to be constant, the voltage across inductor happens to be Vpulse - Vout. This is not a constant. Then why would the current be ramp? – – Aditya Patil Oct 9 '15 at 3:33
• The inductor current is piece-wise constant, as stated in the answer. The circuit is analyzed in each state individually. The current is a upward ramp when the buck switch is on and a downward ramp when the switch is off and the current freewheels through the diode. – SunnyBoyNY Oct 9 '15 at 3:44