---
title: "Step-down chopper RLE load, 5 kHz"
subject: "Power Electronics"
chapter: 4
key: "semester_7/power_electronics:step-down-chopper-rle-5khz"
source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-rle-5khz
site: BCT Engineering
---

# Step-down chopper RLE load, 5 kHz

Power Electronics · Chapter 4: DC chopper

- Asked in: 2079 Bhadra, 8 marks
- Video: [Step Down Chopper with R-L-E Load : Continues Current Conduction Mode](https://www.youtube.com/watch?v=tCHWBwHRYuY) — Learning Electronics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-rle-5khz
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4

## Question

A step-down chopper is connected to RLE load with R = 10 ohms, L = 15.5 mH and E = 20 V with an input voltage of 220 V, the duty cycle, k = 0.5 and chopping frequency f = 5 kHz. Determine, i) Minimum instantaneous load current ii) Peak instantaneous load current iii) Maximum peak to peak current in the load iv) Average load current

## Short answer

With T=200µs, τ=1.55ms, E/Vs=0.0909 < critical 0.4839, current continuous. Minimum load current I_min=8.645A, peak I_peak=9.355A. Maximum peak-to-peak ripple ΔI_max=0.709A. Average load current I_avg=(kVs-E)/R=(110-20)/10=9A.

## Answer

For a step-down chopper with RLE load in continuous conduction, the load current rises exponentially during $kT$ and decays during $(1-k)T$; the standard steady-state results (Rashid) are used.

**Given:** $V_s = 220$ V, $R = 10\ \Omega$, $L = 15.5$ mH, $E = 20$ V, $k = 0.5$, $f = 5$ kHz.

$$
\begin{aligned}
T &= \frac{1}{f} = 200\ \mu\text{s}, \quad kT = 100\ \mu\text{s} \\
\tau &= \frac{L}{R} = \frac{15.5\times10^{-3}}{10} = 1.55\ \text{ms} \\
z &= \frac{TR}{L} = \frac{200\times10^{-6}}{1.55\times10^{-3}} = 0.12903, \quad kz = 0.064516
\end{aligned}
$$

Check continuity: $E/V_s = 0.0909$; critical value $\dfrac{e^{kz}-1}{e^{z}-1} = \dfrac{0.066643}{0.137727} = 0.4839$. Since $0.0909 < 0.4839$, current is continuous.

#### i) Minimum instantaneous load current
$$
\begin{aligned}
I_1 &= \frac{V_s}{R}\cdot\frac{e^{kz}-1}{e^{z}-1} - \frac{E}{R}
= 22\times\frac{1.066643-1}{1.137727-1} - 2 \\
&= 22\times 0.48388 - 2 = 8.645\ \text{A}
\end{aligned}
$$

#### ii) Peak instantaneous load current
$$
\begin{aligned}
I_2 &= \frac{V_s}{R}\cdot\frac{1-e^{-kz}}{1-e^{-z}} - \frac{E}{R}
= 22\times\frac{1-0.937521}{1-0.878946} - 2 \\
&= 11.355 - 2 = 9.355\ \text{A}
\end{aligned}
$$

#### iii) Maximum peak-to-peak load current ripple
The ripple $\Delta I = I_2 - I_1$ is maximum at $k = 0.5$:

$$
\begin{aligned}
\Delta I_{max} &= \frac{V_s}{R}\tanh\frac{R}{4fL}
= 22\times\tanh\frac{10}{4\times5000\times0.0155} \\
&= 22\times\tanh(0.032258) = 0.709\ \text{A}
\end{aligned}
$$

Here $k = 0.5$ already, so the actual ripple $9.355 - 8.645 = 0.709$ A equals this maximum. (Approximation for $4fL \gg R$: $V_s/(4fL) = 0.710$ A.)

#### iv) Average load current
$$
I_a = \frac{kV_s - E}{R} = \frac{0.5\times220 - 20}{10} = \frac{110-20}{10} = 9\ \text{A}
$$

**Answer:** $I_{min} = 8.645$ A, $I_{peak} = 9.355$ A, $\Delta I_{max} = 0.709$ A, $I_{avg} = 9$ A.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With f = 5 kHz and k = 0.5, what are T and the ON time?
   A. T = 200 µs, ON time = 100 µs
   B. T = 5 ms, ON time = 2.5 ms
   C. T = 200 µs, ON time = 200 µs
   D. T = 100 µs, ON time = 50 µs
   Answer: A. T = 1/f = 200 µs and kT = 100 µs. (Key point: Chopping period)

2. What is the average load current?
   A. 11 A
   B. 9 A
   C. 10.9 A
   D. 8.645 A
   Answer: B. Ia = (kVs - E)/R = (110 - 20)/10 = 9 A. (Key point: Average load current)

3. How are minimum and peak currents related to the ripple?
   A. The ripple equals the average current of 9 A
   B. The ripple is the sum of the minimum and peak currents
   C. Peak 9.355 A minus minimum 8.645 A gives 0.709 A, the peak-to-peak ripple
   D. The ripple is zero because the current is continuous
   Answer: C. Delta I = I2 - I1 = 0.709 A; ripple is small because 4fL >> R. (Key point: Ripple meaning)

4. At what duty cycle is the peak-to-peak ripple maximum, and what is the formula?
   A. k = 1; Delta I = Vs/R
   B. k = 0.1; Delta I = E/R
   C. k = 0.5; Delta I = Vs/(R*E)
   D. k = 0.5; Delta I = (Vs/R) tanh(R/(4fL))
   Answer: D. Ripple peaks at k = 0.5, giving (Vs/R) tanh(R/4fL) = 0.709 A here. (Key point: Maximum ripple condition)

## Related in Power Electronics

- [Step-down chopper RLE load continuity](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-rle-200v) — Power Electronics, chapter 4
- [Step-down chopper: fundamental of load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-fundamental-load-current) — Power Electronics, chapter 4
- [Step-down chopper variable frequency, 3 ms ON time](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-variable-frequency-3ms) — Power Electronics, chapter 4
- [Step-down chopper variable frequency, 5 ms ON time](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-variable-frequency-5ms) — Power Electronics, chapter 4
- [Step-down chopper with switch drop: efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-efficiency-1v) — Power Electronics, chapter 4
- [Type-A chopper RLE load numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Atype-a-chopper-rle-numerical) — Power Electronics, chapter 4

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