For development engineers, reading the technical manual of the chip is a basic skill. In this example, the circuit of the optocoupler is taken as an example to explain the basic circuit design process.
Optocoupler introduction
Optocouplers are often used for electrical isolation, so as not to mix high-voltage, high-current and other pulses in the external incoming signal, and burn out our own circuits or chips. For example, 422, 485, CAN, or various outdoor sensors and lines that may encounter lightning strikes, usually use optocouplers when connecting with equipment. The optocoupler is relatively simple, and there are not many peripheral circuits. The simplest external circuit of the optocoupler only has 2 or 3 resistors, but the value of the 2 or 3 resistors is particular. Many engineers do not know it, so just copy a circuit , although it can be used with a high probability, it is often impossible to make the chip work in the correct state.
Optocoupler circuit
This example takes TLP109 as an example, which is a 5-pin optocoupler. The equivalent circuit is as follows: the
left side is a light-emitting diode, and the right side is a photodiode plus a triode. There are 2 resistors on the periphery. The diode on the right side of some optocouplers needs to add a diode current limiter. That is the peripheral 3 resistors.
Some optocouplers do not have the diode on the right, and are directly a phototransistor to detect current, so there are 2 resistors on the periphery.
The peripheral circuit is very simple, as shown in the figure below:
R16 is the current limiting current of the input diode, and R17 is the load resistance of the output transistor.
The basic performance index of the circuit
First determine the circuit parameters:
the left is the output of the microcontroller, and the power supply is 3.3V.
The right side is the output to the interface chip, and the power supply is 5V.
The left and right sides do not share the same ground, which is realized through the DCDC isolation module.
CAN bus, the theoretical maximum speed is 1M, and the speed used in the actual circuit is 500K
parameter selection
Determine the load resistance
The first thing that needs to be determined is the load of the optocoupler. In this example, the load of the optocoupler is the circuit connected behind R17. This is a CAN bus driver, and its input current is at the level of microamperes. It can be considered that there is no load. Here, the load is 1mA. Consider it.
Let's start to check the manual. The figure is logarithmic coordinates, not ordinary standard coordinates .
This figure shows the relationship between load resistance and rising edge and falling edge time when the output is 5V and the input current is 16mA.
tpLH is the rising edge time, and tpHL is the falling edge time. Since we want to work at high speed, we choose the vicinity of the intersection of the two as the working state. At this time, the corresponding load resistance is 2~4K. Under the premise of meeting the working conditions, the resistance should be as large as possible. Some, save power consumption.
Therefore, the value of the circuit is 3.3K, the corresponding rising edge is 0.35us, and the falling edge is 0.27us. In the CAN specification, the sampling point is about 85%. For the CAN rate of 1M, the value does not exceed half of the signal width. , which can meet the needs of the work.
Determine the load current
Because the subsequent load is negligible, the load current here refers to the current on the load resistor when the optocoupler is turned on.
The conduction voltage drop of the optocoupler is calculated as 0.4V, and the load resistance is 3.3K, so the voltage on the load is 4.6V, and the load current is 1.4mA
This table shows the corresponding relationship between output current and input current. Let's look at the bottom line. When IF=5mA, the output current can support a maximum of 3.3mA, which does not exceed 1.4mA of the load, thus determining the load resistance at the output end.
Determine the input current
In the figure above, the current of the light-emitting diode corresponding to the load current of 1.4mA is more than 3mA, and the safety point is 5mA.
Determine the input resistance
Taking the standard working environment of 25 degrees as an example, at 5mA, the voltage drop of the LED is about 1.5V.
The voltage drop on the resistor is 3.3-1.5=1.8V, and the current flowing through it is 5mA, so the resistance should be 360 ohms, and the value of 360 ohms or 330 ohms is acceptable. Considering that 330 is more commonly used, it is finally determined as R6=330.
Determine the working status within the working area.
Under normal temperature conditions, the circuit can work at this time. But what about the extremes of high or low temperature, will it still work?
Finally, you need to check other parameters, mainly temperature-related current, voltage, power and other parameters, and make sure that the parameters are within the working range. This step is essential, but many engineers will not do this step. As a result, the final designed product is fine during the test, and various problems will occur when it is sent to the user.
For the working environment where the temperature is from 125 to -40 degrees, the voltage drop of the light-emitting diode changes from 1.41 to 1.62, and the corresponding current is 5.7 to 5.1mA.
The corresponding load capacity meets the needs, and the margin left is also sufficient for the normal operation of the circuit when other parameters change.
For temperature-insensitive components such as optocouplers, this step is not very important, but for power components, this step is extremely important, which may lead to re-selection of parameters, or even replacement of components that cannot meet the working environment.
Measured
The CAN bus cannot send and receive data normally. Observing the oscilloscope, it is found that the rising edge is relatively slow, and the rising time is about 5mS. This is definitely not possible. When the CAN bus speed is 500K, the signal period is only 2mS. The table
found that the time test condition of the rising edge and falling edge is that the input current is 16mA. Obviously, the input current is not enough, resulting in insufficient light emission of the light-emitting diode, resulting in slow switching of the optocoupler, so reduce the input resistance of the optocoupler to 120 ohms , the fault is resolved.
epilogue
Although it is only the selection of two resistance values, it is very important for electronic engineers to understand the circuit principle thoroughly. If the parameter selection is not appropriate, it will not work if it is light, and it will be unreliable if it is heavy.
Yes, I didn't say otherwise, failure to work is a minor problem, which can be fixed during the development phase. However, unreliable work is a big problem. It is likely to pay development time and plate making costs, and even discover problems after mass production, so the investment will be large.