Re: Op-amp for power supply
- From: "Ban" <bansuri@xxxxxxxxxxxx>
- Date: Wed, 3 May 2006 16:28:04 +0200
hondgm@xxxxxxxxx wrote:
Ban wrote:
There are several problems with your design. They require
understanding how a transistor works. On the net you will mainly
find the outcome of others in the same situation, so I wouldn't rely
on this.
First, that diode will have a huge diffusion capacitance when
conducting, it is only dependent on the forward current and the
reverse recovery time, because when it is supposed to shut off, the
charge carriers have to be removed. This is a multiple of the
depletion capacitance and can be several nF. Luckily the very low
ohmic junction resistance is parallel to it. The same is true for
the transistor.
But the opamp is *much* faster and sees this big capacitance. It
will start oscillating and turning the diode+transistor on again.
So for the voltage control output this is bad. Better to use a
resistor instead of the diode.
How is this done? If I replace the diode with a resistor, then you
have a different circuit alltogether. The diode scheme allows either
the current OA or voltage OA alone to control the output. Whether
this is a good design, however, is questionable.
Look with a scope at the O/P of the voltage opamp. Whenever you give an
increase in voltage by your D/A, it will also loose regulation. It goes high
and the wimpy 47k resistor has to charge up the base capacity, which is
*very * high when conducting, maybe 0.3uF.
When a resistor instead of the diode is there, the opamp doesn't loose
regulation, but will help giving more current, at least as long as you do
not reach the upper voltage limit. Now you can take a Rail-to-Rail opamp and
increase performance.
And get rid of this capacitor at the voltage divider, It will *never* allow
a decent performance, but always let the opamp overreact. Look into your
textbook about loop stabilisation.
Slow down the response time with a capacitor from the Vreg opamp O/P
to its neg. input, maybe 100p. This will greatly improve stability.
What you were doing was the opposite and aggravates the problem.
I believe I've tried this. It does help some.
Now this "pull-up" resistor will draw a lot of current when the
voltage is low and not enough to pull the darlington up, since the
opamp with the diode cannot source current. So use a current source
made from a resistor and current mirror between the rails and set
it to say 5mA when output is zero. The new resistor on the voltage
regulator has now to pull the darlington base down to 0.9V or less
and has to sink those 5mA, so maybe 0.7V/5mA= 120R. If that is not
enough put a high current diode from output gnd to the negative rail
to improve this margin.
The pull-up isn't much of a problem. Actually, I have a 47K on there
and it has no problem supplying enough current to the transistor. And
the OA has no problem sinking this current.
Really? Have you thought how much current you need on the base to drive the
darlington? With 5A the hFE is hardly more than 5000, maybe when it's hot
you will have 10000. You would need 25V across the resistor to deliver that
current. Ah now I see why this resistor is so big. Because otherwise when
there is a short circuit, you would burn out the transistor, because the
current limiter is also saturated and not regulating at all. Well at least
use a current mirror with those 0.5mA, so you can also deliver some current
when the voltage is maximal.
Ok, so I think you're saying to build a plain old voltage regulator
Now the current limiter can not work any more, since when it kicks
in, the voltage opamp output goes high and delivers another 20mA
until the current limit. There is a possibility: you pull down its
*reference voltage* through a Schottky diode and also a 120R
resistor. But now the regulation goes thru 2 opamps so the current
one needs to have an even slower response, maybe 1n in the same
position as
before with the voltage opamp. Now your voltage opamp will never be
out of regulation.
Don't forget a resistor on the DAC voltage O/P since it gets pulled
down by the
current limiter.
with an OA and output transistor, like I have here, minus the diodes.
But, pull down the voltage reference into the op-amp with my current
limiting OA. I still end up using diodes.
Only one diode for the current limiter at the voltage D/A output.
Better would be another additional transistor as an emergency current
limiter, which is driven by a small sense resistor, you could take your
current shunt for that, to pull down the base of your darlington.
'm confused now.
Also before?
ciao Ban
Apricale, Italy
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