# Tag Info

30

These transistors are devices which are highly optimized for use in ultra-low-cost and high-volume consumer products- off-line fluorescent lamp ballasts. The anti-saturation network (for fast switching) and free-wheeling diode are integrated onto one die, minimizing the component count and silicon area required (the diagram shows two parallel 'fingers' of ...

13

The transistor boasts "high switching speed" it gets this by having the lower transistor turn on and steal base current when the main transistor approaches saturation. Thus it is a type of Baker Clamp

13

LED D2 is green and it needs about about 2.0V to light up. LED D1 is red and it needs about 1.6V to light up. And when the transistor is turned on, it has a Vce drop of about 0.2V. Adding the voltage drops of the transistor and red LED D1, that is still less than what green LED D2 needs to work, so there is not enough voltage over green LED D2 to turn on.

9

That is a strange part. At first I thought it was an integrated Sziklai pair, a type of Darlington with a lower saturation voltage. But your part is different. When the collector goes below the emitter, the first transistor will turn on and short the 2nd transistor's base to its emitter, completing its turn off much faster by sucking out its base charge. ...

7

Revised after some thoughts, "saturation clamping" makes more sense. The way how the PNP works in the circuitry (clamping) is effectively what Baker Clamp is for. Thus, the manufacture's drawing seems agreeable. It would seem to me the moment the base of the NPN goes high, the collector goes low, which in turn pulls the PNP base low, whereby it ...

5

As DKNguyen points out, your second circuit will not work, because you are operating the P-channel MOSFET as a source follower. In this configuration its source potential is always lower than its gate by an amount approximately equal to its gate-source threshold voltage ($V_{GS(TH)}$). Your original (top) design will work quite well, for low load ...

5

I think you have misinterpreted the circuit slightly. It reads on the web page that the transistors are turned on one at a time for driving the cathodes of each layer. The current for anodes for the layer is provided by other chips. Most likely the resistors are just simple pull-ups in the kilo-ohm range to discharge parasitic capacitances and to pull up the ...

4

You will be exceeding the absolute maximum voltage on the datasheet (5V). Unless you feel like doing a lot of testing and characterization on many samples from different manufacturers under a variety of conditions in order to reduce the risk, a diode may be the least expensive safe solution. If the sole pull down is that 1M then you can’t stand much beta ...

3

LEDs (all diodes) are non-linear devices. If you look at a voltage-current curve, you will note that the relationship is not a straight line as it is with a resistor: From 0V, as voltage increases, not much happens until the forward voltage (Vf) of the diode is reached. After that point, current increases exponentially. If there is too much current, the LED ...

3

Success or failure as an engineer is judged by what you can produce and accomplish, not by what you've memorized. If you need to use equations often, and have trouble memorizing them, print out a cheat sheet and post it at your workstation, or go paperless and bookmark a reference page in your web browser. It's more important that you know how and when to ...

3

The only reason I can think of for requiring symmetry in the gates' charging and discharging currents is that you require some corresponding symmetry in the rise and fall (sink/source) characteristics at the output side of the MOSFET(s). In the schematic you provide, for example, R4 is responsible for pulling up the drain voltage of the MOSFET M1, which ...

3

It's not 'bad' to have asymmetrical drive. In some cases it might be beneficial or necessary. While your PNP may have a lower current capability or rating than the NPN, this current flows for only a brief time, so is unlikely to actually damage it. Secondly -- it's not a good idea (not very controllable) to limit a BJT collector current by limiting the base ...

2

"I want to make an MCU operated buck converter with the lowest operating current possible since most electronic shops near me only offered buck converters with an operating current of 10 mA and above." Here is a part that can be used, lower quiescent current. https://www.befr.ebay.be/itm/234047609405?hash=item367e531a3d:g:B6kAAOSwI8Jg0Neo MPM3610 ...

2

The Microchip article must be taken in context which is for a half-bridge driver (lower FET is used as a synchronous rectifier) for a buck switching power supply. They may be interested in controlling shoot through (where both FETs are briefly ON during the switching part). Your circuit needs will dictate what drive requirement is required.

2

Additionally, if the first circuit works, could I remove the the parallel diode and use the body diode of the MOSFET instead? This would involve letting current flow from drain to source when the MOSFET is on though so I'm not sure. Provided the MOSFET has a sufficiently low Gate drive voltage specification it will work fine. Initially the voltage between ...

2

This is one of possible approachs how to drive a motor with reverse capability with 3v3 controll signals. V1 and V2 are your GPIOs. Be sure they never be high together so do a software safety. Drive it with 1ms dead time at least. (Example after V1 become low wait 1ms to set V2 high) When V1 high and V2 low ---> CW direction When V2 high and V1 low --->...

2

Can anyone think of a fix for these issues? The fix is a precision shunt regulator based around an emitter follower PNP power transistor being driven by an op-amp. The op-amp has a precision voltage reference on the non-inverting pin and the inverting input is connected to your raw supply: -

2

You can still be an engineer if you don't memorise all those configurations, and the most important formulae, but you won't be a very good one. There's always Google, but if you have to look stuff up every time you see a transistor or need to design something with one, then it will take you forever to actually complete anything. If you don't have a library ...

2

Often, the whole point in a current mirror is to have identical currents in both sides of the mirror. That is the reason for using an enhanced beta mirror, a Wilson mirror or an improved Wilson mirror (as you have there). The usual reason for putting equal value resistors in the emitters is for degeneration to reduce the effects of any Vbe mismatch in the ...

1

R1 and R2 are not matched. So, if the voltage drop across each is the same, there needs to be R1/R2 times the current in R2 than in R1.

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