Misc Electronic Questions

[Billy Mayo]

Transistor controlling the current?
View attachment 89516

Transistor as a comparator? or a switch
View attachment 89517

Transistor as a comparator? or a switch
View attachment 89518


Why did the designer use transistors instead of a FET?

Using a transistor as a switch is different than using a FET as a switch? what is the difference?

 

[Billy Mayo]

Transistor controlling the current?


Transistor as a comparator? or a switch


Transistor as a comparator? or a switch



Why did the designer use transistors instead of a FET?

Using a transistor as a switch is different than using a FET as a switch? what is the difference?

 

[Billy Mayo]

Power Meters, what have you used them for , what kind of circuits?

What were you measuring when using a power meter?

 

[Billy Mayo]

I also see Diodes from the base of a transistor , going to ground

The Diodes on the base going to ground protects it from negative voltage? what does that mean, because a diode is only 0.5 volts so it any negative voltage will be much more than 0.5 volts but it will conductor the transistor if the diode wasn't there.

They put the diodes on the base going to ground , so negative voltage won't turn on the transistor?

 

[Billy Mayo]

What harmonics are clean and what harmonics are dirty?

Are the clean harmonics 1,3,5,8?
the dirty harmonics 2,4,6,7?

The AC outlet from the wall has harmonics too

The AC outlet is 60 hz

The harmonics should be what for a clean AC outlet 60hz signal?

 

[audioguru]

If you look on the datasheet of any NPN transistor its base-emitter is turned on when its base is positive about 0.7V from its emitter.
When the base voltage is negative it does not conduct but causes it to break down (it conducts like a zener diode) and be damaged at 5V or more.
Therefore the diode is added so that the base does not get a negative voltage that would damage it.

The 60Hz from the mains electricity is a sinewave. A sinewave is not supposed to have harmonics because harmonics are distortion.
I have never heard of "clean" harmonics because all harmonics are "dirty". Light dimmers cause harmonics.

Something that conducts symmetrically at the peaks of the sinewave cause it to be a squarewave that has only odd harmonics: 1, 3, 5, 7, 9 etc.
Something that conducts on only one polarity of the sinewave produces even harmonics: 2, 4, 6, 8, 10 etc.

 

[Billy Mayo]

Thanks for the info.

Something that conducts symmetrically at the peaks of the sinewave cause it to be a squarewave that has only odd harmonics: 1, 3, 5, 7, 9 etc.

What would be conducting symmetrically? that would cause odd harmonics?

Something that conducts on only one polarity of the sinewave produces even harmonics: 2, 4, 6, 8, 10 etc

What would conduct in one polarity? that would cause even harmonics?

 

[Billy Mayo]

Therefore the diode is added so that the base does not get a negative voltage that would damage it.

But the diode only protects the transistor , of how many negative volts? the diode is only 0.5 volts, so is the negative voltage is more than -1 volts it will damage the transistor because the protection diode is only 0.5 volts to turn it on

Any negative voltage more than 0.5 volts will turn on the protection diode?

 

[audioguru]

You are copying your posts from the other website.
Now I am also copying my replies to you from the other website:
"A silicon diode normally conducts with about 0.7V. It is 0.5V when it is hot and when its current is very low. It is 1V when it is cold and when its current is very high.
When the diode conducts at -0.7V then it protects the emitter-base from having a higher negative base voltage.
The emitter-base breaks down and is damaged when the base voltage is -5V or more negative."

 

[Billy Mayo]

So how much negative voltage does the diode protect? Only -0.7 Volts?

 

[kubeek]

anything below -0.7V

 

[Billy Mayo]

Isn't based on the voltage rating of the diode?

 

[Billy Mayo]

How far below -0.7 Volts can it go?

 

[kubeek]

No it is not dependent on the voltage rating of the diode. The diode is forward biased, so there is some current flowing through it in the forward direction.
How much current depends on the voltage and output impedance of the thing that supplies the current.
With higher current the Vf will rise, but not much above 1V, but with enough current flowing the diode will dissipate a lot of heat and will eventually melt.

 

[Billy Mayo]

If any negative voltage below -0.7 Volts will conduct the diode, if there is -6 volts the diode will be what? How does it protect the transistor from negative voltage on the base?

 

[audioguru]

So how much negative voltage does the diode protect? Only -0.7 Volts?

The diode clamps at -0.7V when its current is medium. It clamps at 1V when its current is high. Then the base of the transistor never gets near -5V so it is protected from damage.

 

[kubeek]

Like I said, if you put -5V without any limiting resistor, the current will be very high and the diode will melt very soon. The supply voltage will divide according to the supply resistance and the V-I curve of the diode.

Edit: no idea how true the model is, but for 1N4001 right across an ideal 5V supply the current is 33A, and the diode has forward drop of 2.5V at 12A.

 

[audioguru]

The circuit the OP posted has a 1N4148 high speed diode, not a 1N4001 slow power rectifier.
The circuit might have been posted on the other website's forum.

 

[Billy Mayo]

The diode clamps at -0.7V when its current is medium. It clamps at 1V when its current is high. Then the base of the transistor never gets near -5V so it is protected from damage.

I don't get how you can put a - 6volts or -10 volts on the input to the base of the transistor and the protection diode clamps it at -0.7 volts or 1 volt

It doesn't make sense to me

 

[audioguru]

I don't get how you can put a - 6volts or -10 volts on the input to the base of the transistor and the protection diode clamps it at -0.7 volts or 1 volt

It doesn't make sense to me

You do not connect -6V or -10V at hundreds of Milli-Amps to the base. You connect -6V or -10V through a current-limiting resistor so the current is 5mA to 10mA. Then the diode clamps the voltage to about -0.7V.