TLP521-4 Help with calculations

Hi,I couldnt open that topic to the right place. Sorry for that. There is no "new topic" button on that page.

I have an application that contains 2 ICs. 4094BE 8 Bit Shift register and 2 Optocoupler TLP521-4. I am filling the bits of shift register by arduino. There are 8 leds connected to Optocoupler. On LED side of the TLP521-4, datasheet does not say what voltage rate should I apply. I applied 5 volts to shift register but it outputs 2.5 V. And output legs of the shift register connected to the Anode of the LED on the optocoupler through 470 ohm resistor. Datasheet specifies that there is a 1.15 typical voltage drop at 10 ma forward current. If i calculate the current on the led, it is around 2.8 ma. And this is not enough to connect the emmiter to the collector. Is my calculation true ? I can temper with voltage supply to the shift register. Datasheet specifies that max 20 V can be applied to the shift register voltage supply pin. At 5 V supply, it outputs 2.5 volts and if I apply 12 volts to the shift register, it should output around 6 volts . And by doing the calculation again, forward current would be 10,3 ma. This should be enough to produce enough flow through emitter to the collector. Am I right or something is wrong ?

I have an application that contains 2 ICs. 4094BE 8 Bit Shift register and 2 Optocoupler TLP521-4

What's driving the shift register? If it's a 5v Arduino, you're stuck with 5 volt logic. Next, what's the optocoupler connected to?

Problem number one is that you cannot directly drive an led with a cmos gate, you might be able to get about 2ma out of that ancient series 4000 part. Problem two is you're ignoring the current transfer ratio of the coupler which can be as low as 50% - which means for 2ma in you can get 1ma of current on the output side.

I don't have a fix for you without knowing the bigger picture of what you're trying to do.

Data latch and clock pina are connected to arduino. O1 ... O8 is connected to anode lead of the optocoupler through 470 ohm resistor. Emitter leaf is connected to 12v 40A power source. Collector lead is connected to a voltage divider that outputs 10V out of 12V. This voltage divider connected to a N channel mosfet. N channel mosfet opens and closes solenoid valve. There ara 8 solenoid valfs.

Now here is my calculations.

1.15V drop across the led inside optocoupler. Output voltage of the shift register is 2.6 V to be exact when VDD = 5V. I have 470 ohm resistor at every anode lead of the optocoupler.
(2.6-1.15)/470 = 3 mA this value opens the valves sometimes and sometimes does not open.

My question is the following wiring and calculations are true or not.

Datasheet says that I can apply 18V max to 3V min. Now my wiring is as follows.

VDD --- 12V
OE------ 12V
O1...O8 ---- Optocoupler Anode leads
Data - Arduino digital pin(5V)
Clock -Ardino digital pin(5V)
Latch - Arduino digital pin(5V)

Output will be around 6V bu extrapolation. This gives the following equation.

(6-1.15)/470=11.5 mA.

This current is enough according to datasheet. Did I make a mistake are tanılaan calculations are true?


Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

It is easier than trying to interpret text about circuit details, also reverse engineering your project my help find your problem.

Also your code, I hope that you have just a test code written to activate this part of your project.

Thanks.. Tom... :slight_smile:

I have attached a hand draw for you. I dont have issue with the code and anything else. Take a look at the picture. What I am asking is, if I change the power source of CD4094 while everything else stay same, can I get a forward current ~10 mA?

OPs Diagram,

Thanks.. Tom.. :slight_smile:

Im in a constraction area. I dont have my computer with me right now. What is not clear in diagram ? Besides my ankle is broken so I cant write properly for about 2 months from now.

TLP data, the suggested forward current for the input of the opto is 10mA.

Tom... :slight_smile:

So the O1...O2 pins must be around 6-7 V to accomplish that. This is okey. I have done the math. The problem is, Data, latch and clock pins are connected to arduino. This means that logic level of data,latch and clock pins are 5V. Can I power CD4094 from an external 12V power supply and at the sametime change the data inside the CD4094 with 5V logic level data,latch and clock lines?

Where are the Mosfets and what type have you specified ?
What was the reason for specifying opto-couplers ?

And, of course, get well soon.

As stated previously, the 4093 can only sink or source 1mA.

Tom... :slight_smile:

As stated previously, the 4093 can only sink or source 1mA.

Tom... :slight_smile:

I couldnt see that in datasheet. Can you point it please. 4094 can provide 2.5 mA if VDD is 5V and 9.5 mA if VDD is 10 V typically.

You need to look at 74HC595 as your serial in parallel out device.

An example here, basically whatyou are trying to do.

Hope it helps.. Tom.. :slight_smile:


If you look you will see for 10V Vcc its 6.8mA Source and 2mA Sink.
Tom... :slight_smile:

Why do you do this to me dude ? My problem is not serial in parallel out. My problem is voltage drop and output current. If you dont have answer to my question, why bother. Please just answer my question not something else.

Sorry I gave you 4093 not 4094. :o :o

4094 are just as low, these are digital signal ICs, not intended to drive loads, just other digital IC inputs.

Tom... :slight_smile:

As i said before there is no load actually. This shift register only holds which optocoupled photo transistor will be on or off. Basically i need more current then current system. Now i apply 5v to VDD pin. I have simulated circuit on proteus. It works fine but my last and final problem is why İ am encountering with a voltage drop at the output leads ? I couldnt figure it out and tried almost everyrhing. I get 2.6 volts at outputs of the CD4094 shift register.

Have you put an an oscilloscope on it to prove its DC and not pulsed DC.
Have you pulled the opto out of circuit and measured the output voltage.

Do me a favour, put your DMM in AC mode and measure the voltage on the output pin.

"No load actually" what do you mean, there either is or isn't.

You have got the gnd of the arduino and the 4094 and the opto ALL connected together?

Tom... :slight_smile:

  1. There are no AC output. Checked with dmm
  2. I have wired the CD4094 side of the design on breadboard. It outputs 5V with no load. If i connect opto, it drops to 2.6 V. Opto emitter and collector acts like a closed switch but not completely. Current flow rate is not %100 from emitter to collector.

Take a look at the datasheet of CD4094. Can I connect Data, Clock and Strobe(latch) pins to arduino while VDD and OE pins connected to an external 12 V source?

My circuit works when number of chainned shift register is 3-4. But if I connect more CD4094 together, It doesnt work properly. There is a solution for this. I have calculated the required power to fully close the optocoupler emitter and collector. But I am not sure because i couldnt find required information from datasheet. My solution is to apply 12 V to VDD and OE. But my problem is this " if I connect the VDD and OE to 12 VDC, then High means 12 VDC and Low means 0 VDC. In this situation, i cannot change the register values with arduino. But I am not sure of it. I couldnt find any opposing or supporting information to my theory.

And by the way. There are 3 decoupling capacitor on PCBs. I dont know if it is related