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I'm experimenting with/learning about the use of active-loads with a common emitter and I've hit a wall. When using a resistor in the collector it makes sense how the output voltage bias can be set to somewhere sensible such as half way between VCC and ground.

When using an active load however, how is this typically achieved so that you can retain the high gain that comes with an active load?

I've put together this quick circuit to play with and it shows that the Vout is biased down at 123mV. Can this be brought up to say 5V without losing all the gain?

enter image description here

Many Thanks

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  • \$\begingroup\$ Decrease Rref in value. \$\endgroup\$
    – Andy aka
    Commented Aug 23, 2020 at 10:37
  • \$\begingroup\$ Ad a resistor between collector and active load. \$\endgroup\$
    – LvW
    Commented Aug 23, 2020 at 10:43
  • \$\begingroup\$ Cdevelop, just look at the behavioral viewpoint. You have a current source into the BJT collector. There is no DC path for that current source except for the BJT collector. So the BJT collector must take 100%, not 99% and not 101%, but exactly 100% of the DC current source's current. And it has to do that all the time. But the signal is modifying the base-emitter voltage which controls the collector current. The addition of a resistor from base to collector provides a DC path. But there are other (probably better) design choices to make. \$\endgroup\$
    – jonk
    Commented Aug 23, 2020 at 15:13

1 Answer 1

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With an active bias, it is very difficult to set the working point. Because we are practically connecting two current generators in series. So DC negative feedback can be used to stabilize the working point. Unfortunately, everything is specific In this case, a different value of resistance must be added for another transistor. áram

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  • \$\begingroup\$ Thanks for the help - this has got me moving forward. Is this known generally as 'Series-Shunt Feedback' (so I can look it up in my textbook/google)? \$\endgroup\$
    – Cdevelop
    Commented Aug 23, 2020 at 12:26
  • \$\begingroup\$ Yes - good solution. This is the principle we know from biasing another high-gain device: Operational amplifier. \$\endgroup\$
    – LvW
    Commented Aug 23, 2020 at 13:40

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