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While looking at transistor based circuits, I am always trying to understand it with the transistor operating regions.

Question 1:

While looking at a push pull transistor based circuit, I want to understand in what transistor operating region the NPN and PNP transistor will be working. It can't be working in cut-off. So, is it in saturation state or active/linear state in push pull configuration?

It is always a dilemma between the saturation and active state. I have now figured on how to find out whether a transistor is working in saturation or active state (using the Ic and Ib factors, forced drive or overdrive factors.) But in a simple push pull configuration, there are no resistors. So, I am not able to find the working region of the transistor. Can anyone help with this?

Question 2:

Similarly, in a current mirror circuit, I understand that both the matched transistors need to be in active region for proper working in the current mirror operation. But how do you find out the operating region of the transistor with no Ic or Ib factors?

Please explain

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  • \$\begingroup\$ The bottom line to active mode is the hFE drops rapidly from 0.2 to 2V depending on Ic. You can expect 10% of hFE max during saturation. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 6 at 0:22
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Active elements (transistors) in circuits with negative feedback never saturate since the output quantity opposes the input one and does not allow it to saturate the transistor. This effect is achieved differently in the two types of negative feedback circuits, which differ in the way of subtracting the output quantity from the input one:

  • In circuits with series negative feedback (the emitter follower in a push-pull stage), the input voltage is decreased by subtracting the output voltage from it so the resulting base-emitter voltage is never sufficient to saturate the transistor.

  • In circuits with parallel negative feedback (the "active diode" in the current mirror input), the input current is decreased by diverting a big part of it so the resulting base current is never sufficient to saturate the transistor.

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  • \$\begingroup\$ Q#2 does not use feedback in the mirror. Why did you assume this? It is a forcing function with a lower impedance diode controlling a high impedance collector sharing the same Vbe control voltage in an 'open loop" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 13:25
  • \$\begingroup\$ I would only give half points to this answer as it does not explain the way it works properly and focuses on saturation which is more , the way it can fail. See the simulation again and see how negligible Q2 loads the Q1 current (1/1000) tinyurl.com/y8bl8dhz furthermore , there is no feedback in the mirror. @Newbie be advised. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 13:32
  • \$\begingroup\$ @Tony, you are right - there is no feedback in Q2 of the current mirror; it is simply a common-emitter stage. But I have explained Q1 (the "active diode") that is a stage with negative feedback. I focused my attention on the saturation since OP has a problem with distinguishing between saturation and active mode. \$\endgroup\$ – Circuit fantasist May 5 at 14:13
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1) Push-Pull is complementary Emitter Follower, unity gain , linear mode but hFE is still dependent on Ic and Vce with a wide tolerance.

enter image description here

2) With a BJT current mirror, one acts with BC shorted as a diode so the voltage drop can be estimated and that collector current is thus matched for the same Vbe mirror. Those base voltages are deterimined with R values and Vcc for that diode which is matched to the Vbe of the transsitor.

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  • \$\begingroup\$ Thank you. So, For Push pull - Linear/Active region of operation of both the transistors. And for Current mirror, The BC shorted transistor is Saturation while the other is in active region? \$\endgroup\$ – Newbie May 5 at 6:24
  • \$\begingroup\$ Both Vbe's are conducting but the transistor Vce is not saturated. It must a high impedance current source as defined. like this tinyurl.com/ya253hqy \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 6:27
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    \$\begingroup\$ In the range of VIN = -0.6 V ... +0.6 V, both transistors of the push-pull stage are cut-off. \$\endgroup\$ – Circuit fantasist May 5 at 7:12
  • \$\begingroup\$ @Circuitfantasist Note your assumption is false as there is negative feedback to eliminate that condition. But there is the typical limitation of input that must avoid 2V near the supply rails to have sufficient analog gain. Proof of concept tinyurl.com/ycm2rtkg \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 11:17
  • \$\begingroup\$ @Tony, I meant the humble push-pull (without op-amp). Precisely speaking, the negative feedback only allows the input voltage to quickly jump over this "dead" area... but it still exists... and for a very short time both transistors are cut-off... and the op-amp is saturated. This problem can be solved by adding a biasing. \$\endgroup\$ – Circuit fantasist May 5 at 13:01

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