Current leak in Opto-isolator

I don't quite understand what is going on, here. I have a design that uses a Vishay VO14642AT Solid State Relay to connect and disconnect an external battery to the main circuit. See the link below. The off-state leakage c urrent of this device is supposed to be less than 1 microampere, but when t he device is inserted into the circuit with no bias on the LED, the device is passing about 3ma with a 500mv drop across the output terminal when 9V D C is applied. It is enough so the DC-DC converter on the output of the iso lator is producing a full 5V out with no load. What am I doing wrong? Whe n the isolator is removed, the voltage drops to zero, so the leak does not appear to be anywhere else in the circuit.

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Reply to
Leslie Rhorer
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es a Vishay VO14642AT Solid State Relay to connect and disconnect an extern al battery to the main circuit. See the link below. The off-state leakage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, the devic e is passing about 3ma with a 500mv drop across the output terminal when 9V DC is applied. It is enough so the DC-DC converter on the output of the i solator is producing a full 5V out with no load. What am I doing wrong? W hen the isolator is removed, the voltage drops to zero, so the leak does no t appear to be anywhere else in the circuit.

afaict VO14642AT has more than 4 pins, how is it connected?

Reply to
Lasse Langwadt Christensen

er:

uses a Vishay VO14642AT Solid State Relay to connect and disconnect an exte rnal battery to the main circuit. See the link below. The off-state leaka ge current of this device is supposed to be less than 1 microampere, but wh en the device is inserted into the circuit with no bias on the LED, the dev ice is passing about 3ma with a 500mv drop across the output terminal when

9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing wrong? When the isolator is removed, the voltage drops to zero, so the leak does not appear to be anywhere else in the circuit.

It's always nice to include a link to the part.

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in the DC only mode it has 4 pins. (page 3) I'll guess some AC crud turning the led on. Through Q3? via gpio19?

George h.

Reply to
George Herold
:

t uses a Vishay VO14642AT Solid State Relay to connect and disconnect an ex ternal battery to the main circuit. See the link below. The off-state lea kage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, the d evice is passing about 3ma with a 500mv drop across the output terminal whe n 9V DC is applied. It is enough so the DC-DC converter on the output of t he isolator is producing a full 5V out with no load. What am I doing wrong ? When the isolator is removed, the voltage drops to zero, so the leak doe s not appear to be anywhere else in the circuit.

or it is connected backwards so the substrate diode turns on

Reply to
Lasse Langwadt Christensen

Looks like a neat part.

Have you measured the voltage across R12 when the problem occurs? The switching times are not stellar (likely due to charge on the gates, internally), so I would verify that oscillations are not fooling your measurements, and the SSR really is on. Partially, at least some of the time, but not enough to light your LED's.

BTW: This sure seems complicated. Is this just a battery UPS?

Reply to
mpm

ld:

hat uses a Vishay VO14642AT Solid State Relay to connect and disconnect an external battery to the main circuit. See the link below. The off-state l eakage current of this device is supposed to be less than 1 microampere, bu t when the device is inserted into the circuit with no bias on the LED, the device is passing about 3ma with a 500mv drop across the output terminal w hen 9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing wro ng? When the isolator is removed, the voltage drops to zero, so the leak d oes not appear to be anywhere else in the circuit.

What's the leakage.. 3ma @ 0.5V that does look like a diode.

George H.

Reply to
George Herold

Are you using pins 6 & 4 (pin 5 floating)? That is the normal way to use a part like this. Pin 5 is used only if you need some trick involving one o f the output mosfet's body diode.

Try putting 10K across pins 1 & 2, to ensure it's full off when you want it to be. A tiny amount of LED current can partially turning on the SSR.

Your circuit is asking the SSR to handle high in-rush of charging the dc-dc converter's input capacitance, from the 9V battery. You may have actually damaged your SSR, and now it's gone "leaky". Why not use a P-mos instead? What you are doing is called a "load switch", and nothing about your situ ation needs any opto-isolation. Lot's of high-side load switch chips are available that do controlled turn-on, which avoids high in-rush current, an d those chips can detect overload to auto-shutdown. ~Terry M.

Reply to
Terry

a part like this. Pin 5 is used only if you need some trick involving one of the output mosfet's body diode.

Pin 5 is used as neg output in DC only mode operation where you parallel th e MOSFETs and cut RON by half as well as double current rating.

it to be. A tiny amount of LED current can partially turning on the SSR.

dc converter's input capacitance, from the 9V battery. You may have actual ly damaged your SSR, and now it's gone "leaky". Why not use a P-mos instea d? What you are doing is called a "load switch", and nothing about your si tuation needs any opto-isolation. Lot's of high-side load switch chips ar e available that do controlled turn-on, which avoids high in-rush current, and those chips can detect overload to auto-shutdown. ~Terry M.

High Inrush from a 9V ???- not likely to stress this part.

Reply to
bloggs.fredbloggs.fred

es a Vishay VO14642AT Solid State Relay to connect and disconnect an extern al battery to the main circuit. See the link below. The off-state leakage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, the devic e is passing about 3ma with a 500mv drop across the output terminal when 9V DC is applied. It is enough so the DC-DC converter on the output of the i solator is producing a full 5V out with no load. What am I doing wrong? W hen the isolator is removed, the voltage drops to zero, so the leak does no t appear to be anywhere else in the circuit.

That's a really crazy looking circuit. The leakage is almost certainly thro ugh the 12V supply and Q4. Place a 1K from base to emitter on that transist or, which ought to take care of the leakage. And I really am not getting ho w you turn on the SSR when its input LED current is derived from the 5V DC- DC output???

Reply to
bloggs.fredbloggs.fred

uses a Vishay VO14642AT Solid State Relay to connect and disconnect an exte rnal battery to the main circuit. See the link below. The off-state leaka ge current of this device is supposed to be less than 1 microampere, but wh en the device is inserted into the circuit with no bias on the LED, the dev ice is passing about 3ma with a 500mv drop across the output terminal when

9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing wrong? When the isolator is removed, the voltage drops to zero, so the leak does not appear to be anywhere else in the circuit.

rough the 12V supply and Q4. Place a 1K from base to emitter on that transi stor, which ought to take care of the leakage. And I really am not getting how you turn on the SSR when its input LED current is derived from the 5V D C-DC output???

I assume the hairball is so that it latches ON when you attach 12V and stay ON with either supply, but you can turn OFF under software control

Reply to
Lasse Langwadt Christensen

...@gmail.com:

t uses a Vishay VO14642AT Solid State Relay to connect and disconnect an ex ternal battery to the main circuit. See the link below. The off-state lea kage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, the d evice is passing about 3ma with a 500mv drop across the output terminal whe n 9V DC is applied. It is enough so the DC-DC converter on the output of t he isolator is producing a full 5V out with no load. What am I doing wrong ? When the isolator is removed, the voltage drops to zero, so the leak doe s not appear to be anywhere else in the circuit.

through the 12V supply and Q4. Place a 1K from base to emitter on that tran sistor, which ought to take care of the leakage. And I really am not gettin g how you turn on the SSR when its input LED current is derived from the 5V DC-DC output???

ay ON with either supply, but you can turn OFF under software control

Looks like he can't turn it on with the SSR unless the converter is already on, the only control is to turn it off. But if all the GPIO junk is runnin g off the DC-DC 5V then he possibly has a logic lockup flaw.

Reply to
bloggs.fredbloggs.fred

.@gmail.com:

:

hat uses a Vishay VO14642AT Solid State Relay to connect and disconnect an external battery to the main circuit. See the link below. The off-state l eakage current of this device is supposed to be less than 1 microampere, bu t when the device is inserted into the circuit with no bias on the LED, the device is passing about 3ma with a 500mv drop across the output terminal w hen 9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing wro ng? When the isolator is removed, the voltage drops to zero, so the leak d oes not appear to be anywhere else in the circuit.

y through the 12V supply and Q4. Place a 1K from base to emitter on that tr ansistor, which ought to take care of the leakage. And I really am not gett ing how you turn on the SSR when its input LED current is derived from the

5V DC-DC output???

stay ON with either supply, but you can turn OFF under software control

dy on, the only control is to turn it off. But if all the GPIO junk is runn ing off the DC-DC 5V then he possibly has a logic lockup flaw.

if the battery is only intended to provide power for a safe shutdown it mak es sense that it cannot be turned on if there is no 12V power

Reply to
Lasse Langwadt Christensen

Yeah, bingo. I feel a little stupid. The traces to the relay are backwards. I am going to modify them to make sure that is all that is wrong, but I suspect it is the entire problem. I'll let everyone know.

Reply to
Leslie Rhorer

It has six pins, but in DC operation pins 4 and 6 are shorted together. Pi n 3 is NC.

That makes sense. From now on I will. Thanks.

Well, one can wire it that way, but in that case half of the relay is being unused, and the total current carrying capability is cut by half. With pi ns 4 and 6 shorted, one can enjoy the maximum total abilities of the chip.

No, it's quiet. I checked with a meter and even shorted pin 1 directly to ground with no change. Besides, since the board is not attached to the par ent board and the only source of power is a NiMH battery, and since it is a small printed circuit board (no jumpers), stray AC was pretty unlikely in the first place.

Lasse's idea is correct, at least to the point I did wire it backwards. We 'll see if it is the entire problem.

Reply to
Leslie Rhorer

Lousy schematic rendering. Wire "crossing" wire with a dot :-( ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

snipped-for-privacy@gmail.com:

at uses a Vishay VO14642AT Solid State Relay to connect and disconnect an e xternal battery to the main circuit. See the link below. The off-state le akage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, the device is passing about 3ma with a 500mv drop across the output terminal wh en 9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing wron g? When the isolator is removed, the voltage drops to zero, so the leak do es not appear to be anywhere else in the circuit.

through the 12V supply and Q4. Place a 1K from base to emitter on that tra nsistor, which ought to take care of the leakage. And I really am not getti ng how you turn on the SSR when its input LED current is derived from the 5 V DC-DC output???

tay ON with either supply, but you can turn OFF under software control

isn't all the crossings with a dot connections?

Reply to
Lasse Langwadt Christensen

a part like this. Pin 5 is used only if you need some trick involving one of the output mosfet's body diode.

Not true. See page 3:

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it to be. A tiny amount of LED current can partially turning on the SSR.

That's not a bad idea in general, and I had already considered it, but I th ink it shouldn't be necessary in this case. We'll see.

dc converter's input capacitance, from the 9V battery.

Actually, not, although I must allow it is not necessarily obvious from the schematic. GPIO 19 can never be taken high until after the parent board ( a Raspberry Pi 3 computer) is active, which in turn can only happen by powe ring the DC-DC converter with the main input, nominally 12V or so. Thus th e main supply will always suffer the in-rush current, and it is more than u p to the task. Once the Raspberry Pi is up and running, GPIO 19 is asserte d via software, and the battery enjoys an in-rush current from U1 equal to the bias voltage of D9 (nominally 0.7V) divided by the value of R10, assumi ng the offset voltage of U1's inputs is negligible and its gain is fairly h igh, of course. In this case, that results in a charging current of ~21ma. This assumes the main power supply is delivering more than about 1.4V hig her than the battery voltage. Of course, it is possible a situation could exist where the main power supply could be delivering less than the battery terminal voltage at startup, in which case there could be some inrush to t he DC-DC converter, but it is highly unlikely. Even then, the voltage diff erence between the battery, less the drop from D3, and the main supply is n ot going to be more than 3 or 4 volts, and the input caps on the converter are not very large. The output caps would have to be charged in any case, because the computer has to be functioning in order for GPIO 19 to be asser ted.

In the very rare instance where the main supply is bad but still producing a bit more than 7V and enough current to still power up the RPi, it is some what possible I might have to replace the opto-isolator, but it's still a l ot cheaper than the main supply. An extra $2 won't kill anyone.

You may have actually damaged your SSR, and now it's gone "leaky". Why n ot use a P-mos instead? What you are doing is called a "load switch", and nothing about your situation needs any opto-isolation. Lot's of high-side load switch chips are available that do controlled turn-on, which avoids h igh in-rush current, and those chips can detect overload to auto-shutdown. ~Terry M.

Reply to
Leslie Rhorer

Yes. BAD way of doing schematics... dots have a way of vanishing. I think there's a MIL-standard forbidding such. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

rough the 12V supply and Q4.

Not when te 12V supply is not attached. I should have been clearer in sayi ng the 9.6V battery is the only source of power at this point in testing. I don't see how Q4 could be a culprit in any case. If +12V is on its input , then it will be fully saturated by design, not leaking. If not, then I f ail to see how leakage through Q4 would bias O1's LED.

rent is derived from the 5V DC-DC output???

By design intent O1 should only be active after the +12V supply has booted the computer and the RPi has asserted GPIO 19 in software. Once done, the circuit has latched the SSSR through the RPi. If +12V fails, then by the f act the battery is feeding the input of the DC-DC converter, +5V remains in tact, and the computer remains active, despite having presumably lost AC po wer. GPIO 17 informs the computer of the loss of mains power, and at a lat er time it will affect its shutdown procedure. When the computer shuts dow n, GPIO 19 is de-asserted, O1 is deactivated, and the battery is isolated f rom everything.

Reply to
Leslie Rhorer

on:

snipped-for-privacy@gmail.com:

te:

that uses a Vishay VO14642AT Solid State Relay to connect and disconnect a n external battery to the main circuit. See the link below. The off-state leakage current of this device is supposed to be less than 1 microampere, but when the device is inserted into the circuit with no bias on the LED, t he device is passing about 3ma with a 500mv drop across the output terminal when 9V DC is applied. It is enough so the DC-DC converter on the output of the isolator is producing a full 5V out with no load. What am I doing w rong? When the isolator is removed, the voltage drops to zero, so the leak does not appear to be anywhere else in the circuit.

nly through the 12V supply and Q4. Place a 1K from base to emitter on that transistor, which ought to take care of the leakage. And I really am not ge tting how you turn on the SSR when its input LED current is derived from th e 5V DC-DC output???

d stay ON with either supply, but you can turn OFF under software control

I agree, I thought you meant there was a dot where there shouldn't be one

Reply to
Lasse Langwadt Christensen

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