UTMC APPLICATION NOTE The monolithic UT63M1XX Series Bus Transceivers are complete transmitter and receiver pairs conforming fully to MIL-STD-1553A and B. Transceivers provide the electrical bus interface for subsystems to the MIL-STD-1553 bus. MIL-STD-1553 specifies the data bus signal’s wave shape, amplitude, frequency, and power. This application note will discuss the power consumption and dissipation of the bus transceiver as it transmits signals on the MIL-STD-1553 data bus. Specifically it will differentiate between power consumed by the bus interface (i.e., system power) versus power dissipated within the transceiver. For system designers addressing system power consumption and the sizing of power supplies, the power consumption of the bus interface (transceiver plus load) is important. Transceiver power dissipation is important for evaluating the reliability and thermal characteristics of the transceiver. This document will present examples of calculating the power dissipation and consumption for a direct-coupled transceiver transmitting onto the MIL-STD1553 data bus. System Power Consumption System power consumption is defined as the power consumed by the device plus load. UTMC’s UT63M1XX Bus Transceiver data sheet specifies the maximum current per voltage supply including load. To obtain the system power consumption the voltage is multiplied by the AC current. Figure 1 shows an example of the power the transceiver consumes while driving the bus. For sizing power supplies and characterizing system current requirements, no differentiation between power consumption and dissipation is necessary. VCC ICC VCCA ICCA VEE IEE PSYS = VEE IEE + VCC ICC + LOAD INTRODUCTION UT63M1XX UT63M1XX: P OWER CONSUMPTION VS. DISSIPATION VCCA ICCA Figure 1. System Power Consumption Transceiver Power Dissipation Transceiver power dissipation defines the power dissipated in the device while the transceiver drives the bus load. Power supplies VCC and VCCA dissipate essentially all of their supplied power within the transceiver device. A small amount of the VCC power is dissipated at the receiver output load but is negligible compared to the transmitter output. Therefore the power dissipated by these supplies (VCC, VCCA) is equal to the power consumed by these supplies. PD(VCC) = ICC x VCC = 60mA x 5V = 300mW P D(VCCA) = ICCA x VCCA = 10mA x 15V = 150mW *P D(VCCA) = ICCA x VCCA = 10mA x 5V = 50mW *Note : +5 and -15 volt operation (i.e., VCCA = VCC) The power supplied by the negative source (VEE) is consumed by both the load and the transceiver output stage. To calculate the power dissipated within the transceiver, perform a best- and worst-case analysis. MIL-STD-1553 specifies the output voltage level (VO) and load impedance (ZO) for the direct-coupled transceiver (±15 volt version) shown in figure 2. 1 IB ±15 VDC 55Ω 1.4: 1 UT63M1XX VB IO VA N1 N2 RL = Z O/2 VO 55Ω 700 1000 Where ZO = 70 ohms 700 Figure 2. Transceiver Power Dissipation The following calculations are performed on the circuit using these MIL-STD-1553B specification limits: VO 6VP-P (min) 9VP-P (max) Using the above limits the voltage and current supplied by the transceiver are calculated for a single-ended 1MHz cycle. ZO 70 ohms 70 ohms For an ideal transformer the following equations are written for voltage and current: VB = (N1/N2)VA /2 where N1/N2 = .70 IB = (N2/N1)IO /2 where N2/N1 = 1.43 1ms Positive Excursion Negative Excursion VBP (BasePeak) Figure 3. Single-Ended Waveform (VB= 4 VBP) Negative Excursion: VO = 6VP-P, L-L (P-P, Peak-Peak), (L-L, Line-Line) ZO = 70 ohms IO = 171mA/2 = 85mA VA = 4 x VO = 24VP-P, L-L VBP = (N1/N2)VA/2 = 8.4 VO = 9VP-P, L-L (P-P, Peak-Peak), (L-L, Line-Line) ZO = 70 ohms IO = 257mA/2 = 128mA VA = 4 x VO = 36VP-P, L-L VBP = (N1/N2)VA/2 = 12.6 IB = I O x 1.43 = 122mA DVBP = (VEE - VBP) = (6.6 V) PD = DV BP x IB = 805mW IB = IO x 1.43 = 184mA ∆VBP = (VEE - VBP) = (2.4V) PD = ∆VBP x IB = 441mW Where: VBP = the voltage VB measured base to peak DVBP = the voltage imposed across the transceiver PD = power dissipated in the transceiver due to the loading of the output (negative excursion) Positive Excursion: VBP = 8.4V RIL = 14Kohms (internal resistor) IL = 600mA PR = 5mW Where: 2 VBP = 12.6V RIL = 14Kohms (internal resistor) IL = 900mA PR = 11.3mW PR = power dissipated in the transceiver (positive excursion) Single-Ended Power Dissipation: PT = PD + 2PR PT = 805mW + 10mW = 815mW PT = PD + 2PR PT = 441mW + 22.6mW = 463.6mW Blocking Diode: PD (VD) = VD x IO = .9 x 122mA = 109mW The minus supply (VEE) also supplies power that is dissipated within bias circuitry and is not included with the output power dissipation. That power is defined as: PD(bias) = 40mA x 15V = 600mW transmitting information onto the data bus the transceiver power dissipation increase is proportional to the transmission time. Table 1 shows a breakdown of duty cycle vs. message length. Therefore: Message Word Count Idle Transmitting PD(VEE) = PD(total) + PD(bias) Receive Receive 32 1 97% 97% 3% 3% Transmit Transmit 32 1 3% 94% 97% 6% The total power dissipated in transceiver (worst-case) for a complete cycle is as follows: PCD = PD(VEE) + PD(VCC) + PD(VCCA) + PD(VD) = (.815W + .60W) + .30W + .15W + .109W = 1.974W * The above calculations are based on time period t = 680ms *PCD = PD(VEE) + PD(VCC) + PD(VCCA) + PD (V D) = (.815W + .60W) + .30W + .05W + .109W = 1.874W *Note : +5 and -15 volt operation (i.e., VCCA = VCC) Duty Cycle Figure 4 shows transceiver power dissipation during various duty cycles. While receiving information from the bus the power dissipation is low (i.e., idle state); the receiver section of the device is active and the transmitter is inactive. When CONCLUSION Calculation of the worst-case power dissipation for the transceiver while it transmits information onto the MILSTD-1553 data bus is straightforward. Use the same analysis method to determine power dissipation for a transformer-coupled stub and for the various voltage versions of the UT63M1XX Transceiver. 3 25% Duty Cycle P 50% Duty Cycle 1.974 W 1.974 W P W1 W1 1.0 W 1.974 W P W1 1.0 W W2 1.0 W W2 T t W1 W2 Wt Wt Wt PD PD 75% Duty Cycle = (.25t) *1.974W = (.75t) * 1.0W = W1 + W2 = .493t + .75t = 1.243t = (Wt/t) = 1.24W W2 W2 T t W1 W2 Wt Wt Wt PD PD = (.5t) * 1.974W = (.5t) * 1.0W = W1 + W2 = .987t + .5t = 1.487t = (Wt/t) = 1.49W 0% Duty Cycle t W1 W2 Wt Wt Wt PD PD = (.75t) * 1.974W = (.25t) * 1.0W = W1 + W2 = 1.48t + .25t = 1.73t = (Wt/t) = 1.73W 100% Duty Cycle P P 1.974W 1.0W W1 W1 T t W1 W2 Wt Wt Wt PD PD = (t) * 1.0W =0 = W1 + W2 = 1.0t + 0 = 1.0t = (Wt/t) = 1.0W t W1 W2 Wt Wt Wt PD PD = (t) * 1.974W =0 = W1 + W2 = 1.974t + 0 = 1.974t = (Wt/t) = 1.97W Figure 4. Duty Cycle Calculation 4 T T