Circuit Description:

I am powering transistor with two power supplies. One supply is providing 5-VDC to its collector load through ammeter (+Ve - ammeter - load - collector - emitter - common/ground.) The decond supply is providing 0.0-VDC to 0.9-VDC to its base (base - emitter) with no resistor between.


A strange (to me at the moment) phenomenon is happening

  1. Start-up is normal - at 0Vbe there's no current through the ammeter
  2. Starting cut off is also normal as up to 0.4Vbe there's no current through the ammeter
  3. Starting active region is also normal as from 0.5Vbe to 0.8Vbe the current flows through 0mA to about 97mA
  4. Strange after getting in active region once, the transistor can't be turned fully off. Once conducted it keeps its collector-emitter junction turned on and about 40mA current keeps on flowing even at 0Vbe
  5. More Strange After getting active once the transistor shows two active regions one between 0Vbe-0.4Vbe and second between 0.5Vbe-0.8Vbe
  6. [0.0Vbe-0.4Vbe | 40mA-50mA Ice] active region is less linear its Vbe to collector current is not the same all the time but is likely (I'm guessing could be limitation of my equipment as my power supply volt meter has only one significant digit after decimal.)
  7. [0.5Vbe-0.8Vbe | 6mA-97mA Ice] is kind of linear
  8. Back to Normal Now even though there's 0Vbe but when I short circuit the base and emitter through a piece of wire then the transistor turns off and there's no collector current or if I pluck out supply probe from base then too the transistor turns off and there's no collector current

My Questions:

  1. Why is the transistor not turning off at 0.4Vbe to 0.0Vbe as this is its cut off region?
  2. Is there a problem with using dual power supply or is my equipment faulty?
  3. Am I still illiterate on this particular phenomenon and all transistors do have two active regions after getting activated once?

Equipment/Component Details:

  1. Transistors: C828 NPN, A1015 PNP
  2. Load resistors: Two 100 ohms (1/4 watts) resistors in parallel
  3. Collector emitter power supply: Custom power supply with 220VAC -> step down transformer 12 + 12VAC (One 12VAC side of transformer) -> bridge -> filter 2200uf/25V -> 7805 -> filter 104pf
  4. Base emitter power supply: 1501D by M&R
  5. Collector ammeter: DT9205M digital multimeter, selector switch is at its 200mA range

Circuit diagram:

enter image description here

  • \$\begingroup\$ It sounds to me like you might have damaged the transistor. They're extremely sensitive to overcurrent on the base-emitter junction. \$\endgroup\$
    – Hearth
    May 28, 2018 at 1:26
  • \$\begingroup\$ Put an ammeter on the base and repeat the experiment with a new (to be destroyed?) device. Check out the base current at 900 mV. \$\endgroup\$
    – jonk
    May 28, 2018 at 1:29
  • \$\begingroup\$ There are two relatively poorly discussed topics regarding BJTs. One is photogeneration of free charges within the BJT and the other is a tunneling/breakdown mechanism because of the modern high doping concentrations used. These are the only two things that come to mind in explanation of #8 under "problems." Your power supply is probably a one-quadrant supply and the wire actually shuts down the process where the power supply does not. \$\endgroup\$
    – jonk
    May 28, 2018 at 2:16
  • \$\begingroup\$ I put ammeter at base on same transistor and it was showing normal linear increase in base current from 0mA to about 1.xxmA till 0.8v and about at 0.9v base current was about 20mA Plus I've figured out the solution to the problem but still don't understand it's theory and I'm discussing this found solution in answer given by @Felthry \$\endgroup\$
    – Danish ALI
    May 28, 2018 at 15:08
  • \$\begingroup\$ @DanishALI I'm assuming this effect is resetable, as you said. And that it's not tied to a single device, but that it is a general observation across multiple devices. I've mentioned the odd resetable mechanisms I can think of. One thing you haven't mentioned is what the base current looks like when you bring the base voltage (with a power supply adjustment) back down to 0V. Does the ammeter show a base current, still? If so, how much and which polarity? \$\endgroup\$
    – jonk
    May 28, 2018 at 16:08

2 Answers 2


Anatomy of a half-fried transistor.

Imax=50mA @ Vce=0.12V at 25'C on a heatsink
Ipk=100mA ( short pulse) Vce(sat) is defined using Ic/Ib=10 for switch mode which or practical design use 10% of max hFE. Motorola used to call this base over drive factor.

If metallization migrates (contaminates) into semiconductor region it becomes a partial conductor so that it starts conducting with strange non-linear + linear current at lower voltage. Eventually it shorts out CE junction.

Unfortunately since diodes have a Neg Temp coefficient, the Vbe vs Ibe which appears to have a high Ohm bulk resistance, rises in current with Vbe voltage and generates heat in Vbe so as it gets hotter then threshold temp drops even faster leading to a possible thermal runaway with a 0 ohm voltage source.


simulate this circuit – Schematic created using CircuitLab

The forward voltage of a Si diode will drop by about 2.1 mV/°C (negative temperature coefficient, NTC) so if junction has risen 100'C Vbe drops 210mV so with 20 Ohms base resistance at Vbe=900mV now what is your base current?

Its a moving target as assumed temp keeps rising and depends how long you run it for.

Which is why you tested the device incorrectly leading to thermal failure by not using a 500 Ohm base R and 50 Ohm Collector to measure hfe saturation vs I.

see Vbe=0.8Vmax at 10mA with Vcb=5V meaning Vce=4.2V and linear hFE is 130 min @ Ic=2mA so Ib could be 10m/130 or less or 0.8mA thus bulk resistance above 0.6V is 200mV/10mA=20 Ohms but as 0.6V threshold drops with temp to 0.5V now the base current rises with T.

Now you are little wiser about reading datasheet specs and consequences of ignoring them.

Advice: consider temp rise at rated continuous current will be at 85'C with ambient cooling or more.

YOu should have stayed with just one 100 Ohm collector resistor and used 10 Ohm base resistor to measure current.

  • \$\begingroup\$ I had decided to take the OP as his word, regarding #8. Read it closely. Assume the OP actually did what was written and that the results were as written, too. Does anything you write fully explain that specific detail? Metal migration doesn't. That's permanent. Note the OP said they'd used the power supply and lowered its voltage to 0... still collector current. Then they shorted it with a wire, instead. It reset the situation. (As I read it.) Just curious how you might explain that specific detail. \$\endgroup\$
    – jonk
    May 28, 2018 at 4:31
  • \$\begingroup\$ I'm having difficulty understanding how metalization failure relates to the schematic you provided, but your theory in short "NTC in base-emitter junction lowers the input threshold" kind of make sense to me but the other theory on metal migration has a problem as the damage done is not permanent and my transistor is still working good after reducing the load from 50 Ohms to 100 Ohms \$\endgroup\$
    – Danish ALI
    May 29, 2018 at 1:23
  • \$\begingroup\$ if CE self biases with feedback current, no base current is needed until discharged to ground. There may be spurious non-linear behaviors unpredictable effects of lower R , higher C and this is why Huntron tracker is used for defect scanning using VI tracer. Using 50 Ohm load on 5V at Vbe=0.9 exceeds absolute max. Results will be unpredictable as is damage. stop wasting time ignoring specs \$\endgroup\$ May 29, 2018 at 2:32
  • \$\begingroup\$ You can even turn CMOS into an SCR by ignoring specs until power cycle if not overheated, Did you overheat junction? \$\endgroup\$ May 29, 2018 at 2:35

Looking at the datasheet for the 2SC828 (also called the C828), the maximum collector current for this device is 50mA. If you've passed 97mA through it, you've almost surely broken it; that's almost double the absolute maximum rating.

  • 1
    \$\begingroup\$ I'm betting the base current exceeded the collector current. \$\endgroup\$
    – jonk
    May 28, 2018 at 1:32
  • \$\begingroup\$ I can agree that the base of the transistor was taken out on the OP's first cycle of taking the base up to 0.9V with no current limiting. Unfortunately bipolar transistor data sheets will rarely show the max base current but will instead hint at it in the spec for the VceSAT specification. It is not uncommon that a base current over 5 to 10mA for a small signal device like the C828 can damage the device. \$\endgroup\$ May 28, 2018 at 3:08
  • \$\begingroup\$ @jonk Wouldn't be surprised, but the collector current is what the datasheet rates and what the asker said (though they might have meant the emitter current; \$I_{ce}\$ isn't very specific but the ammeter in the diagram is shown on the collector only) \$\endgroup\$
    – Hearth
    May 28, 2018 at 15:25
  • \$\begingroup\$ Fortunately (Device Wise) and Unfortunately (Understanding Wise) the transistor is still heart and healthy. First I changed the circuit with taking out old transistor and putting in new C828 and reducing collector load from 50 Ohms to 100Ohms and also put a 10 Ohms resistor in series with base. Now in first cycle I tested voltage at base as I was considering a voltage drop due to series resistor but the voltage at base were kind of same as voltage supplied from power supply from 0.0v to 0.7v and in second cycle I put ammeter between load and collector again and started varying voltage at base \$\endgroup\$
    – Danish ALI
    May 28, 2018 at 15:28
  • \$\begingroup\$ found active region b/w 0.5v to 0.7v and when I dropped voltage back to about 0.4v the collector current stopped hence rendering normal transistor operation then I even bypassed that 10 Ohms series resistor and still found transistor normal operation after that I put my old transistor in it's place and that transistor too showed normal operation which leaves me with only one theory that is "If we use transistor above Rated Continuous Collector Current then transistor may hold the base junction internally if once turned on" but is this theory real or not I'll have to do more research to find ou \$\endgroup\$
    – Danish ALI
    May 28, 2018 at 15:29

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