Diode voltage vs Forward voltage

[JackJones99]

Hey guys. I'm new here! Anyway, I'm trying to stock up on some commonly used diodes for repairs & different projects, and I'm a bit confused on the VF [voltage forward] part. For instance, on the following datasheet:
http://www.kdiode.com/admin/diotech_file/SF31~SF38.pdf
It says that SF38 [600V 3A] has a VF of 1.7V, and that SF35 [300V] has 1.25V. Now, my question is, say I take SF38 but provide it 300V @3A, would the max VF of it be 1.25V (like SF35) or would it still be 1.7? I'm wondering if I should get two values from each series of diodes or just get the highest value one of each and reduce costs, and this VF thing is not entirely clear to me. Hope someone would be able to clarify this for me.

[rstofer]

Vf is the voltage across the diode where significant current starts to flow.  It is clearly a function of the material, doping, junction size, etc.

There are Vf curves that relate Vf to current, it's not a fixed point.  We usually use 0.7V for rectifier applications and pretty much all diodes when we're working on the back of a napkin.

Germanium diodes (if they're still around) have a Vf of around 0.2V.

You are using power diodes that handle significant current and voltage.  Vf won't be the same as a 1N914.

Here are a couple of important curves.  Note how the Schottky diode turns on abruptly whereas the standard PN diode ramps up in Vf versus current

https://www.powerelectronictips.com/less-famous-attributes-schottky-diodes/

Even the 1N914 can have some variance in Vf, it's spec'd at a max of 1V at 10 mA forward current.
Figure 1 shows it is usually a little better than that at 0.7V 10 mA.  Note that most of the specs are at 10 mA - there's a theme.  It is a low current fast switching diode, not an industrial rectifier.

https://www.vishay.com/docs/85622/1n914.pdf

[rstofer]

Vf * If => Power as in heating.  For rectifier applications, this might be a factor.

Put 5V into the anode of a 1N914 diode and connect a 470 Ohm resistor from the cathode to ground/common.  You should be able to measure 0.7V across the diode and about 10 mA through the circuit.  Change to other valuse like 220 Ohm and 1k Ohm and see what you get.

[JackJones99]

I appreciate your quick response rstofer! I was also quite surprised by your explanation, as I was under the impression that the forward voltage is in fact the power loss that occurs when it's under a specific load [volt/amp]. There's definitely some conflicting information out there. So it's basically just the minimum, or starter voltage of it then?

[Kleinstein]

The curve shown for the shottky diode is a bit odd - at least in the relatively low current range both types of diode behave about exponential and thus not very much difference in the curve for a schottly and normal diode, except for the lower voltage.

With a exponential curve ( I ~ exp(V_f / 26 mV) there is no real starting point, but Vf depends on the current. For every factor of 10 in the current the voltage changes by about 60 mV. If a rectifier like 1N4007 starts at 700 mV for 100 mA, expect some 580 mV at 1 mA, 400 mV and 1 µA and some 220 mV at 1 nA.
With a linear scale for the current it looks like a starting point for the current, but where the bend is depends on the scale. With a log scale for the current the curve usually looks relatively close to a line.
The simple exponential curve is an approximation only, but a relativeyl good one. At high currents one sees some series resistance to actual diode part and there are also other details that effect the curve a little.

The quoted number in the datasheets are often curde numbers and different batches / manufacturers may be a bit different.

[ledtester]

It says that SF38 [600V 3A] has a VF of 1.7V, and that SF35 [300V] has 1.25V.  Now, my question is, say I take SF38 but provide it 300V @3A, would the max VF of it be 1.25V (like SF35) or would it still be 1.7?

The 300V / 600V numbers refer to reverse voltages -- i.e. when the cathode voltage is greater than the anode voltage.

If you apply a 300V reverse voltage across a SF38 you should be fine since the datasheet says it can handle 600V. In this situation very little current flows through the diode.

As for the forward voltage drop, the only guarantee you have from the manufacturer is that the Vf for the SF38 at 3A will be <= 1.7V.

Don't apply a 300V forward bias across either the SF35 or SF38 -- you'll just let the magic smoke out.

[Terry Bites]

I hope this makes sense https://www.allaboutcircuits.com/video-lectures/diode-characteristics-circuits/

[floobydust]

A diode's forward voltage drop is dependent on current flow, temperature, and the reverse-voltage rating of the diode. I think this is the confusion here.
Higher voltage (PIV) diodes have larger junctions and so generally greater conduction losses in the forward direction.
So the 200V part Vf=0.95V compared to 600V part Vf=1.7V at 3A. That is almost (179%) twice the losses and almost twice the heat. This is a big deal at 3A, so the approach to just put in (buy) the highest voltage part is not always best.

I haven't used these super-fast recovery parts 35nsec, it's fine to use UF5408 75nsec but still the Vf is jumps higher past the 400V league.

[JackJones99]

Thanks for all help guys! To sum up what I've understood from all of this:

My initial understanding of what VF is was correct (it's the voltage drop, or power loss) that occurs, and is different for each diode (on some diodes, such as the 1N4001-1N4007 series, it's the same for all of em, but for a lot of fast rectifiers, it does vary between the values). What I was unsure about is if the VF is affected strictly by the current that's applied to it, or a combination of the voltage+current, and turns out it's just the current that affects it, so stocking up on the highest value diode from each series isn't always the best approach it seems. I'll prob get two values from each (for those that have different VF levels)

[Cerebus]

It's all governed by the "Diode Equation":
\[I_D = I_s \{e^{\frac{qV_D}{nkT}}-1\}\]

Where:
\$I_D\$ is the current passing through the diode
\$V_D\$ is the voltage across the diode
\$e\$ is Euler's number (2.71828...)
\$q\$ is the elementary charge (the charge on a single electron in Coulombs) = \$1.602176634×10^{−19}\$
\$n\$ is the ideality factor for the semiconductor in question (1 for silicon)
\$k\$ is the Boltzmann constant (\$1.380649×10^{-23} JK^{-1}\$)
\$T\$ is the thermodynamic temperature in degrees Kelvin
\$I_s\$ is the saturation current for the diode in question. Datasheets don't give this. Basically it's proportional to the junction area of the diode.  For a 1N914 the saturation current is 2.52nA.

At room temperature \$\frac{nkT}{q}\$ comes out at approximately 25.9mV for a silicon diode which means that for every increment of 25.9mV more voltage across a silicon diode it will pass \$e\$ (or 2.72) times more current. So, the current through the diode is controlled by the voltage across the diode, and the relationship is exponential.

[Kleinstein]

The super fast diodes seem to have an unusual high forward voltage drop. For more normal silicon diode the difference with the withstanding voltage is usually not that large. For higher voltages (e.g. > 600 V) an internal PIN construction is more common and this leads to a slightly higher voltage and also slight deviations from the simple Shockly diode equation.
If lower V_f  is wanted it is a good idea no to use the diodes to the full rated current. The curves usually show the pulsed V_f. With more comtineous operation the diode will get hot and this will reduce the votlage drop. At the nominal current and some 100 K temperature rise this can make a difference of some 200 mV.

[Siwastaja]

So, the current through the diode is controlled by the voltage across the diode, and the relationship is exponential.

While all of this is completely true, I know from experience this is extremely confusing to beginners because this is all upside down from practical viewpoint; where diodes usually are used, known voltage is not deliberately applied over a diode, but the opposite is true: current which is defined by the other circuitry flows through the diode and causes an often unwanted drop i.e. loss of voltage.

So you can turn the exponential equation upside down to get Vf on the left side of the equation: voltage drop seen over the diode is a logarithm of the current through the diode. And, logarithm being what it is, a pretty good bang-for-a-buck approximation over a limited range is a constant, hence we see this advice that Vf of a diode is some 0.7V, depending on the diode type of course. In reality, it's not 0.7V but depends on current, but if current is within one order of magnitude from the diode rating, Vf doesn't change that much.

What comes to OP's question, I would try to simplify it this way:

Vf is a voltage drop or loss when the diode is conducting, for ideal diode this would be 0. This causes conduction loss P = Vf * I.

Vf is a per part number parameter; just simply look it up in the datasheet instead of confusing yourself with generalizations of gazillion of diode types and conditions. It's given in certain conditions like temperature and current. Like all parameters, it has variation between parts and hopefully manufacturer gives you minimum, typical and maximum  instead of just "typical" so you have guarantees to work with.

Lower Vf is most often desirable to reduce the power loss. But sometimes circuit designers take advantage of the diode's voltage drop when they need a specific (not highly accurate though) voltage drop somewhere in their circuit.

On the other hand, Vr is the voltage in the front page name (300V, 600V...) and it tells you how much the diode can take reverse voltage when it's not conducting. In this reverse direction, the diode drops all of the voltage available up to a limit where it blows up.