PS2621 datasheet(19/38 Pages) NEC

19 / 38 page

VCC - VCE

5 - 0.8

⊕

= 8.2 (mA) ................(9)

0.51 (k

Ω)

Therefore I3 = I2 + I4 = 8.2 + 1.6 = 9.8 (mA) ................(10)

Let's assume that hFE of transistor Tr1 is 40 (worst). Ib must be as follows:

9.9 (mA)

⊕

= 0.247 (mA) ................(11)

hFE

Similarly, let's assume that VBE of transistor Tr1 is 0.8 V (worst), I1 must be as follows:

VBE

0.8

I1 =

= 0.4 (mA) ................(12)

2 (k

Ω)

Therefore, the output current I0 of the optocoupler is

⊕ I1 + Ib = 0.647 (mA) ................ (13)

If forward current IF is 3 mA (worst) (normally IF = 5 mA), the CTR is calculated as follows:

0.647(mA)

CTR =

x 100 =

x 100 = 21.6% ................(14)

3 (mA)

CIRCUIT DESIGN EXAMPLE (USING THE PS2601)

Fig. 4-1 shows a design example of an optocoupler circuit having a base-emitter resistor for improvement of response ability.

The minimum current transfer ratio (CTR) required for TTL operation is calculated as follows:

Current I4 must be 1.6 mA to drive the TTL and the collector-emitter voltage of transistor Tr1 must be 0.8 V or less. Accordingly, I2

must be as follows:

A resistor of 510 k

Ω

is inserted here.

PS2601

IF = 5 mA

Vcc = 5 V

VOUT

Tn1

R2 = 510

Ω

TTL

R0 = 1 k

Ω

= 2 k

Ω

Figure 4-1. Circuit Design Example

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