Electronics > Beginners
Speed up BJT switching
David Hess:
Baker clamping the transistors will remove the storage time but raise the saturation voltage. A small signal schottky diode should work great but a better switching transistors than the BC327 would help also; the collector-emitter saturation voltage of the BC327 may be a little too high.
Another thing to do is capacitively couple the turn off signal into the base to pull the carriers out more quickly like C1 and C2 do for the level shifter transistors. Unfortunately, since you are using a level shifter which only pulls the base down, there is no suitable signal.
What I would do is configure drive transistors Q1 and Q2 in common base mode, probably by tying the bases to +5 volts, (1) and switch the emitters with the 5 volt logic signal through a resistor which sets the current so R12 and R13 are no longer needed although they might be kept at a lower value and bypassed. This configuration also switches the drive transistors very very quickly. Now the logic signal is the correct phase and a coupling capacitor between the logic signal and the base of Q4 and Q5 provides the correct drive to remove charge quickly.
(1) +5 volts will not leave much compliance for Q1 so a lower base voltage should be used.
Zero999:
--- Quote from: npelov on October 02, 2018, 08:48:58 pm ---@Hero999
Why does Ib go negative and how is R2 helping with it.
--- End quote ---
The base-emitter junction behaves like a tiny battery and R2 helps it discharge more quickly. The sign changes because current flows out of the base, through the base-emitter resistor. Also note that the transistor doesn't turn off, until the current has stopped flowing out off the base.
xavier60:
This photo shows the BCX52-16's Collector waveform in a Buck converter with 5v input at 160ma draw.
The Base current is now 20ma.
The choke is 50uh powdered iron running well into continuous current mode.
See how the transitions now look faster compared to when the load was resistive.
This is deceptive, there are actually hidden losses. The transistor will still be passing some of the tail current caused by the Storage delay after full voltage has developed across C-E. This is a common cause of loss when BJTs are switching inductive loads at high voltages.
In this case, because the voltages involved are low, the losses aren't that much.
When I put the Schottky diode between C-B, the efficiency drops from 78% to 67%.
Extra: The transistor also still has the sluggish turn on although it's now not visible. What's actually happening is because the inductor current is keeping the flywheel diode in conduction, the voltage stays clamped at near 0v until the rising transistor current matches the inductor current at which point the voltage suddenly rises.
xavier60:
Q5 has about 8v available to it, more than enough to drive a P-ch MOSFET.
This waveform is with an AO3407 MOSFET drawing 240ma from an 8v rail. The efficiency is 80%.
The Gate divider is 680\$\Omega\$/1K, upper/lower. Gate threshold is 1.2v.
Don't judge the switching losses by the slope of the voltage transition. Although the transitions look faster for the BJT, it was drawing current for some time before the On transition and after the Off transition.
T3sl4co1l:
Now put a complementary emitter follower after the resistors, and compare losses. ;)
Tim
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