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APPLICATION DATA FOR OPEN COLLECTOR DEVICES<br />

Logical 0 (on levell Circuit Calculations (see<br />

figure BI<br />

The current through the resistor must be limited to<br />

the maximum sink-current capability <strong>of</strong> one output<br />

transistor. Note that if several output transistors are<br />

wire-OR connected. the current through RL may be<br />

shared by those paralleled transistors. However. unless<br />

it can be absolutely guaranteed that more than<br />

one transistor will be on during logical 0 periods.<br />

the current must be limited to 16mA. the maximum<br />

current which will ensure a logical 0 maximum <strong>of</strong><br />

0.4 volts.<br />

Also. fan-out must be considered. Part <strong>of</strong> the 16mA<br />

will be supplied from the inputs which are being<br />

driven. This reduces the amount <strong>of</strong> current which<br />

can be allowed through RL.<br />

Therefore. the equation used to determine the minimum<br />

value <strong>of</strong> RL would be:<br />

Vee -<br />

ISINK capability -<br />

Vout(O) required<br />

ISINK from TTL loads<br />

TTL LOADS<br />

C.I,ul.lion,<br />

___ ~Yc~c~-__ Y;~~t~lo~l_req~u_ire_d __ __<br />

RL(minl = I'i,* 'op.bilily - I'i,* from TTL lo.ds<br />

S - 0.4 4.6<br />

R - = -- = 4100<br />

Llminl - 0.016 _ 0.0048 0.0112<br />

I<br />

.1<br />

~<br />

N=3<br />

N • lin(O) = 3 • 1.6 rnA<br />

MAXIMUM I sink CAPABILITY<br />

OF ONE ON OUTPUT = 16 mA<br />

t Current into OFF outputs i. negligible at logical O.<br />

Calculation:<br />

Vee - V~ut(O) required<br />

RL(min) = ISINK capability - ISINK from TTL loads<br />

RL(min)<br />

5 - 0.4<br />

0.016 - 0.0048<br />

4.6<br />

0.0112<br />

= 41000<br />

FIGURE B -<br />

LOGICAL 0 CIRCUIT CONDITIONS<br />

3-13

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