# Zener regulator : load and zener current

Started by August 18, 2013
```On 8/19/2013 6:59 AM, sridhar09.cherukuri@gmail.com wrote:
> On Monday, August 19, 2013 3:11:58 PM UTC+5:30, Tauno Voipio wrote:
>> On 19.8.13 9:47 , sridhar09.cherukuri@gmail.com wrote:
>>
>>> On Monday, August 19, 2013 3:19:49 AM UTC+5:30,
>>> bloggs.fred...@gmail.com wrote:
>>
>>>> On Sunday, August 18, 2013 8:12:11 AM UTC-4,
>>>> sridhar09...@gmail.com wrote:
>>
>>>>
>>
>>>>
>>
>>>>
>>
>>>>> 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?
>>
>>>>
>>
>>>>>
>>
>>>>
>>
>>>>
>>
>>>>
>>
>>>> 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 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 pow
>>
>> 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?
>>
>
> 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

Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net
```
```On Sun, 18 Aug 2013 05:12:11 -0700 (PDT), sridhar09.cherukuri@gmail.com wrote:

>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

That's a classicly risky circuit. If Vin drops a little, or load current
increases a little, the zener current might go to zero and you lose regulation.

In the other direction, if Vin goes up a little, you can fry the resistor. Or if
the load reduces or is disconnected, you can fry the zener.

It's more reliable if the relative drop across the resistor is large and the
current is weighted more towards the zener, but then it becomes very
inefficient. It's OK for things like voltage references where Vz might be half
of Vin and load current is low, where an efficiency of zero (or less!) is OK.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
```
```On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote in
<kuta6k\$3gg\$1@dont-email.me>:

>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.)

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.

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

Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).
Asymmetric circuit really.
```
```On a sunny day (Mon, 19 Aug 2013 15:38:13 GMT) it happened Jan Panteltje
<pNaonStpealmtje@yahoo.com> wrote in <kute56\$vso\$1@news.albasani.net>:

>On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
><pcdhSpamMeSenseless@electrooptical.net> wrote in
><kuta6k\$3gg\$1@dont-email.me>:
>
>>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.)
>
>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.
>
>
>
>
>>The fundamental control mechanism of a BJT is the base-emitter voltage,
>>which provides pretty tight voltage feedback in an emitter follower.
>
>Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).

Correction, I ment Zout is Zload when going 'down', but Zi / beta when going 'up'.

Asymmetric circuit really.
```
```On Mon, 19 Aug 2013 15:38:13 GMT, Jan Panteltje <pNaonStpealmtje@yahoo.com>
wrote:

>On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
><pcdhSpamMeSenseless@electrooptical.net> wrote in
><kuta6k\$3gg\$1@dont-email.me>:
>
>>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.)
>
>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.

It's an amplifier that has an input impedance. Whether you elect to call that a
voltage amp or a current amp is up to you.

>You want a MOSFET if you want a voltage amplifier.

The classic AC-coupled single bipolar transistor amplifier has a voltage gain
that's independent of beta.

>
>
>
>
>>The fundamental control mechanism of a BJT is the base-emitter voltage,
>>which provides pretty tight voltage feedback in an emitter follower.
>
>Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).

Can you explain that?

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
```
```On Mon, 19 Aug 2013 15:44:17 GMT, Jan Panteltje <pNaonStpealmtje@yahoo.com>
wrote:

>On a sunny day (Mon, 19 Aug 2013 15:38:13 GMT) it happened Jan Panteltje
><pNaonStpealmtje@yahoo.com> wrote in <kute56\$vso\$1@news.albasani.net>:
>
>>On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
>><pcdhSpamMeSenseless@electrooptical.net> wrote in
>><kuta6k\$3gg\$1@dont-email.me>:
>>
>>>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.)
>>
>>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.
>>
>>
>>
>>
>>>The fundamental control mechanism of a BJT is the base-emitter voltage,
>>>which provides pretty tight voltage feedback in an emitter follower.
>>
>>Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).
>
>Correction, I ment Zout is Zload when going 'down', but Zi / beta when going 'up'.

Not true. Even when Zi = 0, an emitter follower's output impedance isn't zero.

>Asymmetric circuit really.

Only if it's slewing large-signal, fast.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
```
```On 08/19/2013 11:38 AM, Jan Panteltje wrote:
> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
> <pcdhSpamMeSenseless@electrooptical.net> wrote in
> <kuta6k\$3gg\$1@dont-email.me>:
>
>> 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.)
>
> 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.
>

I didn't say what I wanted.  The base current comes from recombination,
which in a transistor is a loss mechanism.  That's why beta is not a
good design parameter in most cases.
>
>
>
>> The fundamental control mechanism of a BJT is the base-emitter voltage,
>> which provides pretty tight voltage feedback in an emitter follower.
>
> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).
> Asymmetric circuit really.

In a voltage regulator, one hopes that the base voltage only goes down
much when power is removed.  And the input impedance when the voltage is
going down is much larger than Zload until the base-emitter junction zeners.

Cheers

Phil Hobbs

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

Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
```
```On Mon, 19 Aug 2013 15:38:13 GMT, Jan Panteltje
<pNaonStpealmtje@yahoo.com> wrote:

>On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
><pcdhSpamMeSenseless@electrooptical.net> wrote in
><kuta6k\$3gg\$1@dont-email.me>:
>
>>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.)
>
>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>

```
```On Mon, 19 Aug 2013 12:07:44 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

>On 08/19/2013 11:38 AM, Jan Panteltje wrote:
>> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
>> <pcdhSpamMeSenseless@electrooptical.net> wrote in
>> <kuta6k\$3gg\$1@dont-email.me>:
>>
>>> 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.)
>>
>> 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.
>>
>
>I didn't say what I wanted.  The base current comes from recombination,
>which in a transistor is a loss mechanism.  That's why beta is not a
>good design parameter in most cases.
>>
>>
>>
>>> The fundamental control mechanism of a BJT is the base-emitter voltage,
>>> which provides pretty tight voltage feedback in an emitter follower.
>>
>> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).
>> Asymmetric circuit really.
>
>In a voltage regulator, one hopes that the base voltage only goes down
>much when power is removed.  And the input impedance when the voltage is
>going down is much larger than Zload until the base-emitter junction zeners.
>
>Cheers
>
>Phil Hobbs

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

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
```
```On Mon, 19 Aug 2013 09:26:23 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 19 Aug 2013 12:07:44 -0400, Phil Hobbs
><pcdhSpamMeSenseless@electrooptical.net> wrote:
>
>>On 08/19/2013 11:38 AM, Jan Panteltje wrote:
>>> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs
>>> <pcdhSpamMeSenseless@electrooptical.net> wrote in
>>> <kuta6k\$3gg\$1@dont-email.me>:
>>>
>>>> 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.)
>>>
>>> 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.
>>>
>>
>>I didn't say what I wanted.  The base current comes from recombination,
>>which in a transistor is a loss mechanism.  That's why beta is not a
>>good design parameter in most cases.
>>>
>>>
>>>
>>>> The fundamental control mechanism of a BJT is the base-emitter voltage,
>>>> which provides pretty tight voltage feedback in an emitter follower.
>>>
>>> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload).
>>> Asymmetric circuit really.
>>
>>In a voltage regulator, one hopes that the base voltage only goes down
>>much when power is removed.  And the input impedance when the voltage is
>>going down is much larger than Zload until the base-emitter junction zeners.
>>
>>Cheers
>>
>>Phil Hobbs
>
>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

Works but, "IRL", you'd use a current probe and voltage probe and plot
the load line on your o'scope.  That's how I developed the snubbing
for the SMPS I've commented about here before... "cold but hot"... the
technician's delight watching the boss jump over the back of the lab
stool ;-)

In the Spice world it's just an EVALUE (or B-source)...

EPOWER PCALC 0 VALUE = {VDS(M1)*ID(M1)} ; "PCALC" a node name

Or you can similar calculations in a post-processor, as I do in PSpice
Probe, to check CMOS gates for hot-electron violations...

SOAG(M,VDS,VGS)=(SGN(VD(M)-VS(M)-VDS)+1)*(SGN(VG(M)-VS(M)-VGS)+1)/4

...Jim Thompson
--
| James E.Thompson                                 |    mens     |