On Wed, 29 Dec 2004 13:37:17 GMT, Lancer wrote in
htb5t058gbfee852kcbtht6vcul2jud5lu@2355323778:
snip
Ok Lancer (whichever one you are);
The same Lancer that has been here all along..
In your headers I saw several different servers, two different
newsreaders and two domain addresses (rock.com & ock.com). With all
the forgeries in this group it's easy to see how one might think your
posts are being made by more than one person.
Please explain, (CE bias can be as high as 3-4 volts under heavy load.
Isn't that set by the operating point set by bias you put on the base?
A bipolar transistor requires both a BE bias -and- a CE bias.
Ok, thats he part that I'm not understanding. (We are talking common
emitter, right?) Transistors don't require a CE bias. In the case of
an NPN just Pos voltage on the collector and Neg on the emitter. The
CE voltage is set by the bias on the BE junction.
Remember that a bipolar transistor is a -current- amplifier, not a
voltage amplifier. Saturation is a characteristic of the collector's
-current-; the CE bias is a characteristic of it's voltage. The two
terms are often used synonymously and saturation curves are really CE
bias curves, but that's because 'saturation' has two definitions:
First, it is the point where a device will no longer respond to an
increase at the input. This can happen for many reasons. But in this
case it's because the output has hit the rail, and the rail is the CE
bias (explained later).
Second, it's the point where a transistor is driven so hard that it
causes a forward bias of the BC junction (bad news).
Now..... I'm not trying to dumb this down, but simplification might
help to consolidate our differences:
Consider a transistor configured as a DC constant-current source; i.e,
base bias is fixed and therefore the collector current is constant.
Let's say the collector current is fixed at 1 amp. That current is
constant regardless of the collector voltage..... to a point: Let's
also assume that the CE bias is 2 volts when collector current is 1
amp. If the collector voltage drops below 2 volts then the collector
current will drop. Therefore, the voltage required to put the
transistor into the constant-current part of the curve is the CE bias.
Now let's configure the amplifier for Class B, use a 12 volt supply
and feed it some AC. If the peak output current is 1 amp then the
maximum possible voltage output of the amp will be 12 volts minus the
CE bias of 2 volts, or 10 volts. If the CE bias increases to 4 volts
at 10 amps (realistic value) then the peak voltage can be no larger
than 8 volts, or 2/3 of the power supply voltage.
Ok, now let's double the power supply voltage. Since the collector
current doesn't change then the CE bias doesn't change. The output can
now swing 20 volts, or 24 volts minus the CE bias. And this is 83.3%
of the ps voltage. Since the current is the same regardless of ps
voltage, the efficiency is 25% better with the -higher- ps voltage.
snip
Or are you refering to the losses in the transistor when its fully
turned on?
Not necessarily. CE bias increases with collector current regardless
of saturation (and RF bipolars don't saturate easily). But if the
supply voltage can be increased while maintaining the same collector
current (by changing the BE bias), the loss due to CE bias is not
changed, and that loss is therefore made to be a smaller percentage of
the output power. IOW, the transistor is more efficient with a higher
supply voltage.
If you don't change the CE current, the actual loss in the transistor
hasn't changed. If you increase the supply voltage it will be a
smaller percentage than it was before.
Exactly. A smaller percentage of the input power, and therefore more
efficient.
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