# Instantaneous forward voltage

#### dick_girard

Joined Jun 22, 2007
3
Hi,

I am looking at a power diode that only shows the "instantaneous" forward voltage drop across the diode at a specified current. Most other diodes show the "forward drop" (not the "instantenous" forward drop) with current.

My question is: How do I determine the "constant" forward voltage drop? Is this the same as the instantaneous drop?

I am looking at the Vishay FP30AB dual-diode. They show an "instantaneous" drop of 0.95V drop at 15A per leg. My circuit [a DIY UPS for computer systems] will draw a steady-state current, not instantaneous. Anyone know about this?

Thanks,
Dick

#### Papabravo

Joined Feb 24, 2006
13,956
Hi,

I am looking at a power diode that only shows the "instantaneous" forward voltage drop across the diode at a specified current. Most other diodes show the "forward drop" (not the "instantenous" forward drop) with current.

My question is: How do I determine the "constant" forward voltage drop? Is this the same as the instantaneous drop?

I am looking at the Vishay FP30AB dual-diode. They show an "instantaneous" drop of 0.95V drop at 15A per leg. My circuit [a DIY UPS for computer systems] will draw a steady-state current, not instantaneous. Anyone know about this?

Thanks,
Dick
The two terms don't really go together. The term constant forward voltage drop has no standard interpretation that I am aware of. The term instantaneous forward voltage drop is applied to a circuit where one or more signals has a time varying component. This would be true of a power supply or an inverter. If a pair of values is listed, eg 0.95V and 15A, then it is an average over many devices and represents a typical value. Minimum and maximum values are presumed to represent the three sigma points on a normal distribution. A graph showing the typical and extreme I-V characteristic may provide additional insight.

With that data point and some other parameters from the datasheet it should be possible to come up with an analytical expression for the current as a function of forward voltage drop.

#### n9352527

Joined Oct 14, 2005
1,198
Instantaneous in this case means that the measurement was performed using pulsed, instead of continuous excitation (current or voltage). This is not a new measurement technique and is regularly used in semiconductor measurements and also creeps in to the datasheets regularly.

The main difference between pulsed and continuous measurements is that in pulsed measurements the thermal effect of applying voltage or current (increase in junction temperature) to the device under test (DUT) is minimised. Increase in temperature would change the DUT characteristics and also destroy the DUT (if measuring sufficiently large voltage or current).

The relationship between instantaneous and continuous parameters is governed by the specific temperature coefficient of the parameter. For example, a diode has instantaneous Vf of 1V at If 10A at Ta of 25degC. Now, the continuous Vf at If 10A at Ta 25degC would not be 1V, because the junction temperature would be different due to P = Vf*If. This is similar to temperature derating, only now the temperature increase comes from Tj and not Ta. The calculations are pretty similar, except that power is given by Vf and If, and both are dependent on each other. The easiest approach is probably by using the Vf If graphs at different temperatures on datasheet.

Most pulsed measurements in semiconductor is performed with 300us pulse and less than 1% duty cycle, which are commonly quoted in datasheets. These are the values where the thermal contribution to the Tj can be safely ignored. Either that, or because all of us use Tektronix curve tracers which only support 300us and 80us pulses, and the 80us pulses on early models had rising time problem • Quang Thiện Bùi

#### dick_girard

Joined Jun 22, 2007
3
OK, I understand what you say.

I want to use a full-wave bridge rectifier. The output is smoothed with a large cap. So the output voltage is the peak-to-peak value of the RMS voltage into the bridge, minus the drop across two diodes at 16A. Since the output is smoothed, there are no pulses except when you first turn it on.

So I need to know what the constant [i.e. steady-state] voltage drop will be with a constant current. But the curves only show instantaneous values.

So, how do I find a diode that can handle a constant 16A with Vf = 0.95 if they only show the instantaneous drop? That's my problem. All the diodes I found that can handle that current only show instantaneous voltage.

Thanks,
Dick

Instantaneous in this case means that the measurement was performed using pulsed, instead of continuous excitation (current or voltage). This is not a new measurement technique and is regularly used in semiconductor measurements and also creeps in to the datasheets regularly.

The main difference between pulsed and continuous measurements is that in pulsed measurements the thermal effect of applying voltage or current (increase in junction temperature) to the device under test (DUT) is minimised. Increase in temperature would change the DUT characteristics and also destroy the DUT (if measuring sufficiently large voltage or current).

The relationship between instantaneous and continuous parameters is governed by the specific temperature coefficient of the parameter. For example, a diode has instantaneous Vf of 1V at If 10A at Ta of 25degC. Now, the continuous Vf at If 10A at Ta 25degC would not be 1V, because the junction temperature would be different due to P = Vf*If. This is similar to temperature derating, only now the temperature increase comes from Tj and not Ta. The calculations are pretty similar, except that power is given by Vf and If, and both are dependent on each other. The easiest approach is probably by using the Vf If graphs at different temperatures on datasheet.

Most pulsed measurements in semiconductor is performed with 300us pulse and less than 1% duty cycle, which are commonly quoted in datasheets. These are the values where the thermal contribution to the Tj can be safely ignored. Either that, or because all of us use Tektronix curve tracers which only support 300us and 80us pulses, and the 80us pulses on early models had rising time problem #### John Luciani

Joined Apr 3, 2007
477
As was mentioned the instantaneous measurements are used so that temperature
change is not a factor in the measurement.

For a diode the forward voltage changes by -2mV/DegC. As the temperature increases the power you are dissipating in the diode decreases. Your maximum power dissipation
will occur at your lowest operating temperature. There are probably specifications for instantaneous Vf at 25DegC and 125DegC. The 125DegC value should be apx 200mV lower.

Determine the maximum power dissipation and use the thermal resistance value
to calculate operating temperature.

(* jcl *)

#### Ron H

Joined Apr 14, 2005
7,014
OK, I understand what you say.

I want to use a full-wave bridge rectifier. The output is smoothed with a large cap. So the output voltage is the peak-to-peak value of the RMS voltage into the bridge, minus the drop across two diodes at 16A. Since the output is smoothed, there are no pulses except when you first turn it on.
Au contraire! In a rectifier with a smoothing capacitor, the current flows through the diodes in pulses. Better smoothing (bigger capacitor) makes the pulses shorter and higher in amplitude. The reason for this is that the diodes only conduct at the peaks of the sine wave. Current flows out of the cap at a relatively constant rate, but the charge that was lost between peaks all has to be replaced during the short conduction time. Look at the waveforms below.
So I need to know what the constant [i.e. steady-state] voltage drop will be with a constant current. But the curves only show instantaneous values.

So, how do I find a diode that can handle a constant 16A with Vf = 0.95 if they only show the instantaneous drop? That's my problem. All the diodes I found that can handle that current only show instantaneous voltage.

Thanks,
Dick
BTW, I couldn't find FP30AB (or FP30 anything) at Vishay.

#### n9352527

Joined Oct 14, 2005
1,198
The way I usually pick a diode is to choose an acceptable maximum junction temperature first. Calculate delta T with maximum air temperature, then calculate the maximum power using the thermal resistance. Then, pick a point on the Vf If graph that gives lower power dissipation than the calculated maximum power dissipation. If there is no point that satisfies the requirement, then get another beefier diode.

The above is for continuous current, for pulses, we have to look at the appropriate graph (usually thermal dissipation/resistance against pulses width and duty cycle. Or rough approximation with RMS power and appropriate safety margin, as cooling effect is not usually linear with pulse width or duty cycle.

Or just pick one that looks beefy enough if you're not cost sensitive and manufacturing millions of them 