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I have a load powered by a 12V source. The load can draw up to ~2A:

12V Load

The load has "on and off" modes, but it still draws ~40mA when in "off" mode. I want to switch the load from an MCU to save power when the whole system is off, so I'm looking at adding a transistor:

12V Load with transistor

I'm researching what transistor would be suitable. The TIP31 looks like a suitable candidate, but I believe I need to calculate the thermal requirements. I think I can probably follow this guide for the calculations, but I need to know what the voltage across the transistor (Vce) will be, and that's where I'm stumped.

How can I determine what the voltage will be at the transistor's collector? It seems so simple but my electronics knowledge is fairly basic. How can I calculate or measure this (before buying the transistors)?

Apologies if I've left off any necessary details. Just comment and I'll add them if I have.

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  • \$\begingroup\$ Why haven't you connected a resistor to base(B)? This will set the current I_B which, with using beta(DC Current gain), will give you the current I_C. This way you can easily find V_C. \$\endgroup\$ – Dor Sep 29 '15 at 22:34
  • \$\begingroup\$ Thanks Dor, but Im afraid I didn't quite follow that. Would the collector current not just be the same as the current drawn by the load? And when I do know I_C, how would I find V_C? Thanks \$\endgroup\$ – parrowdice Sep 29 '15 at 22:50
  • \$\begingroup\$ I'm not sure what you mean by "but it still draws ~40mA when in "off" mode". Is this an operational requirement, i.e. you want this to happen, or is just something you've observed? \$\endgroup\$ – Fizz Sep 30 '15 at 0:38
  • \$\begingroup\$ @RespawnedFluff: That's correct, it's something I've observed. The load/device can be driven to what might be described as an "off state", but it still draws ~40ma in that state. I want to be able to put the MCU into sleep mode and cut the power to external devices to save power when not in use, but still be capable of restoring power from an interrupt. \$\endgroup\$ – parrowdice Sep 30 '15 at 7:27
  • \$\begingroup\$ I still don't follow. Do you need to provide those 40mA to the load in your application? Or are you fine with cutting the current to it below that level? The solution you have (in mind) will cut the current well below that when transistor is "off" (in cutoff to be more precise). \$\endgroup\$ – Fizz Sep 30 '15 at 7:39
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Use the following graph in the TIP31 datasheet to determine the voltages you need across your transistor to saturate at your load's max current.

enter image description here

For 2A = 2000mA load (collector current), we have a Vce(sat) of 300mV = 0.3V. Multiply 0.3V by 2A to get a 0.6W dissipation on the transistor when it's "on". That will work ok without a heatsink for a TO-220 package, although a small heatsink wouldn't hurt. Note that if your load draws less than it's max stated 2A, then both the collector current and Vce(sat) are less than what I've calculated above, so the transistor will dissipate less power.

As for the transistor's base, you need to raise its potential to approximately 1.0V (at 2A load), not 0.8 as you have your schematic. (A bit more doesn't hurt.) Otherwise the transistor will not be fully saturated and will use more power as well as exhibit higher Vce than calculated above.

Also, I'm assuming your load still operates properly at just 11.7V, otherwise you need to increase the supply rail slightly. Note that I've ignored the "but it still draws ~40mA when in "off" mode", because I don't really understand what you mean/want by that.

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  • \$\begingroup\$ After doing a lot of extra reading to fill in some areas of knowledge in which I was lacking, your answer makes complete sense. Thank you for taking the time to explain it. Appreciated. \$\endgroup\$ – parrowdice Oct 1 '15 at 6:50
  • \$\begingroup\$ @parrowdice: Well, yes, I've left out how to make those voltages happen in actual circuit (but you didn't ask that). What you have here is called a low-side switch, by the way--useful keyword for finding out more. \$\endgroup\$ – Fizz Oct 1 '15 at 13:13
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If you saturate that transistor, then you will pull its collector low. It will be nearly ground potential. If the base current times the transistor gain is greater than the current that gets through that resistor, then its collector should be "pulled" within millivolts of ground.

An easy way to understand it, is to understand voltage dividers. Voltage dividers work on the principle that the sum of the voltages is zero. Another way to say it is, all of the voltage "drops" across the loads if you follow any given path from positive to negative supply. The amount that drops across a given load is determined by what proportion of the total resistance that part is. If half the resistance is in one component then half of the voltage will drop across that component.

In your circuit, some of the voltage will drop across that resistor, and the rest drops across the transistor. When the transistor is on, it effectively becomes lower resistance, because more current flows through it, so it becomes a smaller proportion of the total resistance, so less voltage difference will be present across the collector-emitter junction. This "pulls" the collector to be closer to the emitter potential, which is ground.

When the transistor is off, very little current leaks through the transistor, so it represents a huge resistance. The transistor becomes the majority of the resistance, so, most of the voltage difference will drop across it. This causes the collector to go high, because little of the drop will occur across the resistor, and a lot of drop occurs across the transistor.

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  • \$\begingroup\$ +1 You really helped me to understand some pieces of information I was missing, thank you very much. \$\endgroup\$ – parrowdice Oct 1 '15 at 6:44

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