They are catch-diodes. Whenever you switch OFF a switch that allowed current to flow in a power inductor, you have to think of using catch-diodes or at least a snubber-circuit. A power inductor (choke) can generate very nasty voltage-spikes that can destroy an IC in a split second.
You may have seen such a catch-diode before in reverse-parallel with the coil of an electromagnetic relay.
You may have seen such a catch-diode before in reverse-parallel with the coil of an electromagnetic relay.
Are you talking about the reverse polarity protection diode? If I understand correctly, those diodes are to be shorted to make the power input capacitor more effective. Alternatively, I am thinking of a simple capacitance multiplier.
i am not really sure what you mean. i see it on the board that the upper layer seems to be shortet and if you measure on both pins of the diode you get an short.
kr
chris
My guess is that D8 serves a mechanical insulating purpose as a spacer. The rather large metallic heatsink is hanging not much above several SMD components. If the heatsink touches these components it may destroy the amplifier. And, the heatsink is riding on top of the small TDA7498E with only two elastic fixation points.
My suggestion is that D8 acts as an insulating spacer together with the two polarity diodes (that takes care of three corners of the heatsink) and one of the SMD components in the last corner.
Machine mounting of an extra diode costs nothing and D8 has a high (insulating) profile.
My suggestion is that D8 acts as an insulating spacer together with the two polarity diodes (that takes care of three corners of the heatsink) and one of the SMD components in the last corner.
Machine mounting of an extra diode costs nothing and D8 has a high (insulating) profile.
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Yes, voltage drop across the chip. Example:Typical voltage drop for lm3886 is 2 .5v.
hi
So you mentioned the point with the lowest voltage where the chip is not working?
If i remember right to my first try at this baord it must about 8V -9 Volts DC
Chermann, no I am talking about something completely different. Lm3886 has typical dropout voltage for positive side is 1.5 v and negative side is 2.5 v. So, if the supply voltage is say 24v, you will get maximum undistorted swing of 22.5v. From this value and with the help of total quiescent power supply current, you can calculate the possible power that will be drawn from the psu for undistorted output power. Then multiply it with 1.5 and you have your psu requirement for a given voltage and impedance load. source: neurochrome psu design article.
But I can't find the typical dropout voltage value for TDA7498. But I am guessing from the Trevor marshall's article that it's 3v.
But I can't find the typical dropout voltage value for TDA7498. But I am guessing from the Trevor marshall's article that it's 3v.
For a (linear) class AB amplifier there are limits for how close to the rails the output can go.
For a class D amplifier it is different because the output consist of two power switches driven into saturation. For the TDA7498, the typical resistance of each output switch is 0.2 Ohm. Thus, the output voltage swing from a TDA7498 is the full rail voltage MINUS the drop across each switch that the load current causes.
As a plain example: With a current of 3A (24V rail voltage and 8 Ohm) the drop across each switch is 3A*0.2Ohm=0.6V. Thus, in this case with 24V and 8 Ohm the output voltage swing is 0.6V to 23.4V.
For a class D amplifier it is different because the output consist of two power switches driven into saturation. For the TDA7498, the typical resistance of each output switch is 0.2 Ohm. Thus, the output voltage swing from a TDA7498 is the full rail voltage MINUS the drop across each switch that the load current causes.
As a plain example: With a current of 3A (24V rail voltage and 8 Ohm) the drop across each switch is 3A*0.2Ohm=0.6V. Thus, in this case with 24V and 8 Ohm the output voltage swing is 0.6V to 23.4V.
Yep, that is one of the reasons why class D is more energy efficient than class AB. If the driver circuit (internal in the chip) is designed (and used) correctly, the output switches can be driven to the low impedance specified in the datasheet. Often boot-strap techniques are used to achieve that (see second meaning in Bootstrapping (electronics - Wikipedia)).
For class AB it is more difficult as the output transistors have to be kept in linear operation and boot-strapping is much more difficult.
For class AB it is more difficult as the output transistors have to be kept in linear operation and boot-strapping is much more difficult.
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