Is it possible to connect diodes in series to increase reverse voltage capability? Is it possible to connect them in parallel to increase the forward current?

Yes. Diodes can be connected in series and/or parallel in order to increase the reverse voltage and/or forward current capacities.

There are practical considerations to keep in mind such as thermal management, voltage isolation, and component count, to name a few.

When connecting diodes in parallel it is usually a good idea to match Vf. Differences in Vf between diodes will get worse as they heat up. Diodes with lower Vf will initially conduct more current, increasing their power dissipation and further lowering their Vf. This in turn increases their current share, continuing to heat up the junction, and so on until a catastrophic failure occurs. 

What is the (I^2)T rating and how is it used?

The I2T (Amps2*Seconds) rating is defined as the single cycle surge current, Ifsm, multiplied by a standard pulse width of 8.3mS (sine wave is assumed).


It is used as a rule-of-thumb to gauge surge capability at different pulse widths. It works because at higher currents, V(forward voltage drop), is dependent on the resistive element of the diode. Vf becomes proportional to current in the diode expressed by Vf = Rdiode * If. The I2T calculation yields energy dissipated in the diode during the pulse duration. Power multiplied by pulse time gives the energy pulse. Energy dissipated during a surge current pulse is proportional to I2T and is usually the driving force behind a failure. The energy pulse causes localized heating which induces mechanical fractures or disruption of the silicon crystal structure. Calculating the maximum I2T can help determine if a diode will survive a current surge.

Example: A diode has an Ifsm rating 100A. Will it work at 150A surge for 1uS?



1. Calculate I2T


I2T = (100A)2 * 8.3mS = 83A2S


2. Determine if I2T under the new conditions is much less than the original calculation. Is the I2t calculation at 150A, 1uS much less than 83A2S?

Is 83A2S >> (150A)2*1uS?

= .023 A2S

Yes, .023A2S is << 83A2S


The diode should be able to handle a surge of 150A for 1uS.

Comment: When approaching fast pulse times (i.e. ns range), I2T using Ifsm is used as an upper limit. Operating I2t should not exceed half of the upper limit.

I am trying to determine the polarity of a high voltage diode. My Fluke meter shows an open circuit in both directions for the 10KV Z100UFG diode. Is that normal?

Yes, and no. When trying to identify the cathode and anode end of a high voltage diode, the meter has to have enough voltage to overcome the forward voltage drop, Vf, of the device under test. We use a test setup that includes a 30V current limited power supply. When testing for polarity, the current should be limited to mA going through the diode when it is connected in the forward direction.

What is the most effective way to mount the K25UF epoxy diode to a printed circuit board?

The most effective method of mounting the K25UF diode (and most all axial-leaded epoxy diodes) is to make a cut-out in the board just a tad larger than the dimensions of the diode body and drop the diode in so the leads lay flat across the solder pads. If possible, avoid lead-forming. If space does not permit it, is important to support the diode body while forming the leads. The leads are relatively short and thick, which enhances the current carrying capacity of the device. The epoxy body is easily damaged if the leads are not supported near the body during the lead forming process.