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Seemingly Simple Circuits, User-Proof External Supplies: Circuit 2, The Improved One-BJT Current Limiter

The improved one-BJT current-limiting supply is not much different than the primitive circuit, though its one additional resistor does change the circuit behavior. The improved circuit is shown below.

Analysis quickly proceeds by Thevenizing the base circuit and β-transforming the Thevenin base resistance to the emitter, resulting in the following equivalent circuit.

When the BJT is not saturated, the output current is a maximum of

where VBE (IO ) = VBE (sc). Under light load, the BJT saturates and as with the previous circuit, VCB (oc) ≈ VEB (oc) and

where VEB (oc) corresponds to the minimum-current value. In the previous circuit this was VEB (oc) = 0.50 V at 1.3 μA.

Solve for Ri from the VO equation:

RB is found by substituting for Ri in the IO equation and solving;

To design, choose RB . Its choice is subject to the constraint that RE ≥ 0 Ω. The subtraction of the RE term in RB results in the maximum RB when RE = 0 Ω. Then for larger RE , RB must be less, and

Dependence of IO on β is minimized when δ IO /δβ is minimum, or whenever RE >> (Ri ||RB )/( β + 1). This compares with what was found for the primitive one-BJT circuit. For this circuit, Ri makes the inequality easier to satisfy.

Test Circuit

To test these equations, a design was carried out and built for IO = 20 mA using a 2N2907 BJT selected for a β of 150. The VEB were calculated using IS = 5 fA = 5×10–15 A. IS is difficult to obtain directly from the VBE (IC ) curves in the parts data because those curves are not exactly exponential. They include the additional linear voltage contribution of series resistance – of re ’, the ohmic emitter resistance, and rb ’, the ohmic base resistance, referred to the emitter. Additional calculation from the curves leads to the conclusion that the total effective ohmic emitter resistance is about 0.5 Ω. If the VBE are calculated from the given IS then the values will be somewhat low, especially for the high-current value. The easiest and most accurate procedure for design is to read the two pairs of numbers off the VBE curves.

The values used in this design were

The resistor values calculate from the design formulas to be

The circuit was built with these standard +/-5 % parts values but this time a $1000 DMM (a Keithley 2000) was used to go through the 5 % resistor parts bins and select parts within about 0.1 % of these nominal values. Then

β dependence is reduced with RE six times that of the base-referred resistance, showing the role of Ri . Also,

The base divider was disconnected from the base and the measured voltage agreed with the above value in all digits. Then the $30 (DT-182) DMM was used to measure output current by placing it as an ammeter across the output, effectively shorting it. The DMMs were both used to measure voltages, and the following values resulted.

The measured value of V was 5.00 V. The measured values of IO appear to verify the design equations.

This improved one-BJT current-limited source is not a bad choice for a low-cost, low-parts-count current limiting supply extender. The value of IO , unlike the carefully value-selected prototype above, will have a spread of values corresponding to parts tolerances, including BJT β . Many less-demanding applications can be satisfied by it.

By adding one more transistor, some additional improvement can be realized, the topic we consider next.

7 comments on “Seemingly Simple Circuits, User-Proof External Supplies: Circuit 2, The Improved One-BJT Current Limiter

  1. ziggle314
    March 4, 2016

    Hi Dennis,

    I was able to almost derive your results, but I have a factor in my results that is different than what you obtained and I wonder if you could take a look. For example, when I try to derive the constraint on RB , I obtain the following result.

    RB ≤(β/IO )*(V-VEB(sc) /(V-VO -VEB(oc /V))

    Your result is similar, but I obtain a V instead of VO -VEB(oc) in my denominator.

    RB ≤(β/IO )*(V-VEB(sc) /(V-VO -VEB(oc) /(VO -VEB(oc) ))

    I was wondering if you could give me any insight on where to look for the problem.

    Thanks

     

    Note: Edited to correct an inequality error.

     

     

  2. D Feucht
    March 5, 2016

    I rechecked my math from the original notebook derivation and you are right that it needs correction. Your result is slightly different that what I ended up with, though closer than my original equation. My result retains the inverted inequality sign (R sub B < =) because if you substitute for R sub i and then solve for R sub B, the R sub E term is subtracted. Consequently, if R sub E > 0 ohms, then R sub B is made smaller so that for R sub E = 0 ohms, R sub B is maximum.

    Because the comment editor is not conducive to equation-writing, I have asked the editor to correct the original article. The correction also clarifies how the R sub B equation is derived.

  3. ziggle314
    March 5, 2016

    Thanks Dennis.
    You are correct about the inequality – I just typed it wrong. I am not used to typing equations using ASCII, and I should have checked my input closer.

    Thanks for the interesting article.

     

    Mark

  4. ziggle314
    March 8, 2016

    Sorry to bother you again. Could you look at your expression for RE? You have a term, VO-VEB(SC), that my derivation shows should be V. I get all of your other expressions.

  5. hamiltonraza1
    March 9, 2016

    great post you have poste here

  6. D Feucht
    March 9, 2016

    Mark,

    You're right again. My notebook date for the derivation was 23MAR09 – evidently a “bad-algebra day”. Editor Steve Taranovich will be updating the text.

    There are a couple of other minor errors and an omission following these formulas. The omission is that the circuit values were solved using VO = 4.0 V. The V BE open- and short-circuit values given in the derivation of the parts values are negative for a PNP and should be VEB instead.

    Thanks again for your feedback. It is always gratifying to see a reader actually work out the math! If you use this circuit, let us know if the measurements agree with the calculated values.

     

  7. ziggle314
    March 9, 2016

    Thanks for mentioning the Vo omission. I was not getting your table results and this explains the discrepancy. I have an application for the circuit and I will be building it in the near future. I will post my results as a comment to this blog post when I complete my work.

    I am a big fan of your 4-volume analog design library. Keep writing up your analog work.

    Mark Biegert

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