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ICL8013 Datasheet(PDF) 4 Page - Intersil Corporation |
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ICL8013 Datasheet(HTML) 4 Page - Intersil Corporation |
4 / 8 page 4 Application Information Detailed Circuit Description The fundamental element of the ICL8013 multiplier is the bipolar differential amplifier of Figure 1. The small signal differential voltage gain of this circuit is given by: The output voltage is thus proportional to the product of the input voltage VlN and the emitter current IE. In the simple transconductance multiplier of Figure 2, a current source comprising Q3, D1, and RY is used. If VY is large compared with the drop across D1, then There are several difficulties with this simple modulator: 1. VY must be positive and greater than VD. 2. Some portion of the signal at VX will appear at the output unless IE = 0. 3. VX must be a small signal for the differential pair to be linear. 4. The output voltage is not centered around ground. The first problem relates to the method of converting the VY voltage to a current to vary the gain of the VX differential pair. A better method, Figure 3, uses another differential pair but with considerable emitter degeneration. In this circuit the differential input voltage appears across the common emitter resistor, producing a current which adds or subtracts from the quiescent current in either collector. This type of voltage to current converter handles signals from 0V to ±10V with excellent linearity. The second problem is called feedthrough; i.e., the product of zero and some finite Input signal does not produce zero output voltage. The circuit whose operation is illustrated by Figures 4A, 4B, and 4C overcomes this problem and forms the heart of many multiplier circuits in use today. This circuit is basically two matched differential pairs with cross coupled collectors. Consider the case shown in Figure 4A of exactly equal current sources basing the two pairs. With a small positive signal at VlN, the collector current of Q1 and Q4 will increase but the collector currents of Q2 and Q3 will decrease by the same amount. Since the collectors are cross coupled the current through the load resistors remains unchanged and independent of the VlN input voltage. In Figure 4B, notice that with VIN = 0 any variation in the ratio of biasing current sources will produce a common mode voltage across the load resistors. The differential output voltage will remain zero. In Figure 4C we apply a differential input voltage with unbalanced current sources. If IE1 is twice IE2 the gain of differential pair Q1 and Q2 is twice the gain of pair Q3 and Q4. Therefore, the change in cross coupled collector currents will be unequal and a differential output voltage will result. By replacing the separate biasing current sources with the voltage to current converter of Figure 3 we have a balanced multiplier circuit capable of four quadrant operation (Figure 5). 2IE RL RL VIN VOUT V+ V- FIGURE 1. DIFFERENTIAL AMPLIFIER A V V OUT V IN ---------------- R L r E ------- == Substituting r E 1 g M ------- kT qI E --------- == V OUT V IN R L r E ------- V IN qI E RL kT ------------------- × == I D V Y R Y -------- ≈ 2I E and = V OUT qR L kTR Y --------------- V X V Y × () = 2IE RL RL VIN VOUT V+ V- Q3 RY VY ID VD D1 qRL kTRY (VX x VY) VOUT = K (VX x VY) = - + FIGURE 2. TRANSCONDUCTANCE MULTIPLIER IE + ∆I IE IE - ∆I VIN ∆VOUT V+ V- IE ∆I = VIN RE FIGURE 3. VOLTAGE TO CURRENT CONVERTER ICL8013 |
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