With the circuit setup shown below, if I have available a type of Q2 transistor with a certain maximum dissipated power and need that D1 withstands a large current that would lead Q2 in trouble, it is necessary to share the current among several transistors in parallel to Q2. One can think that this upgrade can be achieved by connecting the additional transistors exactly in parallel to Q2 as per definition, that is bases connected together, collectors together, emitters together. But transistors are identical only in theory, in practice hfe values usually span a large interval and it is typical that one of them draws more current than the others and gets hotter (being aware that the sum of the collectors current is limited by VBE@Q1/R2). By having all the VBE tied together they are kept at the same value with no possibility to vary in case collector current increase, therefore it may happen that the hottest transistor gets in thermal runaway. My question is: am I right with this assumption? Can this issue be circumvented by placing a small resistor in series eg, 0.1ohm to each emitter, so that VBE can decrease in case collector current rises? Thanks in advance.

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@aconcernedcitizen, @Sparky256 I am aware that the topic resembles the one you cited, but my question focuses on the paralellization. Let's put numbers. Supposing I want 500mA flowing throug D1, trying with Q2 only implies to set R2=0.7V/500mA=1.4ohm. Then I am realizing that VCE(Q2)=VCC-V(D1)-0.7V times 500mA gives me a power number that goes beyond the allowable limit of Q2, therefore I decided to add one more transistor, let's call it Q3, in parallel with its emitter resistor R3 in parallel to R2, by changing R2=R3=2.8ohm (the double as previous). In theory the 500mA current should split equally among Q2 and Q3, but in truth one of them will be more demanding, let's assume 300mA flows in Q2 and remaining 200mA in Q3. Q2 will get hotter and its temperature surge will make Q2 getting more and more current out of the overall 500mA. In this case a feedback aimed at reducing VBE(Q2) as much as IE(Q2) rises would be useful, to me here it looks like VBE(Q2) and VBE(Q3) are tied together to a fixed value, therefore I mentioned the two additional resistor between each emitter and R2 and R3, respectively. Am I loosing anything?

  • \$\begingroup\$ You already have R2. Use two transistors each with an R2 of double the value. The feedback of Q1 will stay the same as the current is halved. \$\endgroup\$ – Oldfart May 1 '18 at 7:46
  • \$\begingroup\$ Duplicate? electronics.stackexchange.com/questions/136208/… \$\endgroup\$ – a concerned citizen May 1 '18 at 8:02
  • \$\begingroup\$ Keep R2 above a minimum level like 47 ohms and the circuit will be fine. Assuming Vcc is not over 15 volts, as the transistor Q2 must dissipate the current times the Vcc volts. \$\endgroup\$ – user105652 May 1 '18 at 8:02
  • \$\begingroup\$ @aconcernedcitizen. I agree it is an exact duplicate and will VTC. \$\endgroup\$ – user105652 May 1 '18 at 8:04
  • 3
    \$\begingroup\$ Possible duplicate of Does this linear current source driver deliver constant current? \$\endgroup\$ – user105652 May 1 '18 at 8:06

The simplest extension of your thoughts would be the following schematic:


simulate this circuit – Schematic created using CircuitLab

This doesn't include separate current monitoring for each of the current-sharing BJTs, \$Q_1\$-\$Q_3\$, but given that their bases are two \$V_\text{BE}\$'s above ground there should be a roughly approximate sharing of currents. \$V_\text{BE}\$ variation between BJTs will have the largest impact on the sharing, but \$\beta\$ variation won't matter too much on that score. \$\beta\$ variation will matter more in sizing \$R_4\$.

Keep in mind that there will be differences in temperature and that this will lead to still further shifts (in a bad way) in terms of the sharing. So if you can help by putting them in thermal proximity to each other, it may help a little bit. Also, keep \$Q_4\$ thermally isolated, if possible. It shouldn't need to heat up above ambient that much and the better isolated it is from the other BJTs, the better its regulation of the total current.

The current sharing, ignoring thermal variation for now, is based upon the following:

$$\begin{align*} I_{\text{E}_1} \approx I_{\text{C}_1} &= I_{\text{S}_1}\cdot\left(e^\frac{V_B-I_{\text{E}_1}}{V_\text{T}}-1\right)\\\\ I_{\text{E}_2} \approx I_{\text{C}_2} &= I_{\text{S}_2}\cdot\left(e^\frac{V_B-I_{\text{E}_2}}{V_\text{T}}-1\right)\\ &.\\ &.\\ &.\\ I_{\text{E}_N} \approx I_{\text{C}_N} &= I_{\text{S}_N}\cdot\left(e^\frac{V_B-I_{\text{E}_N}}{V_\text{T}}-1\right) \end{align*}$$

Ignoring the base current of \$Q_4\$ to simplify the problem slightly, we also know:

$$\begin{align*} V_{\text{E}_1} = R_1\cdot I_{\text{E}_1} &\approx R_1\cdot I_{\text{S}_1}\cdot\left(e^\frac{V_B-V_{\text{E}_1}}{V_\text{T}}-1\right)\\\\ V_{\text{E}_2} = R_2\cdot I_{\text{E}_2} &\approx R_2\cdot I_{\text{S}_2}\cdot\left(e^\frac{V_B-V_{\text{E}_2}}{V_\text{T}}-1\right)\\ &.\\ &.\\ &.\\ V_{\text{E}_N} = R_N\cdot V_{\text{E}_N} &\approx R_N\cdot I_{\text{S}_N}\cdot\left(e^\frac{V_B-V_{\text{E}_N}}{V_\text{T}}-1\right) \end{align*}$$


$$\begin{align*} V_{\text{E}_1} &\approx V_\text{T}\cdot\operatorname{LambertW}\left(\frac{R_1\cdot I_{\text{S}_1}}{V_\text{T}}\cdot e^\frac{V_B}{V_\text{T}}\right)\\\\ V_{\text{E}_2} &\approx V_\text{T}\cdot\operatorname{LambertW}\left(\frac{R_2\cdot I_{\text{S}_2}}{V_\text{T}}\cdot e^\frac{V_B}{V_\text{T}}\right)\\ &.\\ &.\\ &.\\ V_{\text{E}_N} &\approx V_\text{T}\cdot\operatorname{LambertW}\left(\frac{R_N\cdot I_{\text{S}_N}}{V_\text{T}}\cdot e^\frac{V_B}{V_\text{T}}\right) \end{align*}$$

Any particular ratio of currents is then:

$$\frac{I_{\text{C}_i}}{I_{\text{C}_j}}=\frac{\operatorname{LambertW}\left(\frac{R_i\cdot I_{\text{S}_i}}{V_\text{T}}\cdot e^\frac{V_B}{V_\text{T}}\right)}{\operatorname{LambertW}\left(\frac{R_j\cdot I_{\text{S}_j}}{V_\text{T}}\cdot e^\frac{V_B}{V_\text{T}}\right)}$$

(Obviously, you will probably want \$R=R_1=R_2=...=R_N\$.)

Variation of \$I_\text{S}\$ for BJTs in the same family might account for a band of about \$30\:\text{mV}\$ in their \$V_\text{BE}\$. If you can keep the thermal variations to within a band of about \$15\:^\circ\text{C}\$, then this would be about another \$30\:\text{mV}\$ of variation in their \$V_\text{BE}\$. So call this a total worst case situation (all things aligning wrongly) of perhaps a band of about \$60\:\text{mV}\$.

Given that the voltage drop across \$R=R_1=R_2=...=R_N\$ is at least 10 times that much and probably still more, the current sharing should remain pretty good (\$\pm 30\:\text{mV}\$ variations vs \$700\:\text{mV}\$ for the base voltage of \$Q_4\$.) Perhaps \$10\$% span of variation in the voltages over each emitter resistor and so the sharing should be close enough to be useful.

So this technique can work, I think. (Not that I've done it for this circumstance.) You could eliminate the thermal variation by adding more BJTs (Sziklai, for example) and therefore more tightly control the sharing. But I don't see a good reason to go that extra mile in the case you pose. So this should be fine.

The remaining problem will be setting the value of \$R_4\$. Clearly, there is also wide \$\beta\$ variation in BJTs and this also depends highly on temperature as well as collector current. So you will need to examine the datasheet to determine the worst case (smallest) value for \$\beta\$ that you expect and make sure that \$R_4\$ can provide at least that much base current to \$N\$ BJTs. You will also need some minimum collector current for \$Q_4\$.

Note that if the \$\beta\$ values are much better than expected, \$Q_4\$ will have to take up the slack in its own collector current. This will impact its \$V_\text{BE}\$ and therefore also the current setting. However, a 10-fold increase in \$Q_4\$'s collector current due to a prediction error caused by assuming worst-case \$\beta\$ values would only mean about \$60\:\text{mV}\$ variation in \$V_\text{BE}\$. It's unlikely that your prediction vs actual will be that bad. But you will have to determine how acceptable this might be.

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