DAC 1408

DAC 1408

Fig. 13.2.1 shows the pin diagram for IC 1408 DAC.

The IC 1408 consists of a reference current amplifier, an R/2R ladder and eight high speed current switches. It has eight input data lines A₁ (MSB) through A₈ (LSB) which control the positions of current switches.

It requires 2 mA reference current for full scale input and two power supplies V_CC = + 5 V and V_EE = - 15 V (V_EE can range from - 5 V to - 15 V).

The voltage V_ref  and resistor R₁₄ determines the total reference current source and R₁₅ is generally equal to R₁₄ to match the input impedance of the reference current amplifier.

Fig. 13.2.2 shows a typical circuit for IC 1408. The output current I_o can be given as

Note Input A₁ through A₈ can be either 0 or 1. Therefore, for typical circuit full scale current can be given as,

It shows that the full scale output current is always 1 LSB less than the reference current source of 2 mA. This output current is converted into voltage by I to V converter. The output voltage for full scale input can be given as

V_o = 1.992 × 2.5 K = 4.98 V

Note The arrow on the pin 4 shows the output current direction. It is inward. This means that IC 1408 sinks current. At (0000 0000)₂ binary input it sinks zero current and at (1111 1111) ₂ binary input it sinks 1.992 mA.

The circuit shown in the Fig. 13.2.2 gives output in the unipolar range. When digital input is 00H, the output voltage is 0 V and when digital input is FFH (11111111)₂, the output voltage is + 5 V. This circuit can be modified to give bipolar output.

Fig. 13.2.3 shows the circuit for giving output in the bipolar range. Here, resistor R_B (5 K) is connected between V_ref and the output terminal of IC 1408. This gives a constant current source of 1 mA. The circuit operation can be observed for three conditions :

Condition 1         : For binary input (00H)

When binary input is OOH, the output current I_o at pin 4 is zero. Due to this current flowing through R_B (1 mA) flows through R_f giving V_o = - 5 V.

Condition 2         : For binary input 80H

When binary input is 80H, the output current Io at pin 4 is 1 mA. By applying KCL at node A we get,

- I_B + I_o + I_f = 0

Substituting values of I_B and I_o we get,

-(1 mA) + (1 mA) + I_f = 0

ஃ If = 0

and therefore the output voltage is zero.

Condition 3 : For binary input FFH

When binary input is FFH, the output current I_o at pin 4 is 2 mA. By applying KCL at node A we get,

-I_B + I_o + I_f = 0

Substituting values of I_B and I_o we get,

- (1 mA) + (2 mA) + I_f = 0

ஃ I_f = - 1 mA

Therefore, the output voltage is + 5 V. In this way, circuit shown in the Fig. 13.2.3 gives output in the bipolar range.