Muhammad Sukry: Tugas 1. Sub Chapter 1.11 Reverse Recovery Time + Soal-soal

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OSCILOSCOPE

BAHAN

Battery

Dioda

SWITCH

RESISTOR

3.Dasar Teori [back]

Ada beberapa bagian data yang biasanya disediakan pada spesifikasi dioda

lembaran yang disediakan oleh produsen. Salah satu kuantitas yang belum dipertimbangkan adalah waktu pemulihan terbalik, dilambangkan dengan trr. Dalam keadaan bias maju, hal itu ditunjukkan sebelumnya bahwa ada sejumlah besar elektron dari bahan tipe-n yang bergerak melalui bahan tipe-p dan a. sejumlah besar lubang di tipe-n — a persyaratan untuk konduksi.

Elektron di tipe-p dan lubang yang berkembang melalui itu bahan tipe-n membangun a sejumlah besar pembawa minoritas di setiap material. Jika yang diterapkan tegangan harus dibalik untuk membentuk a bias balik situasi,kita akan

idealnya ingin melihat dioda berubah secara instan dari konduksi ke

keadaan nonkonduksi. Namun, karena banyaknya jumlah pembawa minoritas masuk. Setiap bahan, arus dioda hanya akan berbalik seperti yang ditunjukkan pada Gambar. 1.39 dan tetap di tingkat yang dapat diukur ini untuk periode waktu ts (waktu penyimpanan) yang diperlukan bagi pembawa minoritas untuk kembali ke status pembawa mayoritas mereka dalam materi yang berlawanan. Di

esensi,dioda akan tetap berada dalam keadaan hubung singkat dengan arus Ireverse ditetapkan oleh parameter jaringan. Akhirnya, ketika fase penyimpanan ini telah berlalu, arus tersebut akan mengurangi tingkat yang terkait dengan keadaan nonkonduksi. Periode waktu kedua ini dilambangkan dengan tt(interval transisi). Waktu pemulihan terbalik adalah jumlahnya

dari dua interval ini: trr = ts + tt. Secara alami, ini merupakan pertimbangan penting dalam aplikasi switching berkecepatan tinggi. Paling komersial tersedia dioda switching memiliki Sebuah trr dalam kisaran beberapa nanodetik hingga 1 s. Unit tersedia, bagaimanapun, dengan trr hanya beberapa ratus pikodetik (10^-12).

A.EXAMPLE [back]

1.I've been reviewing some past exam and stumbled across this question

So I need to find the current I through the middle diode and the voltage across the bottom left 10KΩ resistor.

Solution

I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode $D_{1}$ and the diode in the middle $D_{2}$ .

Case 1: $D_{1}$ off, $D_{2}$ off: Since $D_{1}$ is off there is no current through the top 5k resistor, and since $D_{2}$ is off, there is also no current through the bottom left 10k resistor. So $V = 0$ and the voltage at the anode of $D_{1}$ is 15 Volts. Contradiction! ( $D_{1}$ should be on).

Case 2: $D_{1}$ off, $D_{2}$ on: again no current through top 5k resistor. Voltage $V$ is

V = \frac{15 V \cdot (5 k | | 10 k)}{10 k + (5 k | | 10 k)} = 3.75 V

Contradiction! (Because the voltage across $D_{1}$ would be $15 V - 3.75 V = 11.25 V$ and it should be on.)

Case 3: $D_{1}$ on, $D_{2}$ off: Voltage $V$ is

V = \frac{15 V \cdot 10 k}{5 k + 10 k} = 10 V

The voltage at the anode of $D_{2}$ is $15 V \cdot 5 k / 15 k = 5 V$ . This agrees with our assumption, because with these voltages $D_{2}$ must be off. So your solution is

I=0A,V=10VI=0A,V=10V

2.What is the reverse recovery time in a diode?

Solution

If a diode is conducting in a forward condition and immediately switched to a reverse condition, the diode will conduct in a reverse condition for a short time as the forward voltage bleeds off. The current through the diode will be fairly large in a reverse direction during this small recovery time.

After the carriers have been flushed and the diode is acting as a normal blocking device in the reversed condition, the current flow should drop to leakage levels.

The charge flowing during reverse recovery time is called "reverse recovery charge" and the diode has to extinguish it ("recovery" from reverse-biased to neutral condition) before you can turn it on. In the end, reverse recovery phenomenon depends on silicon doping and geometry and is a parasitic effect in diodes, because energy involved in the process is lost. strong text

B.PROBLEM [back]

I = 0 A, V = 10 V

1. An a.c. voltage of peak value 20 V is connected in series with a silicon diode and
load resistance of 500 Ω. If the forward resistance of diode is 10 Ω, find :
(i) peak current through diode (ii) peak output voltage
What will be these values if the diode is assumed to be ideal ?

Solution :

Peak input voltage = 20 V
Forward resistance, rf = 10 Ω
Load resistance, RL= 500 Ω
Potential barrier voltage, V0 = 0.7 V
The diode will conduct during the positive half-cycles of a.c. input voltage only.
The equivalent circuit is shown in Fig.1(ii)
Fig. 1
(i) The peak current through the diode will occur at the instant when the input voltage reaches positive peak i.e. Vin = VF = 20 V.
(ii) Peak output voltage :Ideal Diode Case:

2. Determine the state of diode for the circuit shown in Fig. 10 (i) and find ID and VD . Assume simplified model for the diode.

Fig. 10

Solution :

Let us assume that the diode is ON. Therefore, we can replace the diode with a 0.7V battery as shown in Fig. 10 (ii). Referring to Fig.10 (ii), we have,
Since the diode current is negative, the diode must be OFF and the true value of diode current is ID =0 mA. Hence our initial assumption was wrong.
In order to analyse the circuit properly, we should replace the diode in Fig. 10 (i) with an open circuit as shown in Fig.10(iii).

Fig.10 (iii)
The voltage VD across the diode is :
We know that 0.7V is required to turn ON the diode. Since VD is only 0.4V, the answer confirms that the diode is OFF.

C.Multiple Choice [back]

1.Apa syarat suatu diode dikatakan "off"?

A.ketika ID = 0

B.Ketika VD > 0

C.Ketika VD <0

D.Ketika ID >0

E.Ketika Iforward > Ireverse

2. Perhatikan Gambar berikut

Bagaimana Kondisi Diode yang benar??

A. Diode 1 OFF Diode 2 OFF

B. Diode 1 OFF Diode 2 ON

C. Diode 1 ON Diode 2 OFF

D. Diode 1 ON Diode 2 ON

Solution

4.Langkah Percobaan[back]

Step 1:SUSUN dan SIAPKAN KOMPONEN

Step 2:RANGKAI KOMPONEN

Step 3: BUAT SIMULASI PADA PROTEUS

Step 4: MENCOBA RANGKAIAN

Step 5: MENERAPKAN RANGKAIAN

5.Gambar Rangkaian[back]

Ketika belum disimulasikan

Ketika rangkaian disimulasikan

Ketika Switch dipindahkan dari "ON" ke "OFF"

Ketika Switch dipindahkan dari "OFF" ke "ON"

6.Prinsip Rangkaian[back]

Reverse Recovery Time dipengaruhi oleh besarnya tegangan sumber. Semakin besar tegangannya maka Nilai dari trr nya juga semakin besar. atau ibaratnya seperti lampu dalam kehidupan kita sehari-hari.ada beberapa kasus dimana lampu yang awalnya hidup, ketika dimatikan ia tidak langsung padam, tapi ia redup sebentar baru padam.begitu pula dengan dioda,dalam dioda ada yang namanya arus forward(Iforward) dan ada yang namanya arus reverse(Ireverse). Iforward (yaitu ketika arus yang mengalir dari kutub positif sumber menuju anode dioda,maka dioda akan "on") dan Ireverse (yaitu ketika arus yang mengalir dari kutub positif sumber menuju katode dioda,maa dioda akan "off").

Reverse Recovery Time terjadi ketika Dioda mengalami If dan Ir secara sekaligus.awalnya dioda ON ketika ia menerima Iforward tapi ketika dioda di OFF kan ada sedikit jeda untuk dioda dapat OFF jeda ini merupakan gabungan dari ts(storage time =waktu yang dibutuhkan oleh sisa arus untuk kembali ke bentuk kebanyakan arus) dan tt(transition Interval = waktu yang dibutuhkan diode untuk kembali ke kondisi nonconduction) sehingga trr = ts + tt.

Muhammad Sukry

Menu

Tugas 1. Sub Chapter 1.11 Reverse Recovery Time + Soal-soal

1. Tujuan [back]

2. Komponen [back]

3.Dasar Teori [back]

1. An a.c. voltage of peak value 20 V is connected in series with a silicon diode and
load resistance of 500 Ω. If the forward resistance of diode is 10 Ω, find :
(i) peak current through diode (ii) peak output voltage
What will be these values if the diode is assumed to be ideal ?

Solution :

2. Determine the state of diode for the circuit shown in Fig. 10 (i) and find ID and VD . Assume simplified model for the diode.

Fig. 10

Solution :

4.Langkah Percobaan[back]

Step 1:SUSUN dan SIAPKAN KOMPONEN

Step 2:RANGKAI KOMPONEN

Step 3: BUAT SIMULASI PADA PROTEUS

Step 4: MENCOBA RANGKAIAN

Step 5: MENERAPKAN RANGKAIAN

5.Gambar Rangkaian[back]

7.Video Simulasi [back]

8. Link Download [back]

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Posting Komentar

Muhammad Sukry

Menu

Tugas 1. Sub Chapter 1.11 Reverse Recovery Time + Soal-soal

1. Tujuan[back]

2. Komponen [back]

3.Dasar Teori [back]

1. An a.c. voltage of peak value 20 V is connected in series with a silicon diode andload resistance of 500 Ω. If the forward resistance of diode is 10 Ω, find :(i) peak current through diode (ii) peak output voltageWhat will be these values if the diode is assumed to be ideal ?

Solution :

2. Determine the state of diode for the circuit shown in Fig. 10 (i) and find ID and VD . Assume simplified model for the diode.

Fig. 10

Solution :

4.Langkah Percobaan[back]

Step 1:SUSUN dan SIAPKAN KOMPONEN

Step 2:RANGKAI KOMPONEN

Step 3: BUAT SIMULASI PADA PROTEUS

Step 4: MENCOBA RANGKAIAN

Step 5: MENERAPKAN RANGKAIAN

5.Gambar Rangkaian[back]

7.Video Simulasi [back]

8. Link Download [back]

Tidak ada komentar:

Posting Komentar

Muhammad Sukry

1. Tujuan [back]

1. An a.c. voltage of peak value 20 V is connected in series with a silicon diode and
load resistance of 500 Ω. If the forward resistance of diode is 10 Ω, find :
(i) peak current through diode (ii) peak output voltage
What will be these values if the diode is assumed to be ideal ?