![]() ![]() Similar to tape, the condition of the tube, its age, and the quality of its components and craftsmanship will all determine the point of saturation and what harmonics are generated once the tube is saturated. Whereas tape saturation may create a warm sound from high-frequency attenuation, tube saturation creates a warm sound from low-frequency amplification. Tubes often sound “warm” as well, but this more likely due to the second-order harmonic, which is closer to the fundamental frequency and lower in the frequency spectrum. ![]() Tube saturation creates a strong second-order harmonics, as well as both even and odd harmonics, depending on how significantly the tube is being saturated. In addition to creating soft-knee compression, saturation also results in harmonic generation harmonic generation depends primarily on the electrical components that are being saturated.Ī positive charge in a tube inhibits the flow of electrons, in turn compressing its electrical output This often results in compression that is less perceivable (depending on how much of the signal is being compressed). ![]() The compression employed by a saturated electrical component is soft-knee in nature, meaning that the signal will gradually be compressed, as opposed to instantly like with a hard-knee setting. How Does Soft-knee Compression Relate to Saturation? This will make the sound of the recording more detailed and complex, as the often masked aspects of an instrument, mix or master will become perceivable. I'm assuming the current in the LEDs are being controlled by a series resistor and not by trying to control the base-current and count of hFE.Limiting, a form of compression with a high ratio, attenuates dynamic peaks while amplifying low-level signals.Īfter compression, if you increase the amplitude of the entire signal, you have in turn increased the amplitude of the quietest aspects of the recording to the greatest degree. You can use the same base transistor for all of your transistors, because if your highest-current transistor can handle that base current, then so can all of the others (assuming the same transistors being used in similar environments, of course). It is far from uncommon for more than 50% of the heat to be due to the base current and not the collector current in switching applications. In hard saturation, an increase in base current causes about three to ten times the increase in power dissipation than the same increase in collector current does, simply because the base-emitter voltage is about three to ten times larger than the collector-emitter voltage. So a reasonable limit to use for the max base current is 10% of this, though you can probably be pretty sure you can push it to 20% without getting into too much trouble (as long as you allow proper heat sinking). Spec sheets usually give a max collector current value, at least for continuous operation. Most small-signal transistors are considered to be in "saturation" when hFE has dropped to 10 (power transistors often use a lower value). You want your circuit to be as insensitive to hfe as possible. Thanks.Ĭlick to expand.If you are trying to do switching applications, forget about hfe - it doesn't apply.Įven if you aren't, you shouldn't design circuits around hfe values since they vary all over the place for all sorts of reasons. That would correspond the the led with the highest Ic. I want to know if I can use the lowest calculated Rb for all the transistors. What I want to do is switch several different LEDs each with a different current Ic but try and use the same base resistance for each transistor. So increasing the base-emitter current will dissipate more power. I think the total current through the transistor is Ie = Ic + Ibe. I think it has to do with max power dissipation the transistor can handle. But I don't see that value on my data sheet. ![]() How far can I saturate a BJT biased as a switch without it blowing up? There has to be a limit to the base-emitter current it can handle otherwise a zero resistance would work to turn on the transistor. If I calculate Rb using hfe(min) to just saturate the transistor but in reality hfe is more, does Ic go up? Common sense would say no because its still limited by the series resistor Rc. I've been trying to work through some switching applications and I'm wondering what limits the collector current more: ![]()
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