# BJT Constant Current Driver

I'm currently working on this circuit and I ran into 2 questions that I'm not sure I understood them correctly ( I have U1 = 15V , URE = 2 V , β = 180 , UCE = 5V and a constant current I = 2mA )

These are my calculations for this circuit $$U_{R1} = 15 – 2.7 = 12.3 V$$

$$R_1 = U_{R1} / I_B = 12.3 / 2 mA = 6.15 kΩ$$

$$U_{RL} = U_2 – U_{CE} – U_{RE} = 15 – 5 - 2 = 8 V$$

$$R_L = U_{RL} / I_C = 4kΩ$$

$$R_E = U_{RE} / I_C = 1kΩ$$

In which range of UCE is current control possible? When UCE is between 0 and 15V ?

How do we determine the internal resistance of the current source? Is R1 supposed to be the internal resistance here ?

• Why not use two transistors instead and skip the diodes? – Anton Ingemarson Dec 23 '20 at 13:52

How do we determine the internal resistance of the current source?

The internal resistance of the current source is the resistance projected by the collector. Internally, that is defined by this H parameter: $$\h_{OE}\$$.

Regarding the resistance it projects, because $$\h_{OE}\$$ is an admittance, the resistance is $$\\dfrac{1}{h_{OE}}\$$.

This is the dynamic resistance between collector and emitter and it might be in the region of several hundred kΩ.

In which range of UCE is current control possible? When UCE is between 0 and 15V ?

You need at least 2.8 volts from base to 0 volts to forward bias the diodes sufficiently but, diodes are pretty "loose" devices and so you can't expect much in the way of accurate current control at any value of UCE between 3 volts and 15 volts. I mean, it may be OK for what you want but, for someone else with much more stringent requirements, it'll be pretty poor.

Is R1 supposed to be the internal resistance here ?

No.

• Thanks for your detailled answer but just to be clear : by 1/hOE you mean UCE/ IC ? – Gaston Dec 23 '20 at 14:35
• @Gaston yes $h_{OE}$ is $\dfrac{I_C}{V_{CE}}$. – Andy aka Dec 23 '20 at 14:48

Just as an additional information, if you are looking for a constant current driver which does not rely that strongly on the supply voltage, you could try the following circuit:

As you can see the current through the diode is limited by the lower-left transistor and is almost independent of the supply voltage.

The calculation is roughly given by:

1. Determine the emitter resistor

$$R_E = \dfrac{V_{B,E}}{I_{e}}\approx\dfrac{600mV}{2mA}=300\Omega$$

1. Determine the minimum supply voltage

$$V_{in,min} > 2\cdot V_{BE} = 1.2V$$

1. Considering a minimum supply voltage of $$\2V\$$, and that it represents the worst case, the base resistance is given by:

$$R_B \le \dfrac{(2V - 1.2V) \cdot \beta}{I_C} = \dfrac{(2V - 1.4V) \cdot 110}{2mA} = 44k\Omega$$

I hope it helps.

Besides the precision of the circuit using diodes, Vce can vary with no specific problem.

If the diode current is around 3mA you can expect each diode to have a voltage drop of about 0.7V.

For the transistor when the collector current is 2mA the needed base current is under 10uA so the choice of the base resistor R1 is mainly constrained to the diodes current (4K is more convenient than 6.1K).

However, the limitation is on the resistance of the resistor RL which can vary from 0 to not more than 6.5K in your case which is (15V - 1K x 2mA)/2mA .

The internal resistor is the output voltage divided by the current with no load so it is defined by the slope of the Vce(Ic) and it is as mentioned in another answer approximately 1/hoe (it is 1/hoe+1k) but 1k is too small compared to 1/hoe. R1 has nothing to do with it.