**Constant Current Source**

A voltage source that has a very huge internal*impedance as compared with external load impedance or resistance is known as a constant current source. In that case, the load current approximately remains the same when the output voltage changes. Fig. 1.12

(i) illustrates a constant current source. It is a direct current source of 1000 V having internal resistance Ri = 900 kΩ. Here, load RL changes over 3: 1 range from 50 k Ω to 150 k Ω. Over these changes in load RL, the circuit current I is important constant at 1.05 to 0.95 mA or nearly 1 mA. It may be pointed that output voltage V varies nearly in the same 3: 1 range as RL, although load current essentially remains **constant at 1mA. The best example of a constant current source appears in vacuum tube circuits where the tube acts as a generator having internal resistance as large as 1 MΩ. Fig. 1.12 (ii) shows the graph of a constant current source. It is visible that the current remains constant even when the output voltage changes substantially. The following points may be noted regarding the constant current source :

(i) Due to the huge internal resistance of the source, the load current remains essentially constant as the load RL is changed.

(ii) The output voltage changes nearly in the same range as RL, although current remains constant.

(iii) The output voltage V is much smaller than the produced voltage Eg because of the large I Ri drop.

Fig. 1.13 shows the symbol of a constant current source.

Example 1.2. A d.c. source generating 500 V has an internal resistance of 1000 Ω. Find the load current if load resistance is (i) 10 Ω (ii) 50 Ω and (iii) 100 Ω.

Solution.

Generated voltage, Eg = 500 Internal resistance, Ri = 1000 Ω

(i) When RL = 10 Ω

It is clear from the above example that load current is essentially constant since Ri>>RL

.

**Conversion of Voltage Source into Current Source**

Fig. 1.14 shows a constant voltage source with voltage V and internal resistance Ri. Fig. 1.15 shows

its nearly equal current source. It can be easily shown that the two circuits act electrically the same method under all conditions.

**(i) **If in **Fig. 1.14,** the load is open-circuited (i.e. RL → ∞ ), then the voltage across terminal A and B is V. If in Fig. 1.15, the load is open-circuited (i.e. RL → ∞), then all current I (= V/Ri) flows through Ri, yielding voltage across terminals AB = I Ri= V. Note that open-circuited voltage across AB is V for both the two circuits and hence they are electrically equally

(ii) If in **Fig. 1.14,** the load is short-circuited (i.e. RL = 0), the short circuit current is defined by:

If in Fig. 1.15, the load is short-circuited or zero path (i.e. RL= 0), the current I (= V/Ri) bypasses Ri in favor of short-circuiting. It is clearly visible that current (= V/Ri ) is the same for the two circuits and hence both are electrically equivalent.

Thus to convert or transfer a constant voltage source into a constant current source, the following procedure may be applied:

- Put a short-circuit across both two terminals in question (terminals AB in the present case) and determine the short-circuit current. Let it be I. Then I is the current provided by the equivalent current source.
- determine the resistance at the terminals with load removed and sources of e.m.f.s replaced by their internal resistances. Let this resistance be R.
- Then equivalent current source can be shown by a single current source of magnitude I in parallel with resistance R.

Note. To convert a current source of magnitude I in parallel with resistance R into the voltage source,

The voltage of voltage source, V = I R

The resistance of voltage source, R = R

Thus voltage source will be represented as voltage V in series with resistance R.

Example 1.3. Convert the constant voltage source shown in Fig. 1.16 into constant current source.

Solution. The solution involves the following steps :

(i) Put a short across AB in Fig. 1.16 and determine the short-circuit current I. Clearly, I= 10/10 = 1 A

That’s why the equivalent current source has a magnitude of 1 A.

(ii) find the resistance at terminals AB with load *removed and 10 V source changed by its internal resistance. The 10 V source has nearly negligible resistance so that resistance at terminals AB is R = 10 Ω.

(iii) The equivalent current source is a source of 1 A in parallel with a resistance of 10 Ω as shown in Fig. 1.17.

Example 1.4. Convert the constant current source in Fig. 1.18 into equivalent voltage source.

Solution. The solution involves the following steps :

(i) To achieve the voltage of the voltage source, multiply the current of the current source by the internal resistance i.e. Voltage of voltage source = I R = 6 mA × 2 kΩ = 12V

(ii) The internal resistance of the voltage source is 2 k Ω.

The equivalent voltage source is a source of 12 V in series with a resistance of 2 k Ω as shown in **Fig. 1.19.**

**Note.** The voltage source should be represented with the +ve terminal in the direction of the current flow.

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Reference : Principles Of Electronics Multicolor Illustrative Edition By V K Mehta And Rohit Mehta