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Design Idea
Signal-powered linear optocoupler provides isolated control signal
An optoisolator provides isolation for a control voltage and supplies loop forever.
Mitja Rihtarsic, kofja Loka, Slovenia; Edited by Martin Rowe and Fran Granville -- EDN, 7/23/2009
The circuit in Figure 1 provides an isolated control voltage, such as 0 to 10V. In the low part of the range, 0V to approximately 2V, the controlled device is off. Therefore, the upper part of the range must be as linear as possible. You can meet this requirement using a linear optocoupler, such as Vishay’s IL300 or Avago Technologies’ HCNR200 or HCNR201.
These optocouplers each comprise an LED and a photodiode on the transmitting side and an identical photodiode on the receiving side. Because of this construction, the emitted light from the LED should cause the same current to flow in both photodiodes. The current through the photodiode on the receiving side, feedforward current IFF, is the output current, and you must set this current in proportion to the transmitted signal voltage, V1. This current equals the feedback current, IFB, through the transmitter-side photodiode. A feedback loop around the emitting side of the optocoupler keeps the feedback current in proportion to the transmitted signal. When the feedforward current and the feedback current are equal, the output current is proportional to the transmitted signal.
The hidden cost, however, is a power supply. You need some power on both sides of the signal path. The circuit in this Design Idea uses power from signal voltage V1 to supply a feedback loop in the transmitting side similar to the way some circuits in a 4- to 20-mA loop get power from the loop current. Both photodiodes operate in reverse-biased, photoconductive mode. The currents through them are proportional to incident-light flux, which feedback gain K1 and forward gain K2 describe.
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(1) |
where ILED is the LED’s current, and
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(2) |
A description of the circuit begins with a sum of the dc currents at Node A.
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(3) |
The gains of both transistors amplify current IB1 into the base of Q1. The amplified current then flows through the LED.
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(4) |
Equations 1 through 4 yield the output feedforward current:
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(5) |
When the product of feedback gain K1 and transistor gains β1 and β2 is much greater than one, you can cancel out the transistors’ gains, yielding a characteristic that is linear:
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(6) |
The ratio of feedback gain K1 and forward gain K2 is transfer gain K3. Because K1 and K2 are similar, K3 is approximately one. In reality, K3 may deviate, but it changes less than K1 or K2 alone:
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(7) |
Equation 6 subtracts the base-to-emitter voltage from the input voltage. Although the base-to-emitter voltage is not constant, it is desirable to remove it. You accomplish this task using the emitter follower in the receiving circuit. The output voltage, V2, is a sum of voltage across R3 and the base-to-emitter voltage of Q3.
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(8) |
You can use a different equation to yield the feedforward output current:
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(9) |
You can rearrange equations 8 and 9 as:
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(10) |
In the first term in Equation 10, the ratio of resistors R3 and R1 is approximately 1-to-1. You must be careful with the transfer gain, K3, which is the reason that K3 remains in Equation 11.
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(11) |
When K3 is one, voltages VBE1 and VBE3 cancel each other to some degree. Therefore, Equation 11 omits the second term in Equation 10. Base current IB3 depends on resistor R4 and the output load. When you can set both base currents to be equal, the last term would cancel out, too. The values of resistor R2 and capacitor C1 must be small enough so that transistors Q1 and Q2 don’t saturate. C1 enhances stability.
Figure 2 shows the necessary voltage for the circuit to begin operation. The output voltage (upper trace) has flatness at its lowest voltages as opposed to the input voltage (lower trace). Figure 3 shows the two signals’ linearity. Dividing the measured maximum of voltages V1 and V2 yields 0.91V. A test circuit uses an IL300, which has a gain of 0.851 to 0.955. The measurement meets the requirements of Equation 11 despite the equation’s simplifications.
