electrical logic of current limiting - Page 1
Forum section: electronics / projects

author: john23
author_rank: Frequent Contributor
timestamp: November 10, 2024, 05:23:57 pm
content: Hello,given the schematics below the zener diode gets reversed biased at 4.7 V.so the current is I=(15-4.7)/R .we automatickly assume we have 4.7V on the zenner, and we can calculate the current by the resistor.under what conditions the zenner will not enforce the 4.7V threw it?Thanks.https://www.mouser.co.il/datasheet/2/268/1N746A_1_1N759A_1_2c1N4370A_1_1N4372A_1-3442192.pdf

author: Andy Chee
author_rank: Super Contributor
timestamp: November 10, 2024, 05:40:52 pm
content: Quote from: john23 on November 10, 2024, 05:23:57 pmwhen the current is too low (e.g. 1mA)some good explanation here:https://www.sound-au.com/appnotes/an008.htm

author: TimFox
author_rank: Super Contributor
timestamp: November 10, 2024, 05:54:08 pm
content: The behavior at very low currents, as mentioned above, is more pronounced for lower-voltage "Zener" diodes, such as the 1N750, that are actual Zener diodes.For nominal voltages above approximately 6 V, "Zener" diodes are actually "avalanche" diodes where the current increases faster with voltage than lower-voltage units, hence the voltage varies less with current.For example, according to the ancient Motorola book, a 1N750 has 3.0 Vacrossit for 0.1 mAthroughit, and 4.4 V at 10 mA.The avalanche diode 1N754 increases from 6.4 V at 0.1 mA to 6.5 V at 10 mA, a much more constant voltage.See page 4-6 of the 1967 edition of Motorola'sZener Diode Handbook.

author: john23
author_rank: Frequent Contributor
timestamp: November 10, 2024, 06:31:50 pm
content: Hello,There is a logical problem.In order to calculate the zener current threw the limiting resistor we always assume that zenner is working properly.how can i know mathematickly the current without assuming i have proper 4.7V across the zener?Thanks.

author: TimFox
author_rank: Super Contributor
timestamp: November 10, 2024, 06:44:56 pm
content: Quote from: john23 on November 10, 2024, 06:31:50 pmThe data sheet only specifies voltage at a specific test current.If you run the Zener diode at a much different current, its voltage will be different, especially for lower voltage units such as 1N746 through 1N750.To calculate the Zener voltage with a fixed source and series resistance, you need to start with a graph of Zener voltage vs. current and draw a “load line” (determined by the source voltage and series resistance) to find the intersection with the non-linear Zener characteristic.The Zener equations are more complicated than merely a “proper” voltage in series with a resistance: that model is only useful for small variations in current around the specification point.The load line for a source and resistor is the locus of voltage after the resistor as a function of the current flowing into the load (your Zener).There is no logical problem here, just a very non-linear practical device.

author: john23
author_rank: Frequent Contributor
timestamp: November 10, 2024, 07:02:47 pm
content: Hello ,I have made such a plot, I see that in 9mA current I have good 4.7V output as shown below.In the lab I have a voltage source of 12V , how this plot will help me decide what current limiting resistor to choose?Thanks.

author: Andy Chee
author_rank: Super Contributor
timestamp: November 10, 2024, 07:18:29 pm
content: Quote from: john23 on November 10, 2024, 07:02:47 pmYou need to provide information about the load.

author: inse
author_rank: Super Contributor
timestamp: November 10, 2024, 07:27:15 pm
content: What are you using the Zener diode for, to generate a kind of reference voltage or as shunt regulator to supply a circuit?As a reference generator, e.g. TL431 would be much more stable, also as shunt regulator.

author: TimFox
author_rank: Super Contributor
timestamp: November 10, 2024, 07:40:28 pm
content: Quote from: john23 on November 10, 2024, 07:02:47 pmYou mean, (12 - 4.7) V / (9 mA) = 811?As others have discussed above, there is more to specify to get a useful answer.The plot I discussed adds a linear functionVload=Vsource- (IloadxR) and finds the intersection with the diode's I-V curve.Often, the two plots are on a graph with voltage on the horizontal axis and current on the vertical axis, but it works either way.The 1N750 is not designed as a reference diode.

author: john23
author_rank: Frequent Contributor
timestamp: November 10, 2024, 08:46:54 pm
content: Hello, There is a similar case in NPN,I swept the base current.I want to decide what base limiting resistor to use.I made plots shown below.I have 12V voltage source i want base current 37uA.that meens my resistor is 12/37uA=R which is very very large.Is there a way to reduce the need for such a large limiting resistor?Thanks.

author: TimFox
author_rank: Super Contributor
timestamp: November 10, 2024, 09:54:15 pm
content: Your drawing has emitter and collector labels interchanged.In general, you don't specify the base current for a BJT circuit: you specify the collector (or emitter) current and use the manufacturer's wide range of "beta" to calculate the correspondingly large range of base current, and make sure your driving circuit can supply that current.If you look at the datasheet for a 2N3904 (for example), the manufacturer specifies (at a given collector current) a 3:1 range of "beta" (hFEfor parts that pass the production test and are shipped to the user.Seehttps://www.onsemi.com/download/data-sheet/pdf/2n3903-d.pdf12 V/37 uA = 324 k\$\Omega\$, which is not "very very large".In a real circuit,Vbedoesn't vary much across that range of base current, so the actual base current is very close to (12 V - 0.7 V)/Ib.In your posted circuit, much of the "base" voltage is actually across the emitter resistor.

author: john23
author_rank: Frequent Contributor
timestamp: November 11, 2024, 06:32:25 pm
content: Hello TimFox,I have build NPN which base is being biased with a current source ,the problem that My collector current is very low no matter the current i put in the base as shown in the photo and simulation.Why i cant see the Ic/Ib=beta ratio?Thanks.

author: TimFox
author_rank: Super Contributor
timestamp: November 11, 2024, 06:49:06 pm
content: You have applied +5 V to the collector.At 20 mA emitter current, the voltage on the emitter is +20 V, so the base is positive with respect to the collector and emitter.You are only seeing the base current flow into the collector, without current gain.You say you built this: is your graph from measurement or only simulation?Why do you want to bias this stage with external base current instead of somehow stabilizing the emitter current?What collector current do you want to achieve?Why are you using a 1 k\$\Omega\$ emitter resistor?

author: john23
author_rank: Frequent Contributor
timestamp: November 12, 2024, 09:11:40 pm
content: Hello,the purpose of biasing NPN threw current source is to calculate the limiting resistor needed for supplying the base of the NPN.Why in saturation Ic/Ib is very very small?Thanks.

author: TimFox
author_rank: Super Contributor
timestamp: November 12, 2024, 10:38:52 pm
content: Why do you have that absurd 1000 ohm resistor in the emitter?With a +5V supply, you can only get 5 mA through it: everything else in the emitter circuit is wasted base current.

author: john23
author_rank: Frequent Contributor
timestamp: November 13, 2024, 07:55:19 pm
content: Hello , I biased my NPN to design the limiting resistor.by your advice my maximal current should be 1000*(5/200)=25mA which is exactly the saturation current i got in the plot below.I have made a simulation shown below by your advice where I lowered the current going into the base.Three questions:1.In the plot I saw the when the collector current is saturated Vce is very low .Why the Vce is very low in this state?2. suppose I have a 3V voltage source instead of the current source.I want to stay in linear state where Ib=45uA Ic=9mA as shown in the plot below.How do i calculate the needed limiting resistor if I replace the curent source with 3V voltage source?Thanks.

author: TimFox
author_rank: Super Contributor
timestamp: November 13, 2024, 08:02:36 pm
content: For a BJT, "saturation" literally means thatVceis low, such that the B-C junction (normally reverse-biased) starts to forward bias.What did you think it meant?If you want a simple bias network for a NPN, withIc= 9 mA, start with an emitter resistor that gives, say, +2 V at 9 mA.Then put a voltage divider at the base that establishes the base voltageVb(not the current) at that voltage plusVbe= 0.7 V.With the manufacturer's minimum value ofhFE, you will get a maximum value ofIbthat loads that divider; the variation ofhFEwill give a variation in thatVb, which can be made small with reasonably small resistors in the voltage divider.One should not design a bias network that depends directly on the base current for linear operation: that is the design procedure for a switching circuit, where you establish sufficient base current to ensure saturation.

author: bson
author_rank: Unknown
timestamp: November 14, 2024, 09:31:05 pm
content: Quote from: john23 on November 10, 2024, 06:31:50 pmThe datasheet for the zener diode will have a zener current. At this current, and no other, will it have a 4.7V drop. You need to size the resistor to provide this zener current, assuming a 4.7V drop over the diode.