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Zener regulator : load and zener current

In the simple zener regulators like the one below, If the load current is 100ma and zener current is 10ma we calculate the resistor to allow 110ma.

If Vin is above Vz and the load draws 50ma, the remaining 60ma flows through zener. What makes it that the remaining current after the load current gothrough the zener? Whay can't it be the otherway? i.e after the zener current the remaining to the load. Is it due to the load impedance being less than the zener impedance?

Vin------/\/\/\/\--------|-----------|
| |
| |
| |
| / RL
/--- |
/\ |
/ \ |
------ -----
| ---
| -
|
-----
 
S

Spehro Pefhany

Jan 1, 1970
0
On Sun, 18 Aug 2013 05:12:11 -0700 (PDT), the renowned
In the simple zener regulators like the one below, If the load current is 100ma and zener current is 10ma we calculate the resistor to allow 110ma.

If Vin is above Vz and the load draws 50ma, the remaining 60ma flows through zener. What makes it that the remaining current after the load current go through the zener? Whay can't it be the otherway? i.e after the zener current the remaining to the load. Is it due to the load impedance being less than the zener impedance?

Vin------/\/\/\/\--------|-----------|
| |
| |
| |
| / RL
/--- |
/\ |
/ \ |
------ -----
| ---
| -
|
-----
---
-

Thanks
Sridhar


Analytically, you know from Kirchoff's current law that the sums of
the currents into the common juction of the zener, resistor and load
must equal zero.

You can write an equation for each of the currents as a function of
the voltage at that junction point (it will be nonlinear in the case
of the zener diode, and perhaps the load, but assume a resistor Rl).
It may be easier to think of the zener as an ideal zener in series
with a bit of resistance Rz*.

Simply solve for a voltage V at that point such that Kirchoff's law is
satisfied.

i1+12+i3=0

i1 = (Vin - V)/R

i2 = (Vz - V)/Rz , V>= Vz; i2 = 0 otherwise
(assume V >= Vz)

i3 = -V/Rl

=> Vin/R - V/R - V/Rz + Vz/Rz - V/R1 = 0

Vin/R + Vz/Rz = V(1/R1 + 1/Rz + 1/R)

V = (Vin/R + Vz/Rz)*(R1||Rz||R)

As Rz -> 0, V -> Vz, obviously.

For example, Rz = 1 ohm, Vz = 10V, Vin = 20V, Rl = 100 ohms R = 90
ohms

Rp = R1||Rz||R = 0.979325 ohms

V = (20/90) * Rp + 10 * Rp = 10.011V.


* this is a bit closer to the real behavior of a zener, but the
dynamic resistance Rz will vary with current so it's more of a
small-signal approximation.


Intuitively, remove Rl, and you should be able to see that it behaves
wrt Rl as a voltage source equal to Vz (assuming an ideal zener)
provided the zener is conducting. The current through the load is
simply Vz/Rl.

The point at which regulation cannot occur, even with an ideal zener,
is when there is no current left for the zener, so when (Vin - Vz)/R =
Vz/Rl.

So, if Rl < (Vz * R)/(Vin - Vz), what is the load voltage?


Best regards,
Spehro Pefhany
 
A

Artemus

Jan 1, 1970
0
In the simple zener regulators like the one below, If the load current is 100ma and zener current is
10ma we calculate the resistor to allow 110ma.

If Vin is above Vz and the load draws 50ma, the remaining 60ma flows through zener. What makes it
that the remaining current after the load current go through the zener? Whay can't it be the
otherway? i.e after the zener current the remaining to the load. Is it due to the load impedance
being less than the zener impedance?

Thanks
Sridhar

******************************

Think of a zener as a voltage source that can only sink current.
Art
 
If Vin is above Vz and the load draws 50ma, the remaining 60ma flows through zener. What makes it that the remaining current after the load current go through the zener? Whay can't it be the otherway? i.e after the zener current the remaining to the load. Is it due to the load impedance being lessthan the zener impedance?

It has nothing to do with impedances, it has to everything to do with the zener and its terminal voltage. In the first order idealization of zener operation, current in the direction of the diode arrow produces 0V terminal voltage, and current in the opposite direction of diode arrow produces a terminal voltage of Vz volts. In both cases the terminal voltage is constant and independent of the actual magnitude of the current. So to answer your question, if there is any current at all flowing through the zener in your circuit, then the zener cathode node must be /fixed/ at Vz volts. If the node is fixed at Vz volts, then Rload has Vz volts across it and so its current has to be Vz/Rload. Similarly, the input resistor has a fixed Vin-Vz acrossit, so the current through it has to be (Vin-Vz)/Rin. Then since charge isconserved, the zener has to take the excess of (Vin-Vz)/Rin - Vz/Rload. Ifthere is zero or negative excess, this means (Vin-Vz)/Rin does not supply enough current to power Rload at Vz/Rload, obviously, which means there is no Vz because there is not enough current to flow through the zener to develop this Vz. So the zener is in effect out of the circuit, it is open, and the circuit reduces to a simple voltage divider with Rin in series with Rload and Ohm's law sets the output voltage and current.
 
It has nothing to do with impedances, it has to everything to do with thezener and its terminal voltage. In the first order idealization of zener operation, current in the direction of the diode arrow produces 0V terminal voltage, and current in the opposite direction of diode arrow produces a terminal voltage of Vz volts. In both cases the terminal voltage is constant and independent of the actual magnitude of the current. So to answer your question, if there is any current at all flowing through the zener in your circuit, then the zener cathode node must be /fixed/ at Vz volts. If the node is fixed at Vz volts, then Rload has Vz volts across it and so its current has to be Vz/Rload. Similarly, the input resistor has a fixed Vin-Vz across it, so the current through it has to be (Vin-Vz)/Rin. Then since charge is conserved, the zener has to take the excess of (Vin-Vz)/Rin - Vz/Rload. If there is zero or negative excess, this means (Vin-Vz)/Rin does not supply enough current to power Rload at Vz/Rload, obviously, which means there is no Vz because there is not enough current to flow through the zener to develop this Vz. So the zener is in effect out of the circuit, it is open, and the circuit reduces to a simple voltage divider with Rin in series with Rload and Ohm's law sets the output voltage and current.

Thanks for making it clear. I simulated the schematic below in pspice to understand. Current through the base is always 1.3ma irrespective of V1 and R1 with a load of 50 ohms and the rest goes through the zener. Only changingthe load changes the current through the base. Changing load to 100 ohms reduces the base current to approx 600ua and changing it to 25 ohms changes the base current 3.3ma. How are these base current values coming from?

http://www.flickr.com/photos/100471956@N04/


-Sridhar
 
T

Tauno Voipio

Jan 1, 1970
0
er Rload at Vz/Rload, obviously, which means there is no Vz because there is not enough current to flow through the zener to develop this Vz. So the zener is in effect out of the circuit, it is open, and the circuit reduces to a simple voltage divider with Rin in series with Rload and Ohm's law sets the output voltage and current.
Thanks for making it clear. I simulated the schematic below in pspice to understand. Current through the base is always 1.3ma irrespective of V1 and R1 with a load of 50 ohms and the rest goes through the zener. Only changing the load changes the current through the base. Changing load to 100 ohms reduces the base current to approx 600ua and changing it to 25 ohms changes the base current 3.3ma. How are these base current values coming from?

http://www.flickr.com/photos/100471956@N04/


-Sridhar

You are feeding the load with constant voltage = zener voltage -
base-emitter drop, so the load current is constant. The base current is
the load current divided by the DC current gain of the transistor.

I suspect that you're going to blow up your transistor, please check the
power dissipation in it.
 
er Rload at Vz/Rload, obviously, which means there is no Vz because thereis not enough current to flow through the zener to develop this Vz. So thezener is in effect out of the circuit, it is open, and the circuit reducesto a simple voltage divider with Rin in series with Rload and Ohm's law sets the output voltage and current.





You are feeding the load with constant voltage = zener voltage -

base-emitter drop, so the load current is constant. The base current is

the load current divided by the DC current gain of the transistor.



I suspect that you're going to blow up your transistor, please check the

power dissipation in it.



--



Tauno Voipio


What I read about BJTs tell me that we control the base current and the collector current is DC current gain times the base current. Is it something different in this configuration that base current is dependent on the load current and gain? i.e we are not controlling the base current.

Yes, I will check the Pd.

Thanks
Sridhar
 
T

Tauno Voipio

Jan 1, 1970
0
What I read about BJTs tell me that we control the base current and the collector current is DC current gain times the base current. Is it something different in this configuration that base current is dependent on the load current and gain? i.e we are not controlling the base current.

Yes, I will check the Pd.

Thanks
Sridhar


Get a good electronics textbook and read about an emitter follower.
There is feedback from the output voltage
 
Thanks for making it clear. I simulated the schematic below in pspice to understand. Current through the base is always 1.3ma irrespective of V1 andR1 with a load of 50 ohms and the rest goes through the zener. Only changing the load changes the current through the base. Changing load to 100 ohmsreduces the base current to approx 600ua and changing it to 25 ohms changes the base current 3.3ma. How are these base current values coming from?



http://www.flickr.com/photos/100471956@N04/

The voltage across the load resistor is Vz-Vbe, where Vbe is forward bias voltage across the transistor base-emitter junction required for it to conduct. Therefore, the load current is (Vz-Vbe)/Rload, and this is supplied by the transistor emitter current Ie. The transistor base current required foran emitter current of Ie is Ib= Ie/(beta+1) where beta is transistor current gain. So the Ib= ((Vz-Vbe)/Rload)/(beta +1)=(Vz-Vbe)/(Rload x (beta+1)) mathematically describes the base current dependence on Rload and also the variation you are seeing.
 
P

Phil Hobbs

Jan 1, 1970
0
What I read about BJTs tell me that we control the base current and
the collector current is DC current gain times the base current. Is
it something different in this configuration that base current is
dependent on the load current and gain? i.e we are not controlling
the base current. Yes, I will check the Pd.

The base current is really a loss mechanism--an ideal BJT has zero base
current, because all of the emitter current makes it to the collector
and none recombines in the base region. (Recombination is where the base
current comes from.)

The fundamental control mechanism of a BJT is the base-emitter voltage,
which provides pretty tight voltage feedback in an emitter follower.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
S

Spehro Pefhany

Jan 1, 1970
0
Well, eh, I disagree, seems you want infinite beta, I do not want a BJT with infinite beta.
Changes in beta, and its dependence on Ic for example, can be used for gain control,
many other circuits with BJTs use beta in some way or the other,
say for example current limiting comes to mind.

A BJT is NOT a voltage amplifier, it is a CURRENT amplifier.
You want a MOSFET if you want a voltage amplifier.

<getting out the popcorn for this one>
 
P

Phil Hobbs

Jan 1, 1970
0
I "design" around beta a lot more often than around small-signal input impedance
or transconductance. But then, I mostly use bipolars for switching and emitter
followers and simple things, not the sort of thing that you do with laser noise
cancellers and such. I think most people use opamps for precise things nowadays,
and transistors to do dumb stuff.

But I did design the analog multiplier safe-operating-area computer recently,
and it works! That involved most of the low-frequency transistor effects.

https://dl.dropboxusercontent.com/u/53724080/Circuits/Power/SOAR_4.asc
I agree that most discrete circuits are simple things, and that one
normally thinks about beta more than about delta-Vbe.

I still say that the transconductance mechanism is more fundamental,
because you can explain the current amplification by the
transconductance (with recombination), but you can't explain the
transconductance by the current amplification. The reason that beta
varies all over the place is that it's the reciprocal of a small loss
term that depends on a lot of fine processing details.

The OP's confusion seemed to be that he didn't understand the voltage
feedback mechanism in an emitter follower, though, and seemed to think
that current amplification alone was an adequate basis for understand a BJT.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
The base current is really a loss mechanism--an ideal BJT has zero base

current, because all of the emitter current makes it to the collector

and none recombines in the base region. (Recombination is where the base

current comes from.)

That's not quite true. Base current is also due to minority carrier diffusion as well as recombination.
The fundamental control mechanism of a BJT is the base-emitter voltage,

which provides pretty tight voltage feedback in an emitter follower.


Maybe, but you don't drive transistor ports with ideal voltage sources.
 
P

Phil Hobbs

Jan 1, 1970
0
That's not quite true. Base current is also due to minority carrier
diffusion as well as recombination.

True. I think that's a small effect in practical devices, though, since
it wouldn't vary much from device to device whereas beta is all over the
map.
Maybe, but you don't drive transistor ports with ideal voltage
sources.

I'm not sure I'm getting your point here. The OP was thinking of the
BJT solely as a current-gain device. That idea doesn't in itself
constrain the base-emitter voltage at all, whereas thinking about it in
terms of transconductance does.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
True. I think that's a small effect in practical devices, though, since

it wouldn't vary much from device to device whereas beta is all over the

map.







I'm not sure I'm getting your point here. The OP was thinking of the

BJT solely as a current-gain device. That idea doesn't in itself

constrain the base-emitter voltage at all, whereas thinking about it in

terms of transconductance does.



Cheers



Phil Hobbs



--

Dr Philip C D Hobbs

Principal Consultant

ElectroOptical Innovations LLC

Optics, Electro-optics, Photonics, Analog Electronics



160 North State Road #203

Briarcliff Manor NY 10510 USA

+1 845 480 2058



hobbs at electrooptical dot net

http://electrooptical.net

I don't know how gm helps you in this case. For purposes of determining thetransistor loading on the zener, most people agree it will be (beta +1) x Rload. If this estimate is all over the map because of beta, there's nothing you can do to improve on it by using gm, the loading will be still be allover the map. Manufacturers help by providing beta variation with Ic and min and max values, things that help bound the variation.
 
P

Phil Hobbs

Jan 1, 1970
0
I was always under the impression that beta depends on Ic???
Confirms your own statement that beta depends on a lot of things :)

Not in a predictable way, it doesn't. It does droop at very high I_C
due to high level injection, but that's all that's really predictable.
At lower current, different devices of the same type can have beta that
goes up with I_C, goes down with I_C, or is flat to within 10% over two
orders of magnitude of I_C.
Well, it is early morning, I will save the word salad for dinner.

What's so hard to understand? A resistor R in series with the base
provides the same negative feedback as R/beta in series with the
emitter, except for the effects of shunt capacitance and beta
nonlinearity. (*)
Yes, that is what made me say that..

For reasons you already mentioned I try to stay clear of GOhms.

Once at an university designing electronics for some experiments I did notice
that the average? chemist does not have any fears of let's say orders of magnitude...

I give you you are doing beautiful work at the 'edge' of what can be done.
I did study your transistor tester but I still have fear building things with that high resistors,
especially if it needs to be precise.

If normally somebody would show me such a thing for a design review I would suggest looking
if there was an other way..

Moisture, high humidity, you'd need a sealed box too.

Oh, I'm with you there. I certainly wouldn't do it for anything but a
lab one-off, because it takes too much babysitting.

I was trying to measure both log conformance and beta linearity in one
go, to pretty good accuracy. The usual trick of using tee networks in
the feedback loop plus really really low offset amplifiers wouldn't work
because chopamps all seem to have about 200 pA of input bias current. I
could have used a charge dispensing loop, but it's a lot easier to store
samples of a voltage. The right approach would have been to use more
complicated MUXing and a bunch of online calibration, but I'm not as
fast at MCU stuff as you are, and since the job was for a university
group with limited funds, I really wanted to keep the hours down. It
works fine as long as it's me driving it.

Probably most of us have a bunch of those one-offs lying around that
we've built over the years, and they come in really handy sometimes.
(Of course I also have a Keithley 405 Micro-Microammeter that I got for
$5 on eBay--it has a 100 fA FS range, but takes a good two hours' warmup
to get down that low. It's really fun to use, though--electrometer
tubes are actually pretty amazing.)

I bought a Keithley 602 for very cheap, mostly for the meter, range
switch, and box. All solid-state, several years newer, 100 times less
sensitive. I'll put new guts in it one of these days when I'm motivated.


+ | c video -------- b NPN +1 to +2V e | [ ] 75
75 Ohm coax 10 meter
|----------==============================================-----------||---
amp | | |
| [ ] 1k /// /// [ ]
75 | |
/// ///



You don't even need the resistors on the BJT side--it'll drive the
coax fine with all of the DC load at the far end.

The 75 Ohm on the left,.. somehow you need to match the cable
impedance.

Nah, as long as the far end is properly terminated, the near end looks
like a 75-ohm resistor anyway. It's the dual of
series-termination--short circuit on one end, Z0 on the other. Not a
lot of circuit protection, of course.

Will have to play with that is spice one day,
did it but only for frequency sweep, that was OK (used this several times as video output stage).
But when I had to design a big thing to drive RGB into long cables I decided to use very expensive opamps
with huge bandwidth that did drive symmetrically... just to be on the safe side.
No LTspice in those days, IBM PC was a new thing...

IIRC Horowitz & Hill suggest this method, with another resistor in the
collector to provide current limiting. I don't think I've ever actually
used it, but series-termination is dead useful. One good thing is that
it won't oscillate with an unterminated or short-circuited cable,
because the transistor won't be in normal bias.

Cheers

Phil Hobbs

(*) Yes, it's technically R/(1+beta), but up at frequencies where the 1
matters, beta has a serious phase shift, so it's rarely needed.
Designing accurate current mirrors is one of those rare places.


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
P

Phil Hobbs

Jan 1, 1970
0
It is not hard to understand, my error, I was not 100% awake yet...


As to beta versus Ic, I will have to look up some transistor specs,
RF transistor specs...
I see for example the BFR92 has a flat curent gain versus IC,
while the BFR96 falls of below 40 mA, making it more suitable for AGC.
Both are 5GHz wideband small signal NPN...
But I would probably use a dual gate MOSFET for AGC...





Nice, good price too.





I have ordered a Sony super HAL CCD 'starlight' camera module.
what it can do:
http://www.sony.net/Products/SC-HP/effiowld/ 0.001 lux...
ebay:
http://www.ebay.com/itm/15070521460...l?_from=R40&_sacat=0&_nkw=150705214606&_rdc=1

I had almost pinged you for help trying to make sense of the specs...
It is supposed to give a clear picture with just starlight illumination (I mean at 30 fps).
On board is a DSP too that reduces contrast so you can get contrast from areas while in
in other areas there is strong light,
Will go in my 'drone' alias 160 km/h SDcard trafficing ..
This is to bypass NSA (Hello) of course.
It is faster than internet...

I feel your pain. CCD datasheets are uniformly horrible.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
P

Phil Hobbs

Jan 1, 1970
0
Chopamps usually are shooting healthy packets of charge out of both input pins.
Offset voltage and bias current can depend on what impedances the input pins
see, especially whether they see capacitance or resistance.

Some seem to be a lot better than others, e.g. the OPA378 doesn't seem
nearly as bad as the OPA2188.

Cheers

Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
P

Phil Hobbs

Jan 1, 1970
0
I did a 3-opamp 6-pole SK filter with AD8628s. That seems to be an
especially bad configuration... one opamp input has a cap to ground
and the other input connects to the output (unity-gain follower.) I
got something like 5 uv offset per amp. Had to add a tiny trimpot to
tweak the overall offset to zero. This was for some little plug-in
filter boardlets.

https://dl.dropboxusercontent.com/u/53724080/Circuits/Filters/22A451A1.jpg

https://dl.dropboxusercontent.com/u/53724080/Circuits/Filters/22A451A3.jpg


Next time, I'll breadboard anything critical that uses a chopamp; the
data sheets don't generally mention charge injection.

Sometimes they do come clean, e.g.
<https://dl.dropboxusercontent.com/u/105025837/sed/OPA2188ChopperNoise.pdf>

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
 
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