Dual transistor improves current-sense circuit
A two-transistor circuit eliminates variation in output due to temperature.
Robert Zawislak, PE, Consultant, Palatine, IL; Edited by Charles H Small and Fran Granville -- EDN, August 16, 2007
In multiple-output power supplies in which a single supply powers circuitry of vastly different current draws, two perplexing steps are sensing the current that each output draws and deactivating the power supply in the event of an overload on that output. These issues are especially important in protecting the fragile PCB (printed-circuit-board) traces in low-level circuits. A typical circuit would use the base-emitter threshold voltage of approximately 0.6V of a bipolar transistor to trigger the power-supply-protection circuits. Although economical, the transistor’s threshold varies excessively over temperature; hence, the protection level is unstable.
The circuit in Figure 1 essentially eliminates the base-emitter-voltage temperature-variation problem as the derivation of the output voltage and as a function of the load current. By using dual bipolar devices in one case, the manufacturer nearly perfectly matches the two devices. Although this Design Idea describes a positive power supply, you can realize a similar negative-output-supply current-sense circuit using a dual NPN transistor in place of the dual PNP that the figure shows.
The following equations show the derivation of the output voltage as a function of the load current (referring to Figure 1):
VBA+(ILOAD×RSENSE)+(IE×R2)–VBB=0.
[(VBA–VBB)+(ILOAD×RSENSE)]–IER2=0.
IC+IB=IE.
(VBA–VBB)+(ILOAD×RSENSE)–(IC+IB)R2=0.
IB=IC/β.
VBA–VBB+ILOAD×RSENSE–(IC+IC/β)R2=0.
VBA–VBB+ILOAD×RSENSE–[IC×(β+1)/β]R2=0.
VOUT=ICR3.
IC=VOUT/R3.
VBA–VBB+ILOAD×RSENSE–(VOUT/R3)(β+1/β)R2=0.
If VBA=VBB, then VBA–VBB=0, and
ILOAD×RSENSE–(VOUT/R3)(β+1/β)R2=0.
VOUT=ILOAD×RSENSE[R3/(β+1)](β/R2).
If β is high, then β/(β+1)β)≈1, and VOUT=(ILOAD×RSENSE×R3)/R2.
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Maybe I'm missing something, but I don't see how this circuit can work, and it doesn't solve the original problem of shutting down the power supply when an overload occurs (using the regulator IC's current sense input). Part of the problem is a math error in the first equation; it should be
Vba - Iload*Rsense - Ie*R2 - Vbb = 0.
Assuming Vba = Vbb, this requires Ie to be negative - ???
But as Miguel and Jim noted, if you flip the power supply and load and reverse the sense of the load current (or swap Q1A and Q1B and their attached components), then the first equation becomes
Vba + Iload*Rsense - Ie*R2 - Vbb = 0
and the rest of the calculations are correct.
Ron Bauerle - 2007-4-10 10:16:00 PDT -
And the value of R4 is?
Neat circuit. And useful.
Barney Barnes - 2007-10-9 07:32:00 PDT -
by the way, i had to reverse the positions of the power supply and Rload in the simulation. R4 determines the inflection.
miguel.escoto, jr - 2007-20-8 00:55:00 PDT -
There is a sign error on the second line of the equation, the -EeR2 got reversed. Vout has an inverted sense.
I did a simetrix simulation. there is a cut-in problem at low currents and only a small range of linearity to about half of the range.
miguel escoto, jr. - 2007-19-8 23:56:00 PDT -
Great idea--why didn't I think of it--woops looks like polarity across sense resistor is reverse
Jim Keith - 2007-16-8 09:22:00 PDT
