# Opto-characteristic for schematic

I am referring below circuit, where I am using P781 opto-isolator. H1G is I/O pin from LPC2138 controller.( VCC for LPC is 3.3V ) J24: 12V,6W Heater

For above opto, what is minimum If for LED to glow and therefore to conduct transistor at o/p side. I am checking min If in datasheet but couldn't get it.

For 2V or say 3V at H1G pin can transistor conduct?

Thank you.

• Replace 5V with 3.3v, R5 = 150 Ohm, remove R6 (shorted). 14mA sink current lower than maximum rating (20mA). Datasheet nxp.com/documents/data_sheet/LPC2131_32_34_36_38.pdf page 26 – Oka Aug 26 '15 at 12:38
• @ Oka Thanks, if R5=150 Ohms, and Vf=1.3V. How you got value of sink current in this case? Is it (3.3-Vf)/150.? – Electroholic Aug 27 '15 at 6:48
• yes, in datasheet Vf = 1~1.3. Sink current (3.3-1.3)/150=13.3ma ~ (3.3-1)/150=15.3. Average 14.3mA. Recommended If 16-25mA, so it is a bit lower – Oka Aug 27 '15 at 7:05
• Thanks. What is If (Forward current) required to ensure led's conduction? I cant figure out from datasheet. So, that I will increase R5 value which will reduce sink current but will ensure LED's conduction. – Electroholic Aug 27 '15 at 7:26
• Look at figure 4 here macao.communications.museum/eng/exhibition/secondfloor/moreinfo/… look at saturation region. For each different Ib (If for opto), there are some values of Ic that make Vce almost zero, we call it transistor in saturation region. Maybe other member can give us better explaining? – Oka Aug 27 '15 at 8:41

You need to look at this table in the data sheet: - Look at the bottom row - it's telling you that for a forward current of 8.4mA, providing you don't take more than 2.4mA through the collector, you can expect a volt drop from collector to emitter of 0.4V max.

How does this work with your circuit? 2.4mA through R10 will generate a gate voltage of 24 volts but of course this is not possible with a 12V supply - this means the opto's transistor will be even more saturated - that is good normally but, you have another 10k resistor (R6) in series with the collector - this will limit the drive voltage on the gate to about 6V. I'm mentioning this because I don't recognize the MOSFET number and cannot tell you if this MOSFET is suitable with this gate voltage.

If you used a forward current of 1mA and a collector current of 0.2mA the saturation voltage is typically 0.2V but, 0.2mA through R10 will only produce 2V at the gate and this certainly may not be enough to adequately turn it on.

So, given that all we have is the data sheet you are probably looking to drive at least 2mA thru the diode BUT the outcome is pretty much determined by the MOSFET. Also, why are you using R6 - it doesn't seem to serve a purpose.

• You have a 5V supply feeding R5 and with 3V on your IO pin there is the potential to see 2V across the diode. The data sheet says the forward voltage of the diode is about 1.2 volts so, it will always remain on. – Andy aka Aug 26 '15 at 10:53
• Look at fig 1 in the mosfet data sheet - decide what current the load takes, look at the 6V curve for gate voltage, read off drain-source volt drop, calculate power dissipation in mosfet, decide if this is acceptable. – Andy aka Aug 26 '15 at 10:57
• Connect it to 3V3 is my preference. – Andy aka Aug 26 '15 at 11:30
• No, replace the 5V supply with a 3V3 supply - forget about 5V. With 1.3 volts across the diode, 2 volts will be across R5 and, if R5 = 330 oms then a current of 6mA will flow. Maybe R5 could be a bit bigger and reduce drive current to maybe 3mA. – Andy aka Aug 26 '15 at 11:42
• I think that is all I can say about R5 (previous comment). I don't understand about the 560 ohm - it's not in the circuit therefore I can't comprehend what you mean. – Andy aka Aug 26 '15 at 12:19

The IRF1010 specifies its RDS(ON) only for a 10 V gate voltage, so you should remove R6.

To get a high voltage on the MOSFET's gate, you want to drop about 12 V over R10, so the optocoupler needs to pass a current of at least 12V/10kΩ = 1.2 mA.

The optocoupler's minimum CTR is 50 %, so you need an IF of at least 1.2mA/50% = 2.4 mA. However, the datasheet specifies the CTR at a different IF, and you want the phototransistor to saturate, and you should add a few percent to allow for LED degradation, so you should use IF = 5 mA or so.

The maximum forward voltage is 1.3 V, so to get a current of 5 mA, you need to drop 330Ω×5mA = 1.65 V over R5. This means that the voltage of H1G must be pulled down to at most 5V−1.65V−1.3V = 2.05 V, so pulling it down to 0 V should work fine.

To switch off the LED reliably with an I/O pin that cannot actively drive 5 V, use an open-collector output, or add a pull-up resistor to 5 V.

• So does it means Voltage above 2.05 is logic 1 and below that is logic 0? H1G is i/o pin of LPC2138 connected to 3v3 power, So logic 1 am expecting 3V at worst case considering 3v3 power pin voltage. Thanks. – Electroholic Aug 26 '15 at 12:08
• No; the voltage at H1G is not interpreted as logic, it is an analog quantity that drives the LED. When the voltage at H1G is more than 2.05 V, then there is not enough voltage drop for a 5 mA current. When the voltage at H1G is lower than 2.05 V, you get 5 mA current, or more. To switch the LED completely off, you need 5 V at H1G (i.e., a pullup), or an open collector. – CL. Aug 26 '15 at 13:55
• To actually get 5 mA when H1G is at 0 V, replace R5 with (5V−1.3V)/5mA ≈ 740 Ω. – CL. Aug 26 '15 at 13:57
• Oka's 14 mA is more than needed, but would work. – CL. Aug 27 '15 at 7:10
• If H1G is at 3.3 V and the top of R5 at 5 V, there is a voltage drop of 1.7 V over R5 and the LED, which might be enough to switch it on. To ensure the LED is off, the voltage drop over it should be zero; this can be done with a pull-up to 5V, or by using the same power (3.3V) as used by the controller. – CL. Aug 27 '15 at 7:12

## To make sure the photodiode can turn ON/OFF the phototransistor, replace 5V with 3.3V.

As you said, H1G will be connected to I/O pin of the microcontroller, so it is not exactly 0v or 3.3V. Based on the datasheet page 27, low logic can be 0.4v, and high logic can be 2.9v. Connecting with 5V and photodiode in series will make it more difficult for calculating.

If you connect H1G in series with R5=150 Ohm, photodioda, and 3.3v, at worst scenario:

• High logic = 2.9v, If=0, because 3.3-2.9=0.4v is not enough to give the photodioda forward bias. The phototransistor OFF with If=0.
• Low logic = 0.4v, If=(3.3-0.4-1.3)/150=10.7mA (minimum value for this circuit)

With R5=150 Ohm, typical If=(3.3-1.15)/150=14.3mA (typical value for this circuit)

Maximum value of If=(3.3-1.0)/150=15.3mA (maximum value for this circuit)

It is a bit lower then typical value of If=16mA, so it is safe for the optocoupler.

How about for the microcontroller? 15.3mA is safe too, because the microcontroller has the maximum sink capacity 20mA

## Can If=10.7mA turn on the MOSFET?

Let us see the datasheet at page 13. .

From this picture, Ic < 6mA can make the phototransistor inside the optocoupler saturated with Vce < 0.2v. For Ic, lower is better (gives lower Vce). With R6 shorted, Ic = 12v-0.2/R10 = 1.2mA. With Vce ~ 0.2v, Vgs = 11.8v and the MOSFET ON.

Actually based on F1010N datasheet, Vgs threeshold is maximum 4V. So 5.9V or 6V is enough to make MOSFET saturated. You could short R6 or not.