Diode Foward Voltage curve: Pulsed Vs Static or Vishay optimistic datasheet? - Page 1
Forum section: electronics / projects

author: TizianoHV
author_rank: Regular Contributor
timestamp: April 21, 2025, 08:49:20 am
content: Hi everyone Am I doing something wrong?I was looking for a diode with a foward voltage drop of around 0.55 V at 5 mA so that I could use it as a voltage reference (temperature influence and slow variations will be corrected with an autozero function).The BA157 from Vishay (https://www.mouser.it/ProductDetail/Vishay-General-Semiconductor/BA157-E3-54?qs=tdXsjSi%2FIMbKYNQf%252BAKGVA%3D%3D) seemed a good choice from datasheet (570mV@10mA).So I ordered 100 of these from Mouser but after testing them I wasn't impressed: Up to 700mA they show a Vf almost 100mV over the "typical" curve.Am I expecting too much? Usually I found these curves pretty accurate, if not pessimistic.Does it make any difference when they specify "pulsed..." (I thought that it was only for thermal reasons)?Do you have any suggestion for an alternative (or different solution) to it?Thanks!

author: moffy
author_rank: Super Contributor
timestamp: April 21, 2025, 09:02:49 am
content: They are maintaining Tj@25C, with continuous current you are heating the junction so your Tj will be going up with current. About 250mA for the datasheet, the curve goes linear this is when the series resistance of the diode has become dominant. Can't comment on the higher Vf you are seeing.

author: PGPG
author_rank: Frequent Contributor
timestamp: April 21, 2025, 11:11:50 am
content: It is generally not a good idea to use diode as a voltage reference.

author: mawyatt
author_rank: Super Contributor
timestamp: April 21, 2025, 12:53:50 pm
content: If you must use a PN junction, try a Bipolar Transistor (2N3904) connected as a Diode, this might provide better results.Temperature will have a significant effect on a PN junction, and this is often used as a temperature sensor. A typical Bipolar Base Emitter junction will show about -2.2mv/C temperature @ ~1ma.Best

author: TimFox
author_rank: Super Contributor
timestamp: April 21, 2025, 03:49:53 pm
content: Short answer:Manufacturers test the diode voltage using a very short pulse on time, so that the diode guts do not have enough time to heat up.Otherwise, the test time would need to be very long, to allow the device temperature to reach an equilibrium, and the test would not be practical for production QA.

author: DrijSkij
author_rank: Contributor
timestamp: April 23, 2025, 02:04:39 pm
content: Quote from: TimFox on April 21, 2025, 03:49:53 pm"Guts" . I like the technical term.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 23, 2025, 07:57:56 pm
content: Quote from: TizianoHV on April 21, 2025, 08:49:20 amWhat exactly are you trying to do ?You can get 550mV from a BJT, by using a two resistor dropper design : The 5mA feed resistor and then a B-C resistor is chosen to get some mV shift down from VBE,Spice will give you a quick starting point: Eg with a 2N3904 and 5mA, a B-C resistor of 30.255 ohms gives 550.85528mV at 30'CYou do not need 5mA for this, a lower current will also work.eg at 500uA, 178.5 Ohms gives 550.71753mV, or 100uA and 468.6 ohms gives 550.55708mV.If you drop Ic to 16.85uA, no trim R is needed for 550.03864mV, but the 'reference' source impedance is larger at lower currents.You would pick a mainstream vendor and buy a reel of parts to use for this task.