Current through diodes

Discussion in 'General Electronics Chat' started by HarrisonG, Aug 7, 2016.

Aug 1, 2016
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Since I learned about the diode devices so long ago, I have never been able to answer one problem. If we have a zener diode with 5,6v treshold voltage and a transistor with 0,7v treshold voltage, and we connect the negative end of the diode and the base of the npn transistor to a base ressistor directly, then how to find what the current through each of them will be? It's easy to solve this if they were resistors, but the only thing you know about them is that they have some turn on voltage, so I can't use Ohm's law. Except if I gi with the internal ressistance of the devices, but that is too far away from the whole thing and sounds complicated and out of place. Forgot to mention that the emitter is hold at constant 5volts so the transistor can't short the base ressistor. It's confusing me, there is nothing I can hook up to to figure out the zener current or the base current. Sure the zener have some minimum turn on current, but how do I know that this current wont be more. Is it load dependent? Maybe the five volts through the load give some emitter current and thus a certain base current and you can substract that base current from the whole current through the base ressistor to find the zener current Iz = Is-Ib ? I don't know, it's almost like they act by a coincidence, but i'm pretty sure it's load dependant.

2. DickCappels Moderator

Aug 21, 2008
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The short answer is: use SPICE.

It would be helpful if you would draw and post the circuit. I thought I was following you until you said the emitter is held at 5 volts.

Aug 1, 2016
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A simple transistor series voltage regulator

4. Jony130 AAC Fanatic!

Feb 17, 2009
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Aug 1, 2016
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Oh, so that's how it is. One last thing: The zener turn on current is the current at which
I get it. By the way the zener turn on current is the current at which the breakdown occurs and the diode can conduct while in both dirrections right? I haven't heard of turn on current in other semiconductors. If the zener current is less than the turn on current it wont be able to generate the requiered power or heat for the breakdown to occur. Is that right? I might verry well be wrong on that one.

6. hp1729 Well-Known Member

Nov 23, 2015
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Build the circuit. Play with it and see what it does.
Emitter current divided by current gain of the transistor gives you base current. Zener current needs to be 5 or 10 times that to regulate well. Both currents flow through the resistor, Rs.

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7. Jony130 AAC Fanatic!

Feb 17, 2009
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Look here and notice Izk

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Aug 1, 2016
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Okay, i've watched Dave's video, i have readed tons of stuff in google and youtube and i still can't explain myself why the zener won't regulate the voltage if it's current is below Izk. The current below Izk as far as i learned is the leakage current that every diode has because of real life inperfection. I've tried to study the reverse characteristic of the zener diode and the relationship between I and V and graphicly it seems that if you limit the zener current below Izk you wont get the zener voltage but rather a lower voltage. But that i know can't happen in reality because when you apply 5 volts to the diode, you apply 5 volts at the diode. The zener current won't and can't limit that 5 volt to say 3 or 4 volts. The voltage would rather stay the same while the current will reduce . So that basicaly means the voltage is still regulated. And that's weird cause i'm sure i'm wrong here. But here i just got an idea. Maybe if the zener current is below Izk, the diode won't turn on fully and the voltage drop across it won't be 5 volts (according to the reverse characteristic graph), but lower. Cause it won't be turned on fully it would require less voltage to support the current and thereby the voltage drop across it will be less than Vz. And if you have a resistor in series with the diode, a supply voltage of 12volts, and let's say the zener current is reduced below Izk and instead being 5 volts, the voltage drop across the diode will be 3volts. Then the rest 9 volts will fall across the resistor via Kirchoff's law, which won't work for us if we want to get 5 volt output. Don't know if that's the answer, if not, then help me find out why IZK<IZT.

9. #12 Expert

Nov 30, 2010
18,093
9,679
That's exactly where you're wrong. The curve for a zener is not a square wave. The voltage sneaks up as the current increases. If you want perfection, try a TL431.

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10. ian field AAC Fanatic!

Oct 27, 2012
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Vout is pretty much Vz <minus> (about) 0.7V.

The 5.6V Zener came in just nicely for simple emitter follower 5V TTL Vcc regulators.

Usually the current requirement calls for a Darlington emitter follower so the circuit gets a little more complex.

The SMPSU route is easier - you can probably hack a car accessory socket to USB adapter as easy as getting an analogue regulator just right.

In most cases; you can avoid a heatsink making the place look untidy.

11. crutschow Expert

Mar 14, 2008
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It will regulate below Izk. Why do you think it won't?
Izk, is just the current that the zener voltage is rated at.
A zener will still exhibit a knee voltage but as the current goes below (or above) Izk, the voltage will also slightly change (due to the zener impedance Zzt).
There is no sudden drop in the voltage below Izk.

Below is an LTspice simulation of a 1N741 5.1V zener which has an Izt of 20mA.
As you can see, the voltage is 4.7V starting at 1μA current, and goes to 5.2V @ 30mA, giving an average Zzt of about 16.7Ω (specified as 17Ω in the data sheet).

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