8
\$\begingroup\$

I was reading the book Analogue electronic circuits and systems, by Amitava Basak, when I stumbled in this level shifter circuit that uses BJTs instead of the usual mosfets. See the below picture. Please disregard transistor models.

From Analogue electronic circuits and systems, by Amitava Basak

I am puzzled by Q2, the middle NPN transistor: is it always off?

  1. Q5 is connected in diode mode, and its Vbe is fixed at 0.7V
  2. Adding Q2 Vbe plus the voltage drop in R2 should then give 0.7V again
  3. The above imply that Q2's Vbe is less than 0.7 OR the voltage dropo on R2 is zero.

Discarding the null voltage drop on R2, how is it possible that a transistor switched on has a base to emitter voltage below 0.7V?

\$\endgroup\$
  • 1
    \$\begingroup\$ Consider R2 is tied to -VEE and not to GND as R3 is \$\endgroup\$ – po.pe Feb 13 at 13:57
  • \$\begingroup\$ The basic idea here is to use a current generated by \$R_3\$ and diode-connected \$Q_5\$ to set up \$Q_5\$'s \$V_\text{BE}\$ and therefore a mirrored current (but not 1:1 because of \$R_2\$) in \$Q_2\$. \$R_2\$ makes it a much reduced current at the collector of \$Q_2\$, but it's still set indirectly by \$R_3\$. This current creates a knowable voltage drop across \$R_1\$. Since \$Q_1\$'s emitter is a knowable drop from \$V_\text{IN}\$ and since \$R_1\$ has a knowable drop across it, \$V_\text{OUT}\$ will be a knowable drop below \$V_\text{IN}\$ and that's how the shifter works. \$\endgroup\$ – jonk Mar 21 at 18:05
5
\$\begingroup\$

is Q2 always off?

No it is always on. What Q2 does is that it tries to make a (small) current flow into it's collector. That current tries to pull down the voltage at Vout.

Q5, Q2 and R2 are sort of a current mirror but a bad one. The current through R3 isn't copied 1 to 1 as in a "proper" current mirror. Instead since R2 is present the current through Q2 will be much smaller than the current through R3 and Q5. Across R3 we get a fairly constant voltage of VEE - 0.7 V. Since R3 is 1 kohm the current through R3 will be a couple of mA. As said the current through Q2 will be smaller than that, like 100 uA or less (100uA is just my guess, it is too much work to make that a more accurate number, that does not matter anyway for explaining how the circuit works). That 100uA causes the Vbe of Q5 to be not 0.7 V but slightly smaller like 0.6 V (= 0.7 V - 100 mV, as 100 uA through R2 gives 100 mV).

That 100 uA pulls down Vout. Opposite is Q1 pulling the output up (through R1). If Vin is high enough then Q1 can supply so much current that Vout will be pulled up, almost to the value of VCC.

When Vin has a very low voltage Q1 will be opened much less, supplying much less current so Q2 "wins" and Vout is pulled down.

Q5 is connected in diode mode, and its Vbe is fixed at 0.7V

Correct

Adding Q2 Vbe plus the voltage drop in R2 should then give 0.7V again

Indeed Vbe(Q2) + V(R2) = Vbe(Q5) = 0.7 V

What will happen there is that there will be a significantly smaller current flowing through Q2, R2 than there is flowing through Q5.

The above imply that Q2's Vbe is less than 0.7 OR the voltage dropo on R2 is zero.

Both are true, Vbe(Q2) will be slightly less than 0.7 V and there will be a small voltage drop (less than 100 mV) across R2.

\$\endgroup\$
  • \$\begingroup\$ I don't quite see how this works as a level shifter, though. Wouldn't it just act as an emitter follower, with an output about 0.7V below Vin? \$\endgroup\$ – Hearth Feb 13 at 14:13
  • \$\begingroup\$ You're right, it is more a voltage follower (with some voltage drop) and not a real "proper" levelshifter. \$\endgroup\$ – Bimpelrekkie Feb 13 at 14:20
  • \$\begingroup\$ It's not a very good circuit to be advertising as a level shifter, then! \$\endgroup\$ – Hearth Feb 13 at 14:20
  • \$\begingroup\$ @Hearth I fully agree with you, but some people call a circuit which shifts a voltage by some volts a "levelshifter" even though it's more an emitter follower ;-) \$\endgroup\$ – Bimpelrekkie Feb 13 at 14:41
  • \$\begingroup\$ @Bimpelrekkie, thank you for the comprehensive answer. I was assuming incorrectly that the base-emitter voltage drop on a transistor was fixed at 0.7V when conducting. Again, pretty illuminating :) Cheers! \$\endgroup\$ – I DG Feb 13 at 15:05
2
\$\begingroup\$

Resistors \$R_3\$, \$R_2\$ and transistors \$Q_2\$ and \$Q_5\$ form a Widlar current source.

enter image description here

Source: https://en.wikipedia.org/wiki/Widlar_current_source

The current source is always on.

This isn't really a circuit applicable to board-level design. As it requires the transistor be geometrically and thermally matched.

\$\endgroup\$
  • 1
    \$\begingroup\$ +1 for putting a name on it. Interesting read. \$\endgroup\$ – Unimportant Feb 13 at 19:38

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.