I am designing a flyback driver using a salvaged power transistor that was originally the driver transistor of a flyback in a ctr screen.

This power transistor is the 2SC5681

I want to drive this transistor with an IC, and the high base current required by this transitor (1-2A) cannot be directly supplied by the IC. So i need an intermediate stage to drive the transistor base.

First, i planned to put an npn transistor before the 2SC5681 in a darlington configuration to amplify the current gain. Like this

but when I reviewed the 2SC5681 datasheet, i saw that this transistor needs a positive Ib (which is normal) AND a some negative current at the base for switch off (fast) In Red

so my question is, can i use the darlington configuration to achieve the base drive despite this negative current? or what kind of simple base driver can i use for this transistor?

I had thought of using a push pull configuration like a mosfet gate drive system but is it adapted?

push pull gate drive ?


  • \$\begingroup\$ Your proposed drive circuit is not going to be very effective at sucking charge out of the BJT base unless you drive it below ground and connect it to to a negative supply. \$\endgroup\$ Commented Feb 1, 2018 at 21:30
  • \$\begingroup\$ Welcome to EE.SE. Use the push-pull circuit as it is dependable and a common solution. Base drive current should be limited to 5 mA if using 2N3904 and 2N3906. Rgate is usually 22 to 33 ohms. Vcc cannot be more than 15 volts. \$\endgroup\$
    – user105652
    Commented Feb 1, 2018 at 21:30
  • \$\begingroup\$ @SpehroPefhany. OP should put a 1N4148 diode on Rgate for that reason. Anode to mosfet gate. \$\endgroup\$
    – user105652
    Commented Feb 1, 2018 at 21:32
  • \$\begingroup\$ Why does your first circuit have Vcc at 200 volts? Must be a typo. \$\endgroup\$
    – user105652
    Commented Feb 1, 2018 at 21:34
  • 2
    \$\begingroup\$ BTW, searching this part and finding it distributed by Rochester and not Digikey/Mouser makes it likely this part is at or beyond end of life, and if you are planning to be able to reproduce this circuit in the future you might want to find a different transistor. \$\endgroup\$
    – The Photon
    Commented Feb 1, 2018 at 21:48

2 Answers 2


When transistors are used in Darlington configuration they are usually used with resistors R1, R2 to allow the driver to provide the reverse recovery current to Q2 and to prevent leakage (amplified by \$\beta^2\$) from turning on the transistors.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ +1 for pointing out a crucial detail, and why Darlingtons are never used in audio amplifiers. They cause a terrible THD and slew-rate spec. \$\endgroup\$
    – user105652
    Commented Feb 1, 2018 at 23:47
  • 1
    \$\begingroup\$ @Sparky256 I've seen plenty of audio amplifiers with Darlington output stages. Always with negative feedback of course. \$\endgroup\$
    – τεκ
    Commented Feb 2, 2018 at 0:12
  • \$\begingroup\$ @TEK. Yes, and lot of feedback was used in the first generation transistor amps, as they had poor spec's to begin with. A Darlington made with two transistors is very common, especially in high-powered amplifiers. \$\endgroup\$
    – user105652
    Commented Feb 2, 2018 at 0:32
  • \$\begingroup\$ @Sparky256 I agree that Darlingtons are not advised today. Period. Sziklai, yes. Darlington, no. In the past, power PNP wasn't an option. But today, while they still fall short a bit, they at least are no longer unobtainium. So there are options now when there wasn't so much, before. But even then, one might stay with Darlington on one quadrant, but still use Sziklai on the other. The VBE multiplier requirements are much worse for Darlington and unlike the Sziklai where only the much cooler base drive BJTs are included with the VBE mult, the Darlington puts all the VBEs in the multiplier loop. \$\endgroup\$
    – jonk
    Commented Feb 2, 2018 at 9:17

To turn off a bipolar NPN transistor fast, it is useful to have a negative base bias (reverse the base current) to remove the stored base charge. The easiest way to accomplish this is to raise the emitter voltage slightly when the transistor is ON, so that a single-power-supply signal can sink base current.


simulate this circuit – Schematic created using CircuitLab

This presumes that you can find resistor and C1 values suitable for the collector loads being driven by Q1.

  • \$\begingroup\$ thanks for your reply , what is the use of R1 ? that can reduce the current required in Q1 (10A) ? . and where the reverse base current will be sink ? \$\endgroup\$ Commented Feb 2, 2018 at 11:49
  • \$\begingroup\$ @MaximeLoiseau R1 and the emitter current (which presumably is intended to be limited) biases/charges C1. One probably wants circa 2V on C1, so the 'logic' driver sinks base current from Q1. With emitter at 2V, the base-emitter capacitance is discharged through D1 and R2 when logic is LOW. \$\endgroup\$
    – Whit3rd
    Commented Feb 2, 2018 at 21:04
  • \$\begingroup\$ @MaximeLoiseau For 10A current peak, R1 value of a few tenths of an ohm would suffice. \$\endgroup\$
    – Whit3rd
    Commented Feb 2, 2018 at 21:08

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