# Transistors in parallel

I want to use several transistors in parallel to control current through a load. This is to distribute the current through the load across the transistors so that individual transistors with a rated collector current less than that going through the load can be combined to control the load.

Two questions:

1. Would an arrangement such as in the schematic below work well? (Resistor values only very roughly approximated).

2. How should the resistor values be calculated? I was thinking of using the range of hfe values for the transistor as follows: calculate two collector currents: for the minimum value of VR, the minimum and maximum collector current for the minimum and maximum hfe values.

Thanks

simulate this circuit – Schematic created using CircuitLab

Edit: Actually I would remove R-limit, and have VR stretch across the rails, with the wiper connected to R1-R3

• An added bonus to constructing the circuit like this is added redundancy. If you physically construct the circuit so that the parallel resistor/transistors are each part of a removable cartridge (like a vacuum tube/socket), you would be able to pull one out and replace it with an identical one without needing to shut it down (safety would have to be considered, of course, depending on what sort of power supply and load you are driving). – AJMansfield Jul 26 '13 at 16:26

This is actually a very common technique to do, both with BJTs (traditional transistors like drawn above) and MOSFETs. With BJTs, you don't need to bother with separate trimmed base resistors, all you need to do is add current sharing resistors or sometimes called ballast resistors. Look at this page for instance, the first one I found with google that explained this design:

If you use MOSFETs, you don't need the current sharing resistors at all, they can just be parallelled 'out of the box'. MOSFETs have negative feedback 'built in': if one MOSFETs gets a larger share of the current, it gets hotter which in turn increases its resistance and reduces the amount of current going through it. This is why MOSFETs are usually preferred for applications where multiple transistors in parallel are required. However, BJTs are easier to built into current sources as they have fairly constant current gain.

• Fantastic, thank you. How would I calculate the minimum ballast resistor value? (In the datasheets I've found, the only temperature graphs I've found are power derating vs case temperature). Is there a formula that would work across all NPN models? – CL22 Jul 26 '13 at 9:25
• There is no good or bad answer here, it depends on other design choices, generally. The resistor is usually chosen so that the voltage drop across the resistor is approximately one order of magnitude less than the voltage drop over the BJT. However, in some designs this may still yield 10W+ resistors which are unacceptably large, so you might go for even smaller values. – user36129 Jul 26 '13 at 15:20
• Unlike the positive temp coef of Rds that balances current between switched FETs, the negative temp coef of Vth will cause paralleled linear FETs to not share. – gsills Jul 27 '13 at 5:10
• -1 for misinformation current balance in FETs operated in linear mode. – gsills Sep 24 '13 at 20:53
• Well, that depends on what you call misinformation. Yes, trench FETs at high temperatures will have unequal current sharing. But it is good practice to parallel linear-mode FETs. Hot spotting and unequal current sharing is not an issue for most applications, especially if you stay well within the SOA and make sure you derate current at higher temperature you will be perfectly fine. Just don't try to operate the gate with a potentiometer and keep 'em cool. This is employed in many if not all low burden voltage power sinks. – user36129 Sep 25 '13 at 6:26

For an application where you need to parallel transistors and control current in a linear fashion (not switching the transistors fully on and off), BJTs are your best bet. As Olin Lathrop says, the circuit will need to have resistors in series with the BJT emitters to help balance current.

Here is a starting example circuit to show emitter resistor placement.

Re1 and Re2 will help balance current between BJTs. The problem is that Vbe has a temperature coefficient ($\gamma$) of about -1.6mV/C. As the parts heat up, Vbe will reduce allowing more base drive to the transistor from the fixed value of Vc. With a first order model of Vbe's change with temperature, a simple equation for current in Re1 is:

IRe1 = $\frac{(\beta +1) (\text{Vc}-\text{Vbeo} (1-\gamma \text{\Delta T1}))}{\text{Rb1}+\text{Re1} (\beta +1)}$

Of course $\beta$ will vary with temperature too, but that should be much less important.

Careful choice for Re1, and Rb1 would allow the thermal effect on the current to be reduced. We are talking like 20% kind of numbers here. For example, if Vc=2V, Vbeo=0.7V, $\beta$=50, Rb1=10 Ohms, Re1=1 Ohm, and $\text{\Delta T1}$ increased by 100C over ambient; current through Re1 should look approximately like:

So, with Re1 of 1 Ohm, there is about a 10% change with 100 degrees of temp rise. The emitter resistors in this example would have up to about 1.5W in them. Lower values could be used, but then the variation would be greater. Operation of Q1 and Q2 would mostly be independent except for Vc and voltage across Rload.

To really control the current though would require a feedback loop to regulate Vc. And, to really cause current in each transistor to match would require a feedback loop for each transistor.

Don't Try This With MOSFETS. At least don't expect MOSFETs to magically share current.

While MOSFETs are very good for paralleling in switched mode of operation, they will not share current in linear operation. This is because the gate to source threshold voltage ($V_{\text{th}}$) has a negative temperature coefficient. As device temperature rises, $V_{\text{th}}$ becomes less, so the warmer the part the earlier it will start to conduct (Micro Semi has an app note about this). Here is a chart of the transfer characteristic to illustrate.

You can see how $V_{\text{th}}$ lower as $T_j$ rises. This also means that for low drain currents (about 5 amps or so in the chart) transconductance ($g_f$) will effectively be higher for the hotter part. Parallel devices will not start to share current until past the crossover point shown on the chart at about 15 amps. It is unusual for FETs operating in linear mode to ever get to the crossover point.

This is even a problem within a single MOSFET. Hot spotting on a MOSFET die is a well known phenomena. If you pop the top of a MOSFET and pull up a microscope, you will see thousands of cells on the die that are parallel micro MOSFETs. Each micro FET has its own $V_{\text{th}}$. So, for a fixed $V_{\text{gs}}$ and linear operation, the cell with lower $V_{\text{th}}$ will start to conduct first and heat up. $V_{\text{th}}$ will drop and that cell (and those around it) will conduct more. A hot spot will develop. It is possible for the device to be damaged in this way. On-Semi covers this in app note AND8199 (Hat Tip to Phil Frost).

If sharing between cells on a die are poor, imagine how bad sharing will be between separate devices with not well matched $V_{\text{th}}$. Remember how Vbe of the BJT changed by -1.6mV/C? Well $V_{\text{th}}$ of the FET changes by about -3mV/C, about twice as much as the BJT. So, current imbalance between paralleled FETs in linear operation will be much worse than BJTs (and they are bad enough).

To linearly control current with a MOSFET, $V_{\text{gs}}$ needs to be actively controlled by a feedback loop. Here is a recent example of what happens when the MOSFET is not controlled by the feedback loop.

Paralleling linearly controlled MOSFETs for current sharing means having a feedback loop for each device.

• ON Semiconductor AND8199 discusses this in detail. – Phil Frost Jul 27 '13 at 11:15
• @PhilFrost thanks for the link, I like it better than the one I had. Added to answer. – gsills Sep 24 '13 at 20:51

Your circuit as shown is not a good idea because all the transistors won't be equal. There can be significant variation in gain from part to part, and the B-E drops won't match exactly either. To make matters worse, the transistor that ends up taking the most current will get hottest, which makes its B-E drop go down, which makes it take more current ...

The simplest way to get around this with bipolar transistors is to put a small separate resistor in series with each emitter. You have a 50 Ω Load, so 1 Ω emitter resistors should be fine. Now you tie all the bases together direction.

When a transistor carries more current than the others, the voltage accross its emitter resistor will go up. This reduces its B-E voltage relative to the others, which gives it less base current, which causes it to carry less of the overal output current. The emitter resistors basically cause some negative feedback that keeps all the transistors roughly ballanced.

• +1 for adding emitter resistors to balance current between BJTs. – gsills Jul 26 '13 at 16:57