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Zener regulator : load and zener current

Started by Unknown August 18, 2013
In the simple zener regulators like the one below, If the load current is 1=
00ma 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 throug=
h 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 curr=
ent the remaining to the load. Is it due to the load impedance being less t=
han the zener impedance? =20

Vin------/\/\/\/\--------|-----------|
                         |           |
                         |           | =20
                         |           |
                         | /        RL
                       /---          |
                         /\          |=20
                        /  \         |
                       ------      -----
                          |         ---
                          |          -
                          |
                        -----
                         ---
                          -

Thanks
Sridhar
On a sunny day (Sun, 18 Aug 2013 05:12:11 -0700 (PDT)) it happened
sridhar09.cherukuri@gmail.com wrote in
<51cad867-b6d6-4bc6-b542-54cec8145d9e@googlegroups.com>:

>In the simple zener regulators like the one below, If the load current is 1= >00ma 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 throug= >h 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 curr= >ent the remaining to the load. Is it due to the load impedance being less t= >han the zener impedance? > >Vin------/\/\/\/\--------|-----------| > | | > | | > | | > | / RL > /--- | > /\ | > / \ | > ------ ----- > | --- > | - > | > ----- > --- > - > >Thanks >Sridhar
Simplest way to imagine is: remove RL then the output voltage should be Vout = Vin * (RL / (Rs + RL) ) The 'zener' will limit voltage to it's zener voltage however. If you remove the zener only, then the output voltage will always be: Vout = Vin * (RL / (Rs + RL) ) If you remove both zener and RL then the putput voltage will be Vin If you remove ... OK:-)
On Sun, 18 Aug 2013 05:12:11 -0700 (PDT), the renowned
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
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 -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
<sridhar09.cherukuri@gmail.com> wrote in message 
news:51cad867-b6d6-4bc6-b542-54cec8145d9e@googlegroups.com...
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 


On Sunday, August 18, 2013 8:12:11 AM UTC-4, sridhar09...@gmail.com wrote:
=20
> If Vin is above Vz and the load draws 50ma, the remaining 60ma flows thro=
ugh 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 cu= rrent the remaining to the load. Is it due to the load impedance being less= than the zener impedance? =20
>=20
It has nothing to do with impedances, it has to everything to do with the z= ener and its terminal voltage. In the first order idealization of zener ope= ration, current in the direction of the diode arrow produces 0V terminal vo= ltage, and current in the opposite direction of diode arrow produces a term= inal voltage of Vz volts. In both cases the terminal voltage is constant an= d independent of the actual magnitude of the current. So to answer your que= stion, if there is any current at all flowing through the zener in your cir= cuit, 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 deve= lop 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 Rlo= ad and Ohm's law sets the output voltage and current.
On Monday, August 19, 2013 3:19:49 AM UTC+5:30, bloggs.fred...@gmail.com wr=
ote:
> On Sunday, August 18, 2013 8:12:11 AM UTC-4, sridhar09...@gmail.com wrote=
:
>=20 > =20 >=20 > > If Vin is above Vz and the load draws 50ma, the remaining 60ma flows th=
rough zener. What makes it that the remaining current after the load curren= t 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 le= ss than the zener impedance? =20
>=20 > >=20 >=20 >=20 >=20 > 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 o= peration, current in the direction of the diode arrow produces 0V terminal = voltage, and current in the opposite direction of diode arrow produces a te= rminal 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 q= uestion, if there is any current at all flowing through the zener in your c= ircuit, then the zener cathode node must be /fixed/ at Vz volts. If the nod= e is fixed at Vz volts, then Rload has Vz volts across it and so its curren= t has to be Vz/Rload. Similarly, the input resistor has a fixed Vin-Vz acro= ss 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 suppl= y enough current to power Rload at Vz/Rload, obviously, which means there i= s no Vz because there is not enough current to flow through the zener to de= velop this Vz. So the zener is in effect out of the circuit, it is open, an= d the circuit reduces to a simple voltage divider with Rin in series with R= load and Ohm's law sets the output voltage and current. Thanks for making it clear. I simulated the schematic below in pspice to un= derstand. Current through the base is always 1.3ma irrespective of V1 and R= 1 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 r= educes 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
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? > > 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. -- Tauno Voipio
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: >=20 > > On Monday, August 19, 2013 3:19:49 AM UTC+5:30, bloggs.fred...@gmail.co=
m wrote:
>=20 > >> On Sunday, August 18, 2013 8:12:11 AM UTC-4, sridhar09...@gmail.com wr=
ote:
>=20 > >> >=20 > >> >=20 > >> >=20 > >>> 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 curr= ent go through the zener? Whay can't it be the otherway? i.e after the zene= r current the remaining to the load. Is it due to the load impedance being = less than the zener impedance?
>=20 > >> >=20 > >>> >=20 > >> >=20 > >> >=20 > >> >=20 > >> 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 zene= r operation, current in the direction of the diode arrow produces 0V termin= al voltage, and current in the opposite direction of diode arrow produces a= terminal voltage of Vz volts. In both cases the terminal voltage is consta= nt and independent of the actual magnitude of the current. So to answer you= r question, if there is any current at all flowing through the zener in you= r 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 cur= rent has to be Vz/Rload. Similarly, the input resistor has a fixed Vin-Vz a= cross it, so the current through it has to be (Vin-Vz)/Rin. Then since char= ge is conserved, the zener has to take the excess of (Vin-Vz)/Rin - Vz/Rloa= d. If there is zero or negative excess, this means (Vin-Vz)/Rin does not su= pply enough current to pow
>=20 > 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 se= ts the output voltage and current.
>=20 > > >=20 > > Thanks for making it clear. I simulated the schematic below in pspice t=
o understand. Current through the base is always 1.3ma irrespective of V1 a= nd R1 with a load of 50 ohms and the rest goes through the zener. Only chan= ging the load changes the current through the base. Changing load to 100 oh= ms reduces the base current to approx 600ua and changing it to 25 ohms chan= ges the base current 3.3ma. How are these base current values coming from?
>=20 > > >=20 > > http://www.flickr.com/photos/100471956@N04/ >=20 > > >=20 > > >=20 > > -Sridhar >=20 > > >=20 >=20 >=20 > You are feeding the load with constant voltage =3D zener voltage -=20 >=20 > base-emitter drop, so the load current is constant. The base current is=
=20
>=20 > the load current divided by the DC current gain of the transistor. >=20 >=20 >=20 > I suspect that you're going to blow up your transistor, please check the=
=20
>=20 > power dissipation in it. >=20 >=20 >=20 > --=20 >=20 >=20 >=20 > Tauno Voipio
What I read about BJTs tell me that we control the base current and the col= lector 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.=20 Thanks Sridhar
On 19.8.13 1:59 , 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 p
ow
>> >> 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. >> >> >> >> -- >> >> >> >> 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
Get a good electronics textbook and read about an emitter follower. There is feedback from the output voltage
On Monday, August 19, 2013 2:47:27 AM UTC-4, sridhar09...@gmail.com wrote:

>=20 > 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 changi= ng 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 change= s the base current 3.3ma. How are these base current values coming from?
>=20 >=20 >=20 > http://www.flickr.com/photos/100471956@N04/ >=20
The voltage across the load resistor is Vz-Vbe, where Vbe is forward bias v= oltage across the transistor base-emitter junction required for it to condu= ct. Therefore, the load current is (Vz-Vbe)/Rload, and this is supplied by = the transistor emitter current Ie. The transistor base current required for= an emitter current of Ie is Ib=3D Ie/(beta+1) where beta is transistor cur= rent gain. So the Ib=3D ((Vz-Vbe)/Rload)/(beta +1)=3D(Vz-Vbe)/(Rload x (bet= a+1)) mathematically describes the base current dependence on Rload and als= o the variation you are seeing.