What's the optimal way to drive a 2DI50D-100 power module?

Question

I've been messing around for a couple days with this 2DI50D-100 power module that I got out of an old Italian VFD that I got dismissed at work (I'm an electrician) because it stopped working.

This is the schematic that's attached to the device itself:

Basically, I'm trying to couple two of them into an H-bridge to drive a 180 V rated DC motor.

First of all, I tried to make the high side work, with just 24 V, just to do some tests. So I connected the power supply between C1 and E2 and the motor between C2E1 and V-.

To turn it on I just connected C1 to B1 through an ampmeter. The motor ran just fine and the multimeter showed 4 mA so we're all good (are we?).

Then I wanted to do the same test with the lower side, so I powered the module between C1 and E2 like before, latched the motor between V+ and C2E1, and to turn the module on I connected the ampmeter between C1 and B2. I expected it to work exactly like the first test.

Instead, I got some sparks and obviously the current limiter kicked in. I thought the lower circuit was gone, so I did the same test on all three modules; same result.

So, why does this happen? Isn't the lower circuit an exact copy of the upper one? Shouldn't it work the same? Am I missing something or are the lower sides of the modules broken?

Oh, by the way, I fried one of them.

Answer 1

• Make C1 lug +Volts and E2 lug 0V. Wise idea trying with a lower voltage first such as 24V.
• Wire (24V) motor between C2E1 lug and E2 lug.
• To "on" the motor, supply less than 5 Volts, at no more than 10mA, from the B1 post to the E1 post.* Call this the "base current." This "turns on" the top set of transistors, forcing C2E1 to be connected to C1 (+Volts.)
• To "off" the motor, set B1-E1 voltage to zero or slightly negative. This disconnects C1 from C2E1.
• To "actively stop" the motor, supply base current to B2-E2 posts. This connects C2E1 to E2, acting as a "brake" shorted across the motor.
• Never, ever energize both bases at the same time. Doing so connects C1 to E2, or in other words, shorts out the power supply.

*Notice, that as you apply this base current to the upper transistor stage, that both of these base terminals B1 E1 are near +Volts. Therefore this must be isolated from ground and/or E2.

This is essentially using the device like a lightswitch, which can work, but is equivalent to using a chain saw to cut a sandwich.

The module is capable of switching thousands of times per second. In the case of the VFD this was originally in, the controller inside that pulsed these bases in a very precise and fast way to form an average sinusoidal output current to the (AC) motor. The pulse frequency was likely 1, 2, 4, or 8kHz. An approximation to this could originate from an Arduino or similar microcontroller. Microcontrollers are typically used with these because two things are critical to prevent damage:

1. Have a certain amount of "dead-time" between uppper-lower transitions. This is time in which both bases are off, to allow the device to actually stop what it is doing completely.
2. Never, ever allow both bases to be on at the same time, as explained earlier.

More terms to research before applying more than 24 VDC: PWM, pulse width modulation, shoot-through, high-side, isolated high-side-driver.

Answer 2

So, why does this happen? Isn't the lower circuit an exact copy of the upper one? Shouldn't it work the same?

Large power transistors typically have small current gain, so the reason they use 3 transistors, configured like that, for both the high and low side, is to have more gain. To simplify matters, we can replace each set of 3 transistors with a single transistor. So your test setup looks like this (additional diodes and resistors removed for brevity):

^{simulate this circuit – Schematic created using CircuitLab}

A transistor's base-emitter junction acts much like a diode. Current can flow from base to emitter, and that current flow will allow a larger current to flow from collector to emitter, but the amount of B-E current that can flow is unlimited.

So, when you connect +24V to B2 (trough the A-meter) while E2 is connected to V-, you've created a short circuit. You need to limit current externally. For example, by placing a resistor in series with B2.

The reason it did work with the high side is that it's emitter is not connected to V-, but to the load (motor). If you apply +24V to B1 current will flow from base to emitter but then trough the load. The load is acting as a current limit in this case. (And the larger C-E current trough the load will lift the emitter voltage thereby pinching off B-E current even further, but let's keep it simple).

Additionally, when testing either the high or low side, you should not leave the other side's base floating (unconnected). A floating base on a large gain transistor like that will cause it to turn on and off randomly. This can cause a short circuit ("shoot trough") if it causes both transistors to be on at the same time.

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

^{simulate this circuit}

The circuit in Figure A works because the motor is in series with The path from base to emitter of the upper set of transistors.

The circuit in Figue B will not work because the maximum voltage from base to emitter is 3.5V (from the datasheet). By connectinh the base to 24 volts you will damage the lower set of transistors. If they still function then you are fortunate, but there lifetime or performance may decrease.

To test you should have put a resistor (say 4.7k) between the base and 24V to limit the current to about 4mA as shown in Figure C.