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I have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Is there a way to show that the current through \$\text{R}_1\$ is given by \$\text{V}_\text{i}/\text{R}_2\$?

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The math comes from G/(1 + G * H) which is the easily revived transfer function [ math] for a certain type of feedback loop [ formal system concept == math].

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  • \$\begingroup\$ First of all, thanks for your answer. Can this be shown fully, using mathematically? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 15:30
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The OpAmp is trying to bring the potential difference between both inputs to zero.
So assume the OP is regulating the output to such a voltage level, that the negative input has \$V_i\$ as well. Because the negative input is directly connected with the node between T3 and R2 the voltage at this pooint is also \$V_i\$.
The current through R2 is now given by \$V_i/R_2\$. And finally the collector current of T3 is equal to the current though \$R_1\$ and depending on the current gain of T3 (typically around 100) the collector current is almost equal to the emitter current.

Et voila -> \$I_{R_3} \approx V_i/R_2\$

After this you have to do the check if this can possibly be the case, there are obviously some constraints:

  • \$V_b\$ must be larger than \$V_i + V_{CE_{sat}} + V_{R_1}\$
  • Supply voltage of the OpAmp \$V_{supply}\$ must be larger than \$V_i + 2 \cdot V_{BE}\$
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  • \$\begingroup\$ First of all, thanks for your answer. Can this be shown mathematically? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 14:48
  • \$\begingroup\$ Can this be shown fully, using mathematically? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 15:30
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Ignoring the emitter diode drops, the output of opamp \$V_o\$ is fully fedback to the negative input:

\$\begin{align}V_o&=(V_{+}-V_{-})A\\&=(V_i - V_o)A\\\Rightarrow V_o &= \frac{V_iA}{1+A}\\&\approx V_i\end{align}\$

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  • \$\begingroup\$ First of all, thanks for your answer. Can this be shown fully, using mathematically? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 15:30
  • \$\begingroup\$ I don't think you can ignore the diode drops. The whole point of the negative feedback is to correct for them rather than ignore them. \$\endgroup\$ – Transistor Mar 27 '20 at 15:34
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Assuming that the current flowing through R2 equals the current flowing through R1.

$$V_{B,T1}=(V_{NI}-V_I)A$$

$$V_{B,T3}=V_{B,T1}-V_{diode,T1}$$

$$V_{E,T3}=V_{B,T3}-V_{diode,T3}=V_{I}$$

Rearranging the equations yields:

$$V_I = \frac{V_{NI}}{1+(\frac{V_{diode,T1}+V_{diode,T3}}{A})}$$

For a considerable large gain

$$V_I = \frac{V_{NI}}{1+(\frac{V_{diode,T1}+V_{diode,T3}}{\infty})}= V_{NI}$$

Then the current through R2 can be calculated:

$$I_{R2}=\frac{V_{I}}{R_2}$$

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  • \$\begingroup\$ Why is your first assumption right? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 17:55
  • \$\begingroup\$ The emitter current of a BJT equals approx. the base current + the collector current. The base current is much smaller compared to the collector current. $$I_e=I_b + I_c = \frac{I_c}{\beta} + I_c \approx I_c$$ \$\endgroup\$ – vtolentino Mar 27 '20 at 18:40
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Let's make the usual assumptions

  • Op-amp inputs take no current.

  • The output of an Op-amp will do what it can to try and make the inputs equal.

This should give you the the current in \$ R_2 \$.

Now the current in \$ R_1 \$ is this minus the current in the base of \$ T_3 \$ and the gain \$ h_{fe} \$ is large.


Note: \$ h_{fe} = \beta = \dfrac{I_C}{I_B} \$ is typically ≥ 100 so to a first approximation \$ I_C = I_B \$. Now if we know the feedback of the op-amp is such to keep its inputs equal we know the voltage across \$ R_2 \$ and hence the current in it.

This of course assumes \$ T_3 \$ does not saturate.

Does this help?

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  • \$\begingroup\$ I do not see where you're heading, at this moment? Can you help me a bit further? \$\endgroup\$ – Polik Pityl Mar 27 '20 at 17:56
  • \$\begingroup\$ Hopefully I have clarified things a bit. If you are still confused what are you missing. \$\endgroup\$ – Warren Hill Mar 28 '20 at 10:29
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First consider this simpler circuit:-

schematic

simulate this circuit – Schematic created using CircuitLab

If the op amp's inverting (-) input is lower than the non-inverting (+) input then its output voltage will go up, if the inverting input is higher the output voltage will go down, and when they are exactly equal the output voltage on R2 will equal Vi. This is the classic op amp 'buffer' configuration with 100% negative feedback and a gain of 1.

Now add two diodes representing the Base-Emitter junctions of T1 and T3:-

schematic

simulate this circuit

The same thing happens, only now the op amp must raise its output by 2 diode voltage drops (~1.2 V) to make the inverting input equal the non-inverting input.

In the full circuit, if the values of R3 and R1 are not high enough to cause T1 and T3 to saturate then most of the current through R2 will come from them rather than from the op amp output, but the voltage across R2 will always be the same, ie. equal to Vi.

So the current through R1 is not the same as the current through R2. However if the current gain of T3 is high then it will be close, since most of it will come from the Collector and very little from the Base.

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