Re: Zener Diodes / Voltage Regulation
- From: Jonathan Kirwan <jkirwan@xxxxxxxxxxxxxx>
- Date: Sun, 22 Jul 2007 06:18:50 GMT
On Sat, 21 Jul 2007 22:01:28 -0700, FyberOptic <fyberoptic@xxxxxxxxx>
wrote:
<snip>
So while this is possibly out of the scope of asking on a newsgroup,
what sorts of applications could they be used for reliably? I hear
they can be used as regulators, but only if the load is fairly
constant. So, for something like an electronic circuit, with chips
going on and off and constantly changing the load, would this be of
any use there? I've seen them used in situations such as when one is
drawing power from the PC parallel port. I'm assuming this isn't
related to actually getting that power, but keeping it from going too
high.
A zener usually isn't used directly as a power supply regulator, per
se. It may be a good voltage reference, though, as input to a
comparator (which doesn't require much current.) If a zener is used
for supplying some current, a BJT voltage follower is often added (the
base doesn't require much current, once again.)
Take a basic circuit:
R1 / Z1
+Vsupply ----/\/\--+--|<|--------gnd
| /
|
'---- LOAD ---gnd
If you know your load requires something less than 20mA, for example,
you might set things up so that R1 passes that 20mA plus a little more
for the zener, itself. (The zener voltage will not remain the same as
current changes occur, but will stay somewhere in the same area.) So
let's say 25mA. This means that when your load is using 20mA, the
zener itself will get 5mA. But when the load is removed or otherwise
not taking much current, then Z1 will have to pick up the load of the
full 25mA. So say that Vsupply=21V and Z1 is a 6.2V variety. You
compute R1=(21-6.2)/25mA = 592 ohms. Make it 560, for a standard
value and accept the fact that the actual max current will be a little
more than 26mA. Now, the whole thing completely falls apart, though,
if your load requires almost all of that 26mA or more. Once that
happens, R1 will drop too much voltage and the zener won't 'zener' and
your voltage will go completely out of regulation.
Maybe not the very next step, but a small step past it, is to add a
BJT and another diode:
R1 / Z1 D1
+Vsupply ----/\/\--+--|<|---|>|--gnd
| /
|
|
Q1 --- NPN
/ v
+Vsupply -------c e---- LOAD ---gnd
The Q1 emitter will present about the zener voltage to the load. The
addition of D1 does at least two things -- one is that it 'jacks up'
the base voltage of Q1 because there is a base-emitter diode drop
there and D1 helps to set things up so that the emitter is about the
zener voltage. (If you didn't include D1, then the emitter would be
somewhat lower, by one diode drop, roughly); Another that maybe isn't
important for this discussion is temperature compensation.
If the load requires under 20mA, still, then the base of Q1 will
probably only need about 1/100th of that, or about 200uA. Now, if you
set up your zener current to be about 10mA for example, by calculating
R1 accordingly, then the variations over the entire 0mA-20mA load
possibilities will only present a variation of 0uA-200uA shift in the
zener current. That's only 1/50th of the 10mA, so we are talking
about 2% variation. This means the zener voltage will stay put pretty
well. The emitter voltage of Q1 will vary some if the emitter current
gets really tiny (rising by some 60mV per decade reduction in
current), but for anything in the probable range it will be
sufficiently fixed.
Note that for all of the above examples, to calculate R1 you need to
know what your Vsupply is. If that is an unregulated supply and
itself varies a lot, then you may need to calculate R1 on the basis of
the smallest likely voltage there and just realize that at other times
more current will be flowing into Z1 and that will impact the zener
voltage, somewhat. Or, you may just want it to work with any number
of those "wall warts" with different output voltages. So if you want
to set things up well for that more difficult situation, you need to
use a constant current source that doesn't vary over variations in the
unregulated supply. A constant current arrangement requires another
BJT. So it looks more like:
,-----e c--gnd
| v /
| --- Q1
| | PNP
| |
R2 | R1 | / Z1 D1
Vunreg---/\/\-+-/\/\--++--|<|---|>|---gnd
| /
|
| Q2
--- NPN
/ v
Vunreg--------------c e----- LOAD ----gnd
Q1 is now used to generate a constant current through R1, Z1 and D1.
The Vunreg supply can vary a lot, now, and it doesn't have much impact
on that zener current and therefore not much on the zener voltage,
which remains fairly stable regardless of what Vunreg's voltage is (so
long as it is several volts higher than the zener voltage.)
I've kept out things like capacitors and possibly some negative
feedback (which is usually a good idea) to control the output voltage
better, using the zener instead as a voltage reference for comparison
purposes and not as the input to an open-loop driver (no feedback.) If
you added something like that, it would appear more as:
C2
|| .5uF
,--------||-----,
| || |
| | Vunreg
| Vunreg | |
| | | |
| |\ | |
+--------|-\ | |/c Q1
| | >---+-----| NPN
| Vref---|+/ |>e
| |/ |
| | | Vout
| gnd '-------+----+----- LOAD ---gnd
| | |
| | |
| \ --- C1
| R1 / ---
| \ |
| / |
| | |
| | gnd
'-------------------------------+
|
|
\
/ R2
\
/
|
|
gnd
Here, you can see that an opamp is used to compare the Vref (output of
the constant current version of the zener regulator I just mentioned
above) with a portion of the output voltage (Vout.) The opamp then
drives Q1 as needed so that the difference is nil.
[That opamp might only be a differential pair of BJTs with some
additional BJTs for things like a constant current at the pair's
emitter or in the collector of one side as another constant current to
'stiffen' (make it much higher gain) it so that only one diff pair is
needed instead of, say, two pairs.]
Of course, you can just buy an IC that does all this and more for you.
Jon
.
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