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Sometimes after long time working correctly and turning on and off inductive load, Darlington NPN transistor (ONSEMI BD679) which should follow signal (IN4), fails to work and turns on inductive load continuously. It's part of my schematic: NPN transistor switchs inductive load.

It usually works correctly and this problem maybe happen after a while. What issues can do this?

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    \$\begingroup\$ How hot is it running? What's the max switch frequency? Is the diode intact? \$\endgroup\$ – winny Jun 4 '18 at 14:57
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    \$\begingroup\$ Is the BJT thereafter destroyed or can the situation be fixed by powering off something? \$\endgroup\$ – Andy aka Jun 4 '18 at 14:59
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    \$\begingroup\$ @Andyaka yes, after that, BJT is destroyed permanently and is always on. \$\endgroup\$ – Ata Jun 4 '18 at 15:08
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    \$\begingroup\$ @Andyaka it is a simple inductor which i winded wire around a toroid. \$\endgroup\$ – Ata Jun 4 '18 at 15:28
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    \$\begingroup\$ What's the purpose of the inductor? As the circuit doesn't appear to have any output of any sort, does it actually matter whether Q6 works or not? \$\endgroup\$ – Finbarr Jun 4 '18 at 15:43
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Switching an inductive load, even with the flyback diode, is hard on the transistor, and you may be seeing an SOA (Safe Operating Area) issue that is causing the transistor to fail. When the transistor is turning off there will be a time during which the Vce is >30V and the current is still 1.2A. Here is the SOA for a beefier Darlington- the TIP122:

enter image description here

It's also possible that there is stray inductance in your setup (or inadequate bypassing of the 30V rail) that is causing the transistor to see overvoltage.

For the first problem, the most obvious solution is to use a beefier transistor (and, though I don't think it's the problem, you might want to use a 1N5819 rather than a 1N4007 for the flyback diode).

For the second problem, make sure your flyback diode is connected near to the transistor, and that there is sufficient capacitance on the +30V rail to prevent overvoltage. This is a function of the energy stored in the stray inductance. Alternatively, put a TVS directly across the transistor E-C leads.

You can use an oscilloscope and measure the E-C spike at turn off to determine if the second issue is a likely cause.

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  • \$\begingroup\$ Thanks for your recommendations. I actually took in account all the matter you mentions. What i guess is the transistor BD679 is not suitable for switching inductive load. As datasheet said: "This series of plastic, medium−power silicon NPN Darlington transistors can be used as output devices in complementary general−purpose amplifier applications." But some other transistor such as MJE13003 which its datasheet said: "These devices are designed for high−voltage, high−speed power switching inductive circuits where fall time is critical..." \$\endgroup\$ – Ata Jun 4 '18 at 18:03
  • \$\begingroup\$ The lack of an SOA curve is a bit worrisome in an inductive switching application. \$\endgroup\$ – Spehro Pefhany Jun 4 '18 at 18:25
  • \$\begingroup\$ @Ata Those are just marketing phrases, just because it's not designed for switching inductive loads doesn't mean you can't use it to switch inductive loads. But then again, maybe you can't; you'd have to check the actual data to be sure. \$\endgroup\$ – immibis Jun 5 '18 at 1:21
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I guess the problem starts with the lack of base current. A 30 volt supply via a 10 k resistor can provide 3 mA into the base of the BD679 darlington. However, if you look at the Collector-emitter saturation voltage in the data sheet it says: -

2.5 volts (IC = 1.5 A IB = 30 mA)

Because you are only providing one-tenth of that recommended base current to properly turn on the transistor it may be dropping 3, 4 or 5 volts whilst powering your solenoid and this results in maybe 3 to 5 watts of power being dissipated over a period of 5 seconds whilst activated.

there is no heatsink

So it's going to be fairly hot when you turn it off then, if you rapidly switch it on again it is possibly failing due to the junction temperature being exceeded.

Transistor doesn't warm up in that 5 seconds of turning inductor on

I'm not buying that. In the Multicomp data sheet or this one for the device I read: -

Junction to Ambient in Free Air 100 °C/W

So, without a heatsink or other form of adequate cooling you could be damaging the device. I think this is your main problem.

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  • \$\begingroup\$ Thanks for your recommendations. I actually took in account all the matter you mentions. What i guess is the transistor BD679 is not suitable for switching inductive load. As datasheet said: "This series of plastic, medium−power silicon NPN Darlington transistors can be used as output devices in complementary general−purpose amplifier applications." But some other transistor such as MJE13003 which its datasheet said: "These devices are designed for high−voltage, high−speed power switching inductive circuits where fall time is critical..." \$\endgroup\$ – Ata Jun 4 '18 at 18:05
  • \$\begingroup\$ No, the BD679 is quite capable of switching inductive loads. You have the back-emf diode so that should not normally be a problem. It's purely down to burning the junction because you are not getting heat away fast enough (despite you thinking it runs cool) and not driving the base hard enough. \$\endgroup\$ – Andy aka Jun 4 '18 at 18:09
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    \$\begingroup\$ I endorse this analysis - hence my earlier question about a heatsink. Transient stresses caused by short but intense heating cycles will bring about the death of virtually any semiconductor. \$\endgroup\$ – henros Jun 4 '18 at 22:04
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If the op amp acts as a perfect switch, this should work. There is a possibility of heating or oscillation, though, because there is no clear indication of the slew rate of the IN4 input signal, and the LM258 can be a slow-slewing drive to the transistor base.

Since the BD679 can be carrying over an ampere of current, it will self-heat when partly ON. A quick ON-to-OFF transition is important; consider adding some positive feedback, pin 7 to pin 5 with a 500k resistor, and IN4 to pin 5 with a 5k resistor, to make the transitions swift.

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