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I have a circuit using a NPN transistor with a motor before it and a pwm into the base.

The motors draws a current of 600mA if I plug it directly into the power source, just under the maximum allowed by the transistor at 650mA.

The minimum gain of the transistor is listed at 100, with my multimeter reading about 200. The voltage is at 3.7V.

My question is, if I give a base current of lets say 20mA, even with the minimum gain, it is 2A, but will this current be limited by that the motor draws (600mA), or will it push 2A onto the motor?

I have blown 2 of these transistors already, but connecting the base directly to 3.7V. The transistor in question is 2N2222A in the metal case.

Another question is for a simple LED circuit switched by a transistor, is it Ok to depend on the hfe alone? For example, I only have a resistor connected to the base with the collector to HIGH and emitter to the LED?

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  • \$\begingroup\$ You cannot "push" current into something; the current is the function of voltage and resistance, and motor has certain R and will never draw more current than V/R. \$\endgroup\$
    – ilkhd
    May 4 '15 at 22:05
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There are two possible causes of the blown transistors. First, connecting the base directly to 3.7V will cause excessive base current : it makes sense to limit that with a resistor. 20mA is cautious, 100 ohms would allow 30ma which should be safe.

The second in that - while you cannot "push" excessive current like 2A to the motor - if it draws 600mA when running normally, it may draw much higher currents when starting, or when stalled.

Connect the motor to the PSU again, and stall it with your fingers (don't do this with a high power motor! Use some other means of braking it!) Observe the current it draws when stalled ... I'm guessing somewhere over 3 Amps (you may need to fiddle with the current limit to observe this).

It will momentarily draw that current when starting, and that could be fatal to a 650mA rated transistor...

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Since nobody else mentioned it yet, I will mention that there is a concept called forced beta which applies when you use a BJT as a switch (in other words when you want the transistor to be turned on to the max). I think the easiest thing is to just go through an example. If the circuit uses 600mA without the transistor, then you could use that as your maximum collector current.

If beta min is 100, then use a beta of 20 to calculate the base current. In this case, the "forced beta" of the transistor is 20. Now, select your base resistor so that the base current will be Ic/beta(forced). In this case, it would be 600/20 = 30 mA. In order to calculate the base current resistor, we would need to know the driving voltage, and Vbe. With this much current, Vbe will be high. Maybe 0.8V or more. There should be a chart, or you can just measure it. It will decrease as the transistor gets hot.

I am not guaranteeing that this will solve your problem. It sounds like the transistor you chose may not be able to handle this much current. But this basic idea may help you in the future. Note that this only applies to BJT's such as the transistor you are using. FET's behave differently.

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The transistor's Hfe isn't limited by other components, but it represents a maximum value: if it can't conduct enough current from collector to emitter because of other resistances in the circuit, it will simply be 'full on' and conduct as much as possible.

If this weren't the case, you could build a lightning generator with nothing other than a transistor and a 1.5 volt battery. ;)

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    \$\begingroup\$ So lets say I have a 10V supply, a 10 ohm resistor at the emitter, and a 10ohm resistor at the base. This will mean a 1A current into the base, and a 1A current through the rest of the circuit, no matter what the hfe is? \$\endgroup\$
    – Solutions
    May 4 '15 at 22:03
  • \$\begingroup\$ The maximum current the transistor can conduct from collector to emitter is (10 volts - Vce) / 10 ohms, which is about 0.98 amps. The 10 ohm base resistor (from 10V, presumably) will also conduct 0.98 amps from base to emitter, and as long as the Hfe is at least 1 (a safe bet!) then yes, the transistor will conduct the maximum. \$\endgroup\$ May 4 '15 at 22:15
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  1. Never use components close to the maximum parameters; it is not ok to draw 600 ma from a transistor with Imax 650 ma. Motors,for example, draw significant currents when they start.

  2. Why would you connect the base to the power source without a resistor?

  3. No it is neither ok to depend on hFE, because it changes with the temperature and the age of he transistor, nor to connect an LED the way you described; LED should be driven (unless you exactly know why you should do otherwise) by the collector of the transistor, through a resistor.So you need 2 resistors - one in the base of the transistor and one between the collector and LED.

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  • \$\begingroup\$ I didn't account for the excess current at the start. It'd probably be a good idea to use a motor-specific H-Bridge rather than large transistors. They did get really hot after about 30s. My understanding was that the hfe is limited by the rest of the circuit. In which case if the current is limited using a resistor at the collector or emitter, it shouldn't matter the base current as long as hfe*base is over the current of the rest of the circuit. This is appranently flawed. So my current understanding is to limit the base and limit the collector so that their division is less than the hfe? \$\endgroup\$
    – Solutions
    May 5 '15 at 3:40
  • \$\begingroup\$ Yes, this is right. Keep in mind, not only maximum current but also maximum power dissipation is also important. \$\endgroup\$
    – ilkhd
    May 5 '15 at 4:27

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