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For my new amplifier design I need a kind of parallel Norton level shifter for differential signal (to subtract the same externally controlled voltage drop from two signals). One possibility is to use a kind of "multiple output current mirror". With FETs it looks quite simple. Just connect input transistor's gate to any number or output transistors' gates. Like here:

MAX1916

Is it possible to do the same with BJTs?

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    \$\begingroup\$ Yes BJTs will work ,and should your devices be all N chan ? Q2 Q3 Q4 are shown as P chan .Double check your diagram. \$\endgroup\$
    – Autistic
    Sep 7, 2016 at 10:29
  • \$\begingroup\$ Indeed that schematic is wrong Q1 - Q4 should all be NMOS. \$\endgroup\$ Sep 7, 2016 at 10:35
  • \$\begingroup\$ It just an first googled example. I use NPNs. \$\endgroup\$
    – e_asphyx
    Sep 7, 2016 at 10:45

2 Answers 2

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Is it possible to do the same with BJTs?

Yes it is.

But it is advisable to a resistor in series with the emitter of each transistor so that any difference between transistors (including temperature) is somewhat compensated for. Like this:

enter image description here

You can just repeat Q2 + resistor to get multiple outputs, like in the NMOS schematic.

What value for Re (emitter resistors) ?

Use value that drops around 100 - 500 mV across the resistors. The actual amount of voltage drop is not critical.

You will often see NPN current mirrors without the emitter resistors, these will often be on chip and there the transistors are very well matched so resistors are not needed.

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  • \$\begingroup\$ In my current designs I use emitter resistors with about 500mV drops across them. I'm asking my question above because I'm not sure if BJT base current can be an issue with multiple secondary transistors. \$\endgroup\$
    – e_asphyx
    Sep 7, 2016 at 10:43
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    \$\begingroup\$ If it's an issue depends on 1) how accurate do you want the input current to be copied (the summed base current for all NPNs will be extracted from the input current) 2) the beta of the transistors, a higher beta of course means a lower base current so less of what I described in 1) If you're not too fussed about accuracy of current and you're not feeding more than 10 current outputs, just use NPNs with beta more than 100 and it will work. \$\endgroup\$ Sep 7, 2016 at 11:06
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If you have multiple current source paths you can add an additional transistor as beta helper to supply the base currents:

enter image description here

Transistor \$Q_2\$ is then drawing the sum of the base currents of \$Q_1\$, \$Q_3\$ and \$Q_4\$ divided by the beta of \$Q_2\$ from \$I_{IN}\$. A drawback of the beta helper is that the minimum input voltage is \$2\cdot V_{BE}\$ instead of \$V_{BE}\$ in the case of a simple current mirror.

For \$Q_2\$ to operate in the forward active region \$V_{CC}\$ must be at least \$2\cdot V_{BE}\$ (ca. 1.4 V at room temperature).

EDIT:

To avoid additional current drawn from \$I_{IN}\$ a MOSFET can be used instead of \$Q_2\$.

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  • \$\begingroup\$ Interesting configuration somewhere between basic and Wilson current mirror. (The VCC from the schematic looks more appropriate than the \$V_{DD}\$ (\$V_{DD}\$) from the text.) \$\endgroup\$
    – greybeard
    Mar 16, 2023 at 11:08
  • \$\begingroup\$ @greybeard Thank you for the hint, I have changed the text accordingly. \$\endgroup\$ Mar 16, 2023 at 12:00

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