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semiconductors & ohm's law

M

Mofman

Jan 1, 1970
0
Hi All,

I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are involved.
However, as soon as you introduce diodes or transistors, I start having
trouble and I have reached the limits of my research (although not much). I
have the following questions:

How do you calculate the voltage drop across a diode?
How do you calculate the base-emitter voltage drop for a transistor? What
about the collector-emitter voltage drop?

Thanks

M
 
D

Don Pearce

Jan 1, 1970
0
Hi All,

I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are involved.
However, as soon as you introduce diodes or transistors, I start having
trouble and I have reached the limits of my research (although not much). I
have the following questions:

How do you calculate the voltage drop across a diode?
How do you calculate the base-emitter voltage drop for a transistor? What
about the collector-emitter voltage drop?

Thanks

M

The answer is pretty much you don't. You have the options of looking
them up in a data sheet if it is comprehensive enough, or measuring
some real devices.

Spice models for these parameters are generally pretty good, so you
could use these to find the answers.

d

_____________________________

http://www.pearce.uk.com
 
J

John Woodgate

Jan 1, 1970
0
(in said:
Hi All,

I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are involved.
However, as soon as you introduce diodes or transistors, I start having
trouble and I have reached the limits of my research (although not much). I
have the following questions:
How do you calculate the voltage drop across a diode?

There is a formula, V = (kT/q) ln(I/Is), which links the voltage V with
the current I for an ideal junction with no series resistances. To use
it, you need to know Is for the particular device. k = Boltzmann's
constant, T = absolute temperature, q = charge on the electron, Is = a
characteristic of the device, hardly ever given in the data sheet.
Instead, you get curves of typical, and, if you are very lucky, maximum
and minimum forward voltage against forward current, maybe at two
temperatures.

But for normal silicon diodes at low currents (up to a few tens of
milliamps), the value of V is between 600 and 700 mV approximately. For
Schottky diodes, it is between 300 and 450 mV, AIUI. For germanium
diodes, about 200 mV.

If the diode is reverse-biased, the voltage depends on the rest of the
circuit.
How do you calculate the base-emitter voltage drop for a transistor?

Use the above formula, I being the base current. But you may need to add
in I(rbb' + ßree') to allow for series resistances. The point about the
voltages being rather generic to the *material*, not the device,
applies.
What
about the collector-emitter voltage drop?
That depends on the operating conditions, and is likely to be something
you optimise when designing the circuit.
 
O

Olaf

Jan 1, 1970
0
I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are
involved. However, as soon as you introduce diodes or transistors, I
start having trouble and I have reached the limits of my research
(although not much). I have the following questions:

How do you calculate the voltage drop across a diode? How do you
calculate the base-emitter voltage drop for a transistor? What about
the collector-emitter voltage drop?

take a look at http://homepage.ntlworld.com/g.knott/ , the intermediate
section contains answers and explanations for both diodes and 'normal'
transistors (junction transistor) on newbie level. Usually you don't
calculate the voltage drop. Take a default value (see the link) or look in
the datasheet.

bye, Olaf
 
J

John Woodgate

Jan 1, 1970
0
take a look at http://homepage.ntlworld.com/g.knott/ , the intermediate
section contains answers and explanations for both diodes and 'normal'
transistors (junction transistor) on newbie level. Usually you don't
calculate the voltage drop.

One of the things about newbies is that you never know whether they
happen to have a first-class degree in maths, and really WANT the
equations. (;-)
 
O

Olaf

Jan 1, 1970
0
One of the things about newbies is that you never know whether they
happen to have a first-class degree in maths, and really WANT the
equations. (;-)

You're right, a newbie to electronics might be a calculus whizard. So I
didn't answer OP's question and you did. ;-) Maybe M will let us know what
he was looking for after all...

bye, Olaf
 
L

Leon Heller

Jan 1, 1970
0
Mofman said:
Hi All,

I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are involved.
However, as soon as you introduce diodes or transistors, I start having
trouble and I have reached the limits of my research (although not much). I
have the following questions:

How do you calculate the voltage drop across a diode?

It's about 0.7V for an ordinary Si diode. The data sheet should give it
for various currents. Temperature comes into it as well.
How do you calculate the base-emitter voltage drop for a transistor?

As above.

What
about the collector-emitter voltage drop?

You usually calculate this as part of the design process, when biasing
the device.

Leon
 
J

John Popelish

Jan 1, 1970
0
Mofman said:
Hi All,

I am an electronics newbie trying to understand how to calculate power
dissipation/voltage drops/current for simple circuits involving
semiconductor components.

Calculating these values is easy if only "passive components" are involved.
However, as soon as you introduce diodes or transistors, I start having
trouble and I have reached the limits of my research (although not much).

So does everyone else, in certain cases. The nonlinear math can be a
bear. This is one of the reasons circuit simulators were created.
I have the following questions:

How do you calculate the voltage drop across a diode?

There is a logarithmic function that describes the typical current to
voltage relationship, but most diodes drop about 6. to 1 volt in their
normal operating range. To see how a particular diode will act, it is
a good idea to check the data sheet. This usually contains a graph of
voltage drop versus current with three curves for three different
temperatures, room temperature, very cold and very hot, to give you an
idea of the range.
How do you calculate the base-emitter voltage drop for a transistor? What
about the collector-emitter voltage drop?

The base to emitter voltage is a diode junction, so the generalities
above apply. Again, the data sheet often has a lot of detail on this,
including how it changes when the transistor saturates.

If you want to wade in to the math, here is a paper that compared two
models of the voltage to current relationship. Don't be put off by
the greek letters or the fact that they use U for voltage.

http://www.comppub.com/publications/MSM/2000/pdf/T54.04.pdf

The collector to emitter voltage is totally dependent on the base bias
current and the collector load, as well as the current gain and other
details of the particular transistor, so there can be no simple rule
to guess this voltage. You will have to learn a lot more about how
transistors actually work to make good estimates of this.

But the one rule that always works is that the power going into a
device in any given instant is the product of the current through it
times the voltage across it. This includes the case where the current
is going opposite the way the voltage is trying to push it, except,
there the power is negative (applies to batteries and other energy
storage components).
 
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