Misc Electronic Questions

[Billy Mayo]

In most cases the designer will implement some sort of "dead-time" control, so both push-pull drivers cannot be on at the same time.

The On time and off time is controlled by a Microcontroller, so the dead time is also controlled by the microcontroller

In an audio amp, class AB required that one of the bipolar transistors in a push-pull stage to be partially on and thermally compensated. This eliminates "crossover distortion".

What components would thermally compensate ?

How does a transistor always Turns ON eliminate crossover distortion?

I thought the crossover distortion is the Dead time or when both transistors are OFF

DV/DT actually somewhat determines power dissipation

My manager was saying when the PAIRS of transistors are slightly off they both have different DV/DT , which is how many volts it it takes in one seconds

 

[KeepItSimpleStupid]

1. What components would thermally compensate ?

2. How does a transistor always Turns ON eliminate crossover distortion?

3. I thought the crossover distortion is the Dead time or when both transistors are OFF

1. Diodes
2. It's PARTIALLY ON, so a sine wave or any other, can smoothly go through zero.
3. Yes. See 2. Partially ON means never totally OFF, so no crossover distortion.

http://hackaweek.com/hacks/?p=332

 

[KeepItSimpleStupid]

1. What components would thermally compensate ?

2. How does a transistor always Turns ON eliminate crossover distortion?

3. I thought the crossover distortion is the Dead time or when both transistors are OFF

1. Diodes
2. It's PARTIALLY ON, so a sine wave or any other, can smoothly go through zero.
3. Yes. See 2. Partially ON means never totally OFF, so no crossover distortion.

http://hackaweek.com/hacks/?p=332

[latex]\frac{dv}{dt}[/latex] is actually the rise-time which is the amount of time between 10% of the waveform amplitude and 90% of the waveform amplitude. That's how it's defined. Could be say, for example 100V/uS or 100 V per microsecond. That's the rise time of my audio amp.

There is another relationship that I forget that relates risetime to bandwith.

 

[Billy Mayo]

So the diodes thermally compensate how?

I have seen those diodes before so i know what you're talking about

Push pull transistor pairs need to be thermally compensated because why?

 

[Billy Mayo]

DV/DT is actually the rise-time which is the amount of time between 10% of the waveform amplitude and 90% of the waveform amplitude.

Yes, but how does DV/DT apply to Switching Timing ON and off times? because if the push pull pair is slightly off, each transistor will have a different DV/DT rise time, so what will happen?

 

[audioguru]

Push pull transistor pairs need to be thermally compensated because why?

When a transistor heats up then it conducts more. In a push-pull audio amp then both output transistors conduct more which makes them hotter which makes them conduct more which makes them hotter which makes them conduct more which makes them ..... It is called Thermal Runaway.

So the diodes thermally compensate how?

When a diode heats then its forward voltage drops. So it reduces the base-emitter voltage of the output transistors so they do not conduct more when they also heat up.
The diode is silicon and the output transistor are also silicon so they perfectly thermally compensate.
Sometimes a transistor is used to replace the diodes and it is adjustable.

 

[KeepItSimpleStupid]

Vbe is a function of temperaturr. https://www.physicsforums.com/showthread.php?t=322302 If the diodes are at the "same" temperature as the transistors, then Vbe "tracks" or compensates.

If you don't, you get thermal run away. Simply stated: Repeat When a transistor heats up it conducts more. When it conducts more it heats up more. Until

 

[KeepItSimpleStupid]

Post 305: One device might get hotter than the other.

 

[Billy Mayo]

When a diode heats then its forward voltage drops. So it reduces the base-emitter voltage of the output transistors so they do not conduct more when they also heat up.

How ok you mean it CLAMPS the base emitter voltage so when the transistor heats up it doesn't DRIFT up and down in voltage , the diode CLAMPS it

 

[audioguru]

The two diodes are IN PARALLEL with the two base-emitter diodes of the transistors. The diodes CLAMP the voltage that biases the base-emitters.
When the heat from the transistors warm the diodes then their forward voltage drops which reduces the base-emitter voltages so the transistors do not conduct more. The diodes are supposed to be mounted on the heatsink of the output transistors so they sense the heat.

 

[KeepItSimpleStupid]

Without any compensation, "at higher temperature, a p-n junction has more carriers available due to increased thermal energy. Hence at 50 C, the junction is easier to drive than at 25 C. A given current incurs a smaller voltage drop at higher temp" from the earlier link.

 

[KeepItSimpleStupid]

In an audio amp, for instance, you basically have a transistor connected to a power supply rail and a very low ohm fusible resistor, e.g. 0.22 ohms.

Biasing sets some fixed current to flow through this resistor with no signal.

Now, suppose there wasn't any way to limit the the Vbe drop. Once the transistor starts to heat, it conducts more, thus increasing the current through the fusible resistor. Now, there are more electrons available because of thermal energy, it heats up even more. So, there is your thermal run-away problem.

---

A lot of audio amps has what's called a Vbe multipler. This multiplication is adjustable. If it is too high, the current through the fusible resistor will continue to increase until something breaks.

I built this amp (The Leach Amp) https://users.ece.gatech.edu/mleach/lowtim/ in the mid 80's with my own design changes. Here https://users.ece.gatech.edu/mleach/lowtim/2ndstage.html is a discussion of the Vbe multiplier.

My changes were mostly power supply related and component selection and protection. Nearly all the resistors are metal film. The bias adjust is 10 turn. It's packaged in a 2 RU case.

The power supply is 4 x 35 VAC at 3a each with 9600 uf of computer grade caps on each rail. The transformer is a custom Torroidal.

With ~40,000 uF at 50 V capacitance at turn on, there is quite an inrush current. An inrush current limiter was built into the system. The design is such that reversing the NPN and PNP output transistors will result in one fried resistor. With a fancier design, I could even stop that from happening.

There is no inherent VI limiting or other protection mechanisms except a 3A AGX fuse at the output and a four 3A 3AG fuse, one on each power rail. If one rail goes down, all rails go down and that magic resistor pops.

The AMP doesn't have a power switch. It's turned on by another device that uses a triac.

The audio media these days sucks and can't do the system justice.

 

[Billy Mayo]

Thanks for the info on that

I was troubleshooting a power supply , the DC buss would drift from 150v drifting up to 180volts

The MOSFET was the problem, the mosfet causes a Heatsink that screws onto the PCB

When I took the heatsink off and the screw, the DC buss was stable with no drift

I put a new MOSFET with a new heatsink and it works

My question is why did the MOSFET work without the heatsink , but when I put the heatsink on it would drift the voltage up ?

Mosfet is IRFBC30

 

[KeepItSimpleStupid]

I doubt the heatsink was the problem, however, it's possible that the mounting screw created stress somehow. You supposed to use a bevel washer to keep constant tension.

Other issues may have been a bad thermal insulator or too much paste or even an unintentional bridge made by the heatsink. The anodizing on a heatsink is generally non-conductive.

 

[Billy Mayo]

With the heatsink ON the MOSFET it fluctuates the voltage , creeping the voltage up starting at 150volts and ends at 180volts

With the headsink OFF the Mosfet stays at 150volts

 

[Billy Mayo]

What logic signals have an " X " looking signals

What is the " X " part called?

Look at the address signals , it seems like the address signals are 2 signals and the " X" part is when One signal is turning off and 2nd signals is turning ON

The " X " is a rise time and a fall time , crossing each other

What is this section called? when two logic signals cross like this?

How do you get this section very narrow?

Click here to see: The PD state has the " X " cross over section
https://www.google.com/search?q=log...nce-mixed-signal-methodology-guide%2F;800;343

Click to see:
https://www.google.com/search?q=log...2Ftikz%2Fexamples%2Ftiming-diagram%2F;423;500

 

[KeepItSimpleStupid]

X is a "don't care" state. This is best seen in datasheets. See the function table https://www.electro-tech-online.com/custompdfs/2013/05/74F153.pdf on page 3.

 

[Billy Mayo]

What u mean a Don't Care state?

 

[kubeek]

Don´t care means: you can have High or Low on the input and it doesn´t matter. The function of the circuit doesn´t care about the state of that input.
Here is an example from the function table of 74hc595.

 

[KeepItSimpleStupid]

@kubeek
Nice example. Shows other states too.