# What does “dV/dt” mean for TRIACs?

This is probably obvious, but since I still don't have an engineering education, I ran into this problem:

What does dV/dt mean? What does it affect on a TRIAC?

• dv/dt means a voltage derivation over time - slope steepness. A high dv/dt can lead to self triggering. – Marko Buršič Mar 15 '17 at 10:58
• High school calculus - dy/dx. It's the slope in the graph – slebetman Mar 16 '17 at 3:26
• dV/dT is what I always think of, with horror, when I see ill-informed howtos for hobbyist fireworks or rocketry control systems that just put a bare thyristor/triac in series with a switch and battery :) – rackandboneman Mar 16 '17 at 8:50

When the current across the triac falls under $I_H$, which is the holding current, the triac stops conducting. With a pure resistive load this happens at the very end of the sine wave cycle, and voltage and current are in phase. When the load has an inductive component (e.g. a motor) then there is a lag between current and voltage. At the moment when the current drops below $I_H$, the voltage has already risen with the opposite polarity. Therefore when the triac turns off there is a big dV/dt on the triac - "voltage is cut off immediately". This situation can lead to self triggering the triac, and it starts to conduct uncontrolled. The remedy is to use a snubber circuit, i.e. an RC in parallel with the triac.

dV/dt is the derivative of the voltage with respect to time. In other words, it's the change in voltage (delta V, or ΔV) divided by the change in time (delta t, or Δt), or the rate at which the voltage changes over time.

If you had a curve of $y = x^2$ like below: -

The slope at when x = 3 (y = 9) can be estimated by calculating how much y changes divided by how much x changes. The change is called "delta" hence the slope is $\Delta y/\Delta x$.

Taken to infinitesimal changes, mathematically it gets "renamed" to dy/dx. It can even be algebraically proven by adding dy and dx to the original formula: -

$y + dy = (x + dx)^2$

$y + dy = x^2 +2xdx +dx^2$

Subtracting y ( = $x^2$) from both sides gives: -

$dy = 2xdx + dx^2$

Then noting that if $dx$ is very small then $dx^2$ can be ignored hence: -

$\dfrac{dy}{dx} = 2x$

In other words at any point on the curve $y = x^2$ the slope is 2x

In terms of voltage changing with time it is dv/dt. It has significance for triacs and mosfets and can cause such devices to trigger or partially activate if the rate of change of voltage with time is too great.

So far everyone has explained what $\frac{\Delta v}{\Delta t}$ means ( rate of change of voltage, its gradient, the 1st derivative of a voltage w.r.t. time )

But what does this have todo with TRIAC's? Triacs, like Thyristors/SCR's can be re-gated if there is high dv/dt across the device

http://class.ece.iastate.edu/ee330/miscHandouts/AN_GOLDEN_RULES.pdf

This is most likely to occur when driving a highly reactive load where there is substantial phase shift between the load voltage and current waveforms. When the triac commutates as the load current passes through zero, the voltage will not be zero because of the phase shift (see Fig. 6). The triac is then suddenly required to block this voltage. The resulting rate of change of commutating voltage can force the triac back into conduction if it exceeds the permitted dVCOM/dt. This is because the mobile charge carriers have not been given time to clear the junction.

• Marko explained what it has to do with TRIACs – DerStrom8 Mar 15 '17 at 11:51
• yup I spotted someone did state that after I hit submit – JonRB Mar 15 '17 at 12:29
• Been there, done that! =P – DerStrom8 Mar 15 '17 at 13:30

Dv/dt is the expression for charge injected into the Triac's internals (the silicon); the energy mechanism Q = C*V, when we make incremental changes and see what happens, becomes dQ/dT= C * dV/dT + V * dC/dT. After chosing to ignore the 2nd part, and recognizing current = dQ/dT, we are left with

$$I = C * dV/dT$$

whereupon we discover high rates of change of voltage will trigger the Triac.

The charge injection of dV/dT also puts FETs at risk. Unless sufficient Source contacts and Well contacts are in place, the charges will pursue ALL POSSIBLE PATHS; current crowding into contacts may cause I*R drops large enough to turn on emitter-base junctions of parasitic bipolars, in which case the bipolar adds to the current flows. In many cases, that brings about Gain > 1 positive feedback, and the FET/bipolar try to discharge the entire VDD charge storage network down to ZERO VOLTS. With that mere attempt, silicon and aluminum melt.

How to avoid? Design the Source and Well contacts for transient charge tasks, not just for DC leakage control.

Here is microphotograph of high voltage under transient conditions (1volt per nanosecond) injecting charge, with that charge then crowding around a Well Contact.