# Does a BJT optocoupler have an inherently short circuit proof output?

I want to use an NPN optocoupler (ACPL-214) to drive an offboard relay. I know how to size the input resistor to limit the LED current to 20mA. This will cause the output current to be in a certain range (depending on CTR and other factors.)

Question 1: Is this opto output short circuit proof (since the input LED current is limited?)

Question 2: Does it matter if the opto is used in a high side configuration (collector connected to battery, emitter connected to the relay coil) or low side configuration (collector to relay coil and emitter to GND?)

High Side

Low side

Question 3: Do the answers above apply to normal silicon BJT transistors if the base current is limited?

The question might be trivial but I fear that shorts can cause some kind of current transient that could blow the transistor.

• Can you clarify what you mean with "short circuit"? Is the coil going to be short circuited, or is the output of the opto being short circuited? – Vladimir Cravero Aug 31 '20 at 18:59
• the output of the opto. thw wire going out of the board (opto_emitter) could hit the metal chassis due to mis-wiring or something . ( the chassis is grounded) . will this fry the opto ? – Eng Sam Aug 31 '20 at 19:08
• I will diagrams – Eng Sam Aug 31 '20 at 19:15
• but I fear that shorts can cause... You're the typical beginner that worries about everything. Just learn how others to their projects, work neatly and check things so that you avoid shorts etc. Also, don't use a supply that will deliver 5 A when it is shorted, instead, guess what current your circuit needs and use a supply with a current limit and set that limit appropriately. When you short things and too much current/voltage happens then components get damaged. So avoid that situation. Trying to make everything "short circuit proof" is impossible. – Bimpelrekkie Aug 31 '20 at 19:16

Is this opto output short circuit proof (since the input LED current is limited?)

Yes and no. :) Yes for small currents, no for larger currents. At low currents the output impedance is much higher (current source) than at high currents.The models here are accurate only in the low current range.

Does it matter if the opto is used in a high side configuration (collector connected to battery, emitter connected to the relay coil) or low side configuration (collector to relay coil and emitter to GND?)

It doesn't matter here at all, because we don't have to drive the base.

Do the answers above apply to normal silicon BJT transistors if the base current is limited?

Yes, just like here. The output characteristics of the BJT are steeper at higher base currents.

In this respect, PNP is worse than NPN. What is shown here forms a selected 'complementary' pair.

Regardless, linear optocouplers are not really good in switching mode. There are optical relays with MOSFET output for switching purposes.

• thank u this answers my question. So , even BJT transistor needs short circuit protection in the form of something taht will decrease the base current when higher than normal current are passing in the collector – Eng Sam Sep 1 '20 at 15:26
• Yes that's correct. – csabahu Sep 1 '20 at 15:34

The output of the opto-coupler is NOT short-circuit proof, in that if the relay coil is shorted or draws too much current the opto transistor will blow. If you are asking if the LED is short-proof, that would be correct. The series resistor determines the LED current. These are normally IR LEDs with just a 1.1 volt drop so if it is shorted the current changes by a small amount.

By the way, usually 5 mA of current will drive the LED fully ON, so 20 mA is wasting power as heat. Opto's with high gain (400 > 500) will work at 2 mA.

Normally the collector drives the relay with the emitter ground so you take advantage of the transistors current gain, hFE or $$\Beta\$$. You may need a booster transistor as opto-couplers are not known to have a source or sink current over 10 mA, so be sure to read the datasheets.

I added a simple schematic to show the OPTO being used with a NPN bjt to boost the drive current to over 100 mA if required. R3 makes sure relay is OFF if opto-coupler is OFF. R4 limits drive current to base of Q1.

EDIT: Opto-couplers CANNOT supply enough current for common relays, so you will need a transistor with much stronger drive current, and its emitter should be grounded. The wire you have in RED has NOTHING to do with the LED current, which can be 5 mA. The relay may need 10 to 30 mA @ 12 volts, beyond what the opto-coupler by itself can drive. If you do not understand this, please find another hobby or at least study a lot of simple circuits so you are certain of the outcome when you build something.

simulate this circuit – Schematic created using CircuitLab

• thank u . your first sentence is the answer , but can you explain why ? The rest of the answer I do understand . but why is the opto-coupler output not short circuit proof . If the input current is 10mA for example , and CTR is 50% , output should be 5mA approximately . isn't this current the maximum current that will be generated by the optocoupler and no more current will pass even if the relay coil short , where will the excess current come from ? – Eng Sam Aug 31 '20 at 19:22
• I say this because this is to my limited understanding the way linear current limiting works , but the transistor in linear region by limiting base current . something like the analogy of a water "faucet" controls the flow . – Eng Sam Aug 31 '20 at 19:25
• There is NO excess current unless you short the power feed, which should be enough to drive the relay ON. Vcc can be as high as 24 VDC and still work just fine. – user105652 Aug 31 '20 at 19:33
• Vcc can range from 3.3 V to 24 V, but should match the relays coil voltage. Be sure D1 is installed with correct polarity. – user105652 Aug 31 '20 at 19:38
• I will amend my answer. Opto-couplers CANNOT supply enough current for common relays, so you will need a transistor with much stronger drive current, and its emitter should be grounded. The wire you have in RED has NOTHING to do with the LED current, which can be 5 mA. The relay may need 10 to 30 mA @ 12 volts, beyond what the opto-coupler by itself can drive. If you do not understand this, please find another hobby or at least study a lot of simple circuits so you are certain of the outcome when you build something. – user105652 Aug 31 '20 at 23:06