To make a crude voltage regulator, I've made a simple emitter follower using a BD139 transistor. When the 0.5A load is connected, the circuit behaves as expected: the voltage across the load sits at about 0.8V below the voltage at the base. However, if I increase the load impedance, or disconnect the load, the output voltage rises to equal the voltage at the base. That's a problem, because I want the output voltage to be constant despite variations in output current. Any ideas, please?

  • 3
    \$\begingroup\$ Need minimum load current, a resistor works well \$\endgroup\$ – sstobbe May 24 '17 at 15:29
  • \$\begingroup\$ your Ie/Ib ratio is too HIGH, try a current sensing MOSFET with BJT feedback instead or a darlington or really.. an LDO, don't reinvent the wheel.l reverse engineer what is commercially avail. in LDO's then understand before you fail \$\endgroup\$ – Sunnyskyguy EE75 May 24 '17 at 15:29
  • 1
    \$\begingroup\$ Look at the IV curve of a diode with zero current- zero voltage, add a resistor for at least a few mA of constant load \$\endgroup\$ – sstobbe May 24 '17 at 15:43
  • 1
    \$\begingroup\$ Also the shorthand relation for the base-emitter voltage is for every 10x increase in current deltaVbe is +60mV \$\endgroup\$ – sstobbe May 24 '17 at 15:45
  • 1
    \$\begingroup\$ @sstobbe Great stuff added here. Are you going to write a nice answer about it, showing how setting a minimum load can help in the direction of achieving the goals? It seems you are the right one to do it -- you've covered all the right bases in your comments -- and the OP still seems to need the hand-holding. \$\endgroup\$ – jonk May 24 '17 at 15:50

An emitter-follower operates on the principal that the base-emitter voltage of a forward biased transistor is approximately equal to 0.6 or 0.7 V. So for some input voltage on the base, the emitter tracks the input minus one diode drop. $$ V_O = V_I - V_{BE} $$

Of course this is just the shorthand approximation, the actual base-emitter voltage is given by:

$$ I_C = I_S \exp \left( {\dfrac{V_{BE}}{V_T}}\right)$$

A sample IV plot is the following,enter image description here

Note the scale on the x-axis is in 60 mV graduations to illustrate a common rule of thumb for transistors. Every 10x increase in collector current is a 60 mV increase in base-emitter voltage near room temperature.

To answer your first question of,

However, if I increase the load impedance, or disconnect the load, the output voltage rises to equal the voltage at the base.

Looking at the sample IV plot, you can see that when you have zero load current, the base-emitter forward voltage is also ~ 0 VDC. So Vbe = 0, i.e. the base voltage equals the emitter voltage.

As a simple shorthand approximation, if you load the output with 1/10 your max rated current, you will achieve 60 mV of load regulation. A restive load at 1/10 rated current is a simple solution.

Impedance Transformation

When the transistor is operating in active mode (Vcb>0), the transistor also provides an impedance transformation of the source resistance by \$1/\beta\$. A sample schematic of a common-emitter buffer amplifier:


simulate this circuit – Schematic created using CircuitLab

From standard transistor theory, you are aware that \$ I_C = \beta I_B \$. So, as you begin to load the output of Q1, \$ 1/\beta \$ of the load current is sourced through the base. As a result of having a finite source resistance from your reference, the voltage right at the base terminal of Q1 droops by \$ I_B * R_S \$.

As a practical example assume RS = 10 ohms and \$ \beta \$ = 100, the output appears as having an output resistance of $$ R_o = \dfrac{R_S}{\beta} = 0.1 \textrm{ Ohms} $$

Your goal here is to keep the source resistance small enough so the the output voltage only depends on the IV curve of your transistor.


The three most important design choices to make are

  • RE, for minimum load current
  • Rs, lower for lower Ro
  • beta of Q1, must have good beta over the full load range


  • For DC don't worry about the 1/gm you see in typical amplifier notes, its just a linearization of the Ic-Vbe curve (unless you care about its ac output impedance) just look at the IV curve

  • There is also thermal design to consider, Vbe, as rule of thumb has tempCo of -2mv/deg. This does work in your favor as you pull more load current the transistor heats up and thermally pulls Vbe back up

  • \$\begingroup\$ Wonderful. I'd add that there is also a Shockley equation \$\frac{\textrm{d}V}{\textrm{d}I}\$ at the emitter tip, based upon the operational value of \$I_C\$, and that this adds to the effective resistance. In your example, and taking the OP's \$500\:\textrm{mA}\$ peak load example, that would add to make about \$150\:\textrm{m}\Omega\$ instead of \$100\:\textrm{m}\Omega\$ and that it will be worse with lower loading. Every BJT also has some Ohmic RB (20 ohms for 2N3906?) and RE. So the \$R_o\$ result could be "improved," depending on the level of the OP to process such added info. \$\endgroup\$ – jonk May 24 '17 at 18:10
  • \$\begingroup\$ @jonk Thanks for the great feedback, I agree more could be added regarding the effective Ro. My hope was to illustrate the two main points independently Ic(Vbe) and Source Resistance. I will try to incorporate your suggestions into a better post \$\endgroup\$ – sstobbe May 24 '17 at 18:35
  • \$\begingroup\$ I'm not even expecting an addition. It would be interesting only if you can explain it simply enough to reach the OP well. But say looking at small variations around \$I_C=100\:\textrm{mA}\$, with \$R_b=20\:\Omega\$, \$R_S=10\:\Omega\$, \$R_e=100\:\textrm{m}\Omega\$, \$\beta=250\$, then \$R_o= R_e+\frac{R_b+R_S}{\beta}+\frac{26\:\textrm{mV}}{I_C}\approx 500\:\textrm{m}\Omega\$. And here, \$\frac{\textrm{d}V}{\textrm{d}I}\$ is more than half of it. Spice confirms the result, too. \$\endgroup\$ – jonk May 24 '17 at 18:50

The problem is that the 500 to 700 mV or so voltage drop across a silicon junction is just a approximation at reasonable currents. Base to emitter of a NPN transistor is such a junction. You got about 800 mV because the current was relatively high for that transistor. That's all reasonable.

However, when the current is 0, the voltage drop is also 0. The voltage is roughly the log of the current once you get to reasonable levels, so it is relatively constant over a reasonable current range. 0 is well below reasonable for the approximation.

The simplest solution is to always guarantee some minimum load by putting a resistor permanently on the output. Even 1 mA will probably cause over 500 mV of drop. If so, you now only have 300 mV variation between no load and full load.

A even better solution is active regulation. If the B-E voltage drop is inside a feedback loop, the feedback can compensate for it. Here is a example:

Due to the feedback around the opamp, it adjusts the base voltage of the transistor to whatever it needs to be so that its emitter is at the reference voltage.

In the simplistic version of this circuit, C1 and R1 wouldn't exist, and the negative input of the opamp would be directly driven from Vout. However, the overall gain the opamp provides is low, and there is some non-linearity going on in the B-E junction, so the opamp might get unstable. C1 feeds back the high frequencies where the opamp is more likely to be unstable at, directly from its output to its negative input. That slows the opamp response and increases stability. R1 provides some impedance so that the signal thru C1 doesn't get swamped. This example is quite conservative. Many opamps will be stable with well less than 100 pF for C1. Even with these values, the opamp can still respond to load changes up to about 160 kHz.

R2 is just to put some load on the output so that there isn't a large jump needed from the opamp between no load and the first little bit of load.

This circuit should regulate the output to within a few mV from the full range of 0 to maximum load. Eventually either the transistor can't pass more current or the opamp can't supply enough base current.

  • \$\begingroup\$ Thank you very much! My thoughts were beginning to drift in the op-amp direction also, but your answer moves them forward a lot. :-) \$\endgroup\$ – user3910824 May 26 '17 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.