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?
On a sunny day (Sun, 18 Aug 2013 05:12:11 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :
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
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.
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?
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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
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Think of a zener as a voltage source that can only sink current. Art
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?
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, snipped-for-privacy@gmail.com wr ote:
:
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?
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?
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.
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.
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?
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
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.
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?
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.
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?
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= Ie/(beta+1) where beta is transistor cur rent gain. So the Ib= ((Vz-Vbe)/Rload)/(beta +1)=(Vz-Vbe)/(Rload x (bet a+1)) mathematically describes the base current dependence on Rload and als o the variation you are seeing.
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
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Dr Philip C D Hobbs
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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.
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On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs wrote in :
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.
Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload). Asymmetric circuit really.
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.
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
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
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.
formatting link
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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
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...
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