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The simple approximation is to simply subtract the typical forward voltage drop from the power supply then perform the normal RC transient calculation.
The more precise is to use the diode equation in conjunction with the RC transient differential equation. My maths isn't good enough to work it out so I'd turn to SPICE for help.
Hero:
Yes in many of these circuits they are using an ideal diode, with no
voltage drop, but in these circuits they appear to be using a first
approximation to a real diode which is an idea diode with a voltage
drop.
Transgalactic:
In circuit 1 you have to figure out the initial capacitor voltage
for when t<0, and that shouldnt be too hard.
Then you have to find the time response up to the point where
the diode conducts (sometime after t=0) and then the diode
acts as a voltage 'clamp'. When the diode acts as a voltage
clamp, it looks like an ideal diode (no voltage drop) in series
with a small battery equal to the voltage drop v_0 with the (+)
terminal connected to the cathode of the diode.
In circuit 2, the diode is probably the same as in circuit 1, where
it drops v_0 volts. The only difference here is that if it drops
v_0 volts then it will affect the step response by subtracting
v_0 voltage from the source.
It will be short circuit, unless you consider it to be like the diode in the
first circuit when then it will drop a voltage equal to v_0 volts.
For example, if v_0=0.7v then the diode will drop 0.7 volts.
This means 0.7v will be subtracted from the unit step input.
If you consider th diode to be totally ideal, then it will not drop any
voltage.
1.
regarding the first circuit:
"when then it will drop a voltage equal to v_0 volts.
For example, if v_0=0.7v then the diode will drop 0.7 volts.
This means 0.7v will be subtracted from the unit step input."
i know that when we have a positive voltage the diode will be short circuit
so in time t>0 we have a positive voltage
why its still an open circuit??(it should be short circuit)
i cant understand this drop thing
2.in both circuit there are cap's
which in parallel to the diode so the voltage on the diode is the voltage on the cap
and because vc(0-)=vc(0+)
in 1st circuit vc(0-)=-v0
in 2nd circuit vc(0-)=0
so both diodes has to be open circuit in the transition period
[1]
For the first circuit...
For the time between t=0 and t=t1 the diode is an open circuit.
It is open until Vc comes up to equal v_0. This means you
have to solve for the time t1 when this happens.
Since Vs=-Vo for t<0 that means you have to first solve
for Vc(0-) and then take it from there to solve for t1.
Keep in mind that in that drawing v_0 is not Vc(0), it is
the diode drop (assumed to be constant).
If you need to see the math let me know.
[2]
For 1st circuit, you have to solve for Vc and since Vs=-Vo
then the voltage on the cap will be the voltage as if the
diode wasnt there. When Vs goes positive, the cap starts
charging positive and you then solve for t1.
For the 2nd circuit, we assume Vc(0)=0, and for the unit step
the peak voltage is 1-Vdiode, and so instead of the unit
step we have (1-Vdiode)*u(t), which is less than u(t).
For the impulse source however, the diode is insignificant even if
it has a voltage drop, so the diode is a short for the impulse
source. It will also be a short for the unit step response too
but only if you consider the diode voltage drop to be zero (ideal diode).
For the first approximation diode (vdiode=0.7) we would end
up with a 0.7v voltage drop, for example, so the source would
be u(t)-0.7 which is 0.3*u(t).
For the impulse source, delta(t)-0.7=delta(t) which is no change.
Note that in the 2nd circuit there is a postive voltage on
the diode at t=0+ because the unit step or the impulse puts
it there immediately and the cap voltage is zero.
A voltage drop is an increase or decrease in voltage between two nodes in
the circuit.
Code:
o[FONT=Courier New]----R----o
n1 n2[/FONT]
In the above, if n2 is at 5v and n1 is at 3v, there is
a 2v voltage 'drop' between n1 and n2. If n1 is at 3v
and n2 is at 5v, we still say there is a voltage drop
between n2 and n1. Thus, when we say 'drop' the polarity
is obvious from other references.
Yes the diode needs a positive drop (anode positive, cathode negative)
in order to conduct current. There are a few differences with different
diode models however. Here are three types:
['Vanode' is the voltage at the anode referenced to ground, and
'Vcathode' is the voltage at the cathode referenced to ground]
1. Ideal (perfect diode):
This diode conducts perfectly when Vanode>=Vcathode.
In other words, this diode is a short when Vanode>=Vcathode.
and there is no voltage drop.
This diode is a complete open circuit when Vanode<Vcathode.
We might call this diode the 'perfect' diode, but it's often called 'ideal'.
2. First approximation Ideal Diode (perfect diode with voltage drop):
This diode conducts when Vanode>=(Vcathode+vd) where vd is the
voltage drop of the diode when conducting, considered to be a constant
over all time and for all currents.
This diode does not conduct when Vanode<Vcathode.
Once conducting, the diode acts as an opposing voltage source who's
voltage equals vd, the characteristic voltage drop of the diode.
3. Ideal Diode with Exponential Response:
This diode uses the "Ideal diode equation" and that involves either an
exponential or a log to calculate.
This diode is like the first approximation diode, except that it has a voltage
drop that depends on a log and a current that depends on an exponential:
I=IS*(exp^(v/(N*Vt))-1)
V=N*Vt*ln(i/IS+1)
where N and IS are constants, and Vt is the 'thermal voltage' equal to k*T/q
which equals about 26mv at 27 degrees C.
This diode is used in more advanced studies though, and often includes a
series resistance Rs.
Now for these problems they are probably using either the perfect diode
or the perfect diode with voltage drop.
The two different diodes will make a difference when we go to analyze the
response to a unit step because the one that has a voltage drop (like 0.5
or 0.7) will act like a small battery in series opposing the source and so
that subtracts from the unit step to produce a smaller driving voltage step.
The diode that does not have a voltage drop (diode 1) will not subtract
anything from the driving source, so the unit step remains the same as
if no diode were present.
On the other hand, when we go to analyze the response to an impulse it
doesnt matter what diode we choose (1, 2, or 3 above) because the
voltage drop does not matter. Thus, we can short out the diode when we
analyze the response to an impulse, except that we also remove the short
after t=0+. Thus, the diode conducts only around t=0, after which the cap
charges up instantaneously and and when the impulse goes back to zero
the diode goes to an open circuit.
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