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I am designing a circuit where I want to protect Q1 from current overload and popping in the event that Rl is a short. The length of wire in a short circuit situation is 0.7 ohms. Usually Rl is 10 ohms (when it is not shorted) or so and thus 1.25A flows through it. RL has LED's in it so I need to keep 12VDC. I have looked at an LM513 as a current source but it looks like the voltage drop from Vin to Vout is too great.

A microcontroller is driving the base of Q1

I have designed the circuit below using 3A transistors, but I keep popping Q1. I have verified that the circuit works by changing out the Rs resistor value and verifying that it regulates the current correctly.

I chose the Rs to 0.5 so that I could run a max of 1.5A. With the transistors rated at 3A I thought this should be fine.

My thought is that Q1 is popping too quickly before it even has a chance of turning off with the instantaneous current of approximately 10A flowing through it.

Should I replace Q1 with a mosfet transistor?

Schematic is below.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ What is the transistor you're actually using for Q1? \$\endgroup\$ – Spehro Pefhany Aug 28 '15 at 23:08
  • \$\begingroup\$ Your circuit and your circuit description seem to be at odds with each other, so I'm confused. What is it you're really trying to do? \$\endgroup\$ – EM Fields Aug 28 '15 at 23:54
  • \$\begingroup\$ It is not a 3904, but it is a bjt npn rated at 3A. \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:23
  • \$\begingroup\$ @Spehro Pefhany An FJC1963 \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:32
  • \$\begingroup\$ I have calculated RS (sense resistor) power value to be 0.5ohms P=I2R = 1.5*1.5*.5 = 1.125W = 1.5W \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:47
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Please go back and take a look at your data sheet. A 2N3904 is only rated for a continuous current of 100 mA (normal) with an absolute maximum of 200 mA. You are trying to draw way more current than the poor little thing can handle.

Furthermore, you will only be providing ~10 mA of base current, and assuming (by dint of wishful thinking) that the transistor can handle the current, when you use a BJT as a switch, which you are doing here, you should assume a gain of 10 to 20, with an assumption of 10. So you are not providing adequate base drive even if your transistor could handle it.

Just as bad, consider what would happen if your limiter worked as intended. You would be current limited to 1.5 amps. You'd get about .75 amps across your sense resistor, and another 1 volt or so across your short. Then the transistor would see about 10 volts, and at a current of 1.5 amps would dissipate about 15 watts. Now go back to the data sheet, and see just how much power a 2N3904 really can dissipate.

Using a FET in place of Q1 seems reasonable, but keep in mind a few things -

1) If the circuit works, Q1 will dissipate 15 watts, and this will require a much larger heatsink than you are accustomed to using.

2) Without looking closely at the problem, I will note that the FET may not take kindly to your gate-voltage limiting approach, and it's possible that the limiting will take the form of very large-swing current oscillation.

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  • \$\begingroup\$ As noted above it is not a 3904, but a bjt npn rated at 3A \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:24
  • \$\begingroup\$ It is an FJC1963 \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:34
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I have done exactly this, but with ~2.3A instead of 1.5A. I had 12V LED modules, pulsed by PWM, with a hardware circuit current limiter as you show, but with an N channel MOSFET as suggested by WhatRoughBeast.

enter image description here

I had 48 of these running in parallel, to give 100A+ during a pulse for a strobe light on a mobile robot.

They worked fantastic, I recommend them! Allowed me to safely use any supply 15V or above. Note my Rload (actually a current sense resistor) was 5W rated, and my MOSFET was an SOIC-8 package FET with 8A current rating. My application meant I never went above 10% duty cycle though, so if you expect your total pulse ON time to be significant, get a 30A FET. This means their Rds_on resistance is low, and they dissipate less heat during operation.

Be aware though that despite what the "on" resistance is, the FET acts more like a linear voltage controlled resistor during that over-current limiting condition. So -any- extended duration like this will blow it up, especially if the difference in voltage between the supply rail and what the LEDs are using is large. Check power using P = IV, where V is the remaining voltage after the current sense resistor formula, and the drop over the LED load.

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  • \$\begingroup\$ So could I just swap out a mosfet transistor in place of Q1? Would I need to recalculate R1 and R2? Like I said before, the circuit works, it just can not handle the short circuit RL like I am intending it to. \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:39
  • \$\begingroup\$ the circuit would normally be 1.25A when not short circuited. The short circuit is a fault situation. I just want to protect in the case of short circuit. Also in the circuit is a polyfuse, but it is too slow to protect Q1. \$\endgroup\$ – tooclosetosee Aug 29 '15 at 15:42
  • \$\begingroup\$ I suggest that yes, you change Q1 from an NPN transistor to a N-channel MOSFET. The resistors R1 and R2 can stay the same, R2 is far less important when you use a MOSFET rather than a BJT. set the "sense" resistor so that the current limit is reasonable, and above the expected operating current. How do your LEDs normally current limit? \$\endgroup\$ – KyranF Aug 30 '15 at 1:25
  • \$\begingroup\$ The RL is multiple LED's that use resistors to set the current. The normal load of RL is 1.25A @ 12V. My intention is for this circuit to allow a maximum of 1.5A, so it will not current limit RL. It is just to keep the installers from shorting RL and thus popping Q1. The circuit as it currently stands is the same as I have drawn above except RS and Q2 are removed and Q1 is an FJ1963. \$\endgroup\$ – tooclosetosee Aug 30 '15 at 14:20
  • \$\begingroup\$ Yeah, so with the Q2 + RS feedback, it does indeed make a good current limit feature. If the current limit is only there in case the people installing the device may accidentally short circuit the two contacts going into this device to connect the load, then you may want to find an alternative, such as a comparator across the terminals and a simple high side P Fet switch to disconnect V+ if it looks like a short \$\endgroup\$ – KyranF Aug 30 '15 at 16:59
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  1. If the 3.3 volt source is your microcontroller I/O, then with the 10 ohm R1 you describe in your text (which is way different from the 10k you show on your schematic) there'd be 330 milliamperes through it with V1 high, not 1.25 amperes, and that's if the MCU could drive a 330 milliampere load, which it can't.

  2. If the 3.3 volt source is a microcontroller I/O, then every time it goes low it'll appear as if R1 had been shorted to ground. That'll turn OFF Q1, stopping the flow of charge through RL, so even if R1 shorts out full time, it's highly unlikely that Q1 could pop from overcurrent with V1 low.

  3. It appears from what you've said that what you're trying to do is build a gated, current-limiting LED driver. If that's the case, then you need to get more specific about identifying that as a goal.

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That part can only dissipate 0.5W at 25 degrees C (less when warmer). When your circuit is limiting the transistor is dissipating more like 10-15W, so the transistor will expire in very short order.

If you want it to be able to withstand a continuous short circuit you need a bigger package and a large heat sink.

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