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For my internship, I am tasked to design a power supply whose main purpose will be to power a 90 Vdc motor. The motor is used for a short amount of time (roughly 30 seconds), pulling about 1 A during most of this time, but towards the end of its use as torque increases it can pull up to about 3.1 A for a few seconds. Since the operation time is very short and the motor will not be used very frequently, efficiency is not a key issue. Additionally, since the load is a motor not much regulation is required and quite a bit of ripple is tolerated.

To accomplish this, I have decided to use a simple series pass transistor with a 91 V zener diode as a reference. Input will come from mains 120 Vac, 60 Hz, with a rated output of 90 Vdc, 4 A. The design will also be transformerless in order to reduce cost and size. My design right now uses a full bridge rectifier to convert 120 Vac to 170 Vdc, with a capacitor to reduce the amount of ripple. I am wanting to use a npn Darlington configuration for the series pass transistor but this is where I have questions.

When looking at transistor power ratings, is this average power or instantaneous power? Going from 170 V to 90 V requires a Vce of 80 V, so at rated current instantaneous power will be over 300 W. Due to the amount of ripple, average power will be much less than this. Either way, the Darlington will have to have a very high power rating and a large heat sink. The highest power rating Darlington that I have been able to find is the MJ11032 (300W) but I do not know if this is suitable for this application. Are there any other transistors out there that are better suited for this or do you have any suggestions on other ways to go about this?

Lastly, I know there will be concerns about safety, but all of my designing right now is on the computer and when I get to the prototyping stage this will be done with the help of an experienced engineer. Also the power supply will be completely enclosed in the final product, and all interface will be done via a separate 24V side that has already been designed.

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  • \$\begingroup\$ 300 W for 30 seconds will require fan cooling or a massive heat sink. A simple buck converter comes to mind. \$\endgroup\$ – winny Jun 21 '17 at 15:36
  • \$\begingroup\$ Related question: Maximum Transistor current and voltage \$\endgroup\$ – The Photon Jun 21 '17 at 16:08
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Your concerns about power are well-founded, but there are other issues you need to address first.

The most obvious is the question of capacitance, but it's trickier than you might think (maybe not, since you haven't shown your proposed design). Olin Lathrop has argued that you'll minimize power by putting your major capacitance on the output rather than the input. This is true, but there's a catch - there almost always is. Let's assume you do something like

schematic

simulate this circuit – Schematic created using CircuitLab

The problem occurs when the input voltage goes to zero. Then the zener voltage will be pulled to zero (more or less) by the base-collector junction, and this will place about 90 volts across the emitter-base junction, destroying it. While it's true that the bridge will resist major current flow back into the bridge, the issue of leakage currents must be addressed.

Putting the capacitor across the input will avoid this, as well as greatly reducing output ripple. It will, as has been noted, give you power dissipation heartburn.

As to your specific questions, transistor power is generally given as average rather than instantaneous, although a time scale of several seconds may well count as closer to average. You should also look closely at the dissipation rating and notice that it is specified for a case (rather than junction) temperature of 25 C, and this is not likely realistic in your instance.

I personally would not want to depend on a single transistor in this application, but rather two or more in parallel. It complicates the design, but it greatly reduces the pain of thermal management and avoiding hot spots.

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  • \$\begingroup\$ You can't ground the negative output of the full wave bridge. You have to assume one of the AC inputs is already grounded. With this kind of scheme, both outputs of the full wave bridge are "hot", and with significant common mode voltage relative to ground. \$\endgroup\$ – Olin Lathrop Jun 21 '17 at 17:11
  • \$\begingroup\$ Good point about the reverse bias across the B-E junction. However, that can be easily solved with a diode in series with the emitter and a large-valued bleeder resistor to drain off reverse leakage. \$\endgroup\$ – Olin Lathrop Jun 21 '17 at 17:13
  • \$\begingroup\$ @OlinLathrop - Yeah, but in this case the ground symbol is just the circuit return. As we've said so often in other questions, "ground" is just a convenient name for a reference voltage. I assumed the "experienced engineer" would understand. \$\endgroup\$ – WhatRoughBeast Jun 21 '17 at 17:13
  • \$\begingroup\$ In this case where confusion with the line power ground is a very real possibility, you need to go out of your way to make it clear that your "ground" is just the floating negative output of the full wave bridge. I think calling it "ground" at all in this case is asking for trouble. If someone were to connect your ground to the AC line ground, at least one of the diodes would blow up. \$\endgroup\$ – Olin Lathrop Jun 21 '17 at 17:20
  • \$\begingroup\$ @OlinLathrop - In deference to your concern, I've edited. \$\endgroup\$ – WhatRoughBeast Jun 21 '17 at 17:28
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Before you go any further, calculate how much power the transistor needs to dissipate. The answer will probably tell you to find a better design. In this case, efficiency isn't so much about wasting power, but having to deal with the heat from the wasted power. You can buy a lot of electronic parts for the cost of having to remove 100 W of heat from a transistor.

Even with your existing proposal, put the cap after the pass element, not before it. That way you aren't always starting with the absolute worst input voltage. The average will be lower. With the cap immediately after the rectifier, you're always dropping 170 V to 90 V with a linear regulator. Even at the steady current of 1 A that 80 W, and 250 W for a few seconds at the end!

If you want something simple, find a transformer that drops the incoming AC voltage so that you get the desired voltage right out of the full wave bridge followed by a large cap. Experiment with a variac to see what you need. Even just buying a variac and dedicating it to this purpose will be cheaper than keeping a transistor that dissipates over 100 W from blowing itself up. You might be able to find some old surplus auto-transformer with enough fixed taps that one of them gets you what you want.

I just checked, and you can get a brand new 320 VA variac for $65. That's probably cheaper and certainly easier than a massive heat sinking and cooling project.

Again, you need a better design, not some magic answer how to make your current scheme work.

Added

You now say you have a limited volume of 10" x 5" x 3". In that case, you can't use your original scheme. That's not enough space to get rid of the large heat it would produce while still keeping the components cool enough to not fry themselves.

A transformer capable of stepping down the voltage so you can rectify it and use it directly wouldn't take more volume than you have. However, finding one off the shelf that fits your geometry is unlikely.

What you need is a proper switching power supply. If power factor isn't a concern, then you just full-wave rectify the AC into a cap as you intended before. However, now you PWM that to drive the motor. Doing this at 170 V and 300 W is not a beginner project, but it certainly can be done. It would fit into your space with room to spare.

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  • \$\begingroup\$ Forgot to mention in my initial post, I have limited design space (10" x 5" x 3") which is part of the reason I am trying do without a transformer. \$\endgroup\$ – wolfpack1021 Jun 21 '17 at 17:48
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since the load is a motor not much regulation is required and quite a bit of ripple is tolerated.

Since you don't specify how much ripple is "quite a bit of ripple" then I will assume that 100% ripple is within your specifications. Also, it's gonna be cheap. This solution will cost $2.

  • 110VAC has peaks at 155V.
  • Rectify 110VAC with a diode bridge. You get a rectified sine wave with 155V peaks (neglecting diode drops).
  • This has an average DC value of 99 Volts.
  • Close enough. Connect DC motor at output of rectifier. Done.

FYI I have a Skil Energy power screwdriver which works exactly like this. Raw AC, Rectifier, PWM, DC motor. This is excellent, because PWM on a DC motor gives immensely better torque control (via the trigger) than a dumb triac dimmer does. The result is a tool that is light, doesn't run out of batteries, works very well, has massive torque, yet is very easy to control.

Screws don't care about motor torque being chopped at 100Hz. Maybe your load does, but... you didn't specify it ;)

Now, if you want something smarter, here we go:

  • Rectifier
  • Caps
  • PWM
  • Motor

Caps should be large enough so that the voltage doesn't sag under 90V when load is maximum. PWM can be implemented using a specialized controller, or a comparator in self-oscillating hysteretic mode. The idea is to create an average voltage of 90V on the motor by chopping the rectified and smoothed AC. This will be small, light, cheap, and won't dissipate a lot.

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  • \$\begingroup\$ The OP says the input will be 120 VAC, which has peaks of 170 V. That puts the rectified DC average at 108 V, which is 1.2x what the motor wants. That might be OK for 30 seconds at a time, with time to cool down in between, but you'd have to look at that carefully. You also have to make sure the additional torque and no-load speed are acceptable. \$\endgroup\$ – Olin Lathrop Jun 21 '17 at 17:27
  • \$\begingroup\$ PWM is the way to go then. \$\endgroup\$ – peufeu Jun 21 '17 at 20:06

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