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take a look at this circuit:

enter image description here

I have a problem with calculating \$I_{out}\$, or, more likely, understanding how this circuit works. I know that \$I_{out}=I_{C4}-I_{C2}=I_{C1}-I_{C2}\$, but what is the relation between \$I_{C1}\$ and \$I_{C2}\$? In other words: how does the presence of \$V_S\$ and \$R_C\$ influence \$\frac{I_{C1}}{I_{C2}}\$?

We have given: \$V_{CC}=-V_{EE}=15V, \ R_C=1k\Omega, \ V_S=5V, \ I_0=2mA\$

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    \$\begingroup\$ The diff-amp wants to divert current back and forth between its collectors, based upon the input signal differences. The current mirror wants to force both collector currents equal to each other and resists attempts otherwise. The only solution to this irresistible force meeting an immovable object is to provide a wire (\$I_\text{OUT}\$) so that both can be happy. The diff-amp gets to have different collector currents when it wants it, the current mirror gets to have the same collector currents, and the wire squirts out the differences. Pretty simple, once you think about it. \$\endgroup\$ – jonk Jun 27 '18 at 0:12
  • \$\begingroup\$ @jonk but how do I calculate the difference in the given circuit \$\endgroup\$ – SantaXL Jun 27 '18 at 0:22
  • \$\begingroup\$ If balanced inputs are at the bases, then the currents are equal and the difference is zero. Otherwise, it's pretty easy to work out from the simplified equation relating active mode collector current to base-emitter voltage. Have you attempted to use that equation, yet? \$\endgroup\$ – jonk Jun 27 '18 at 0:36
  • \$\begingroup\$ I'll need some kind of response about the equations. Have you attempted anything with them, yet? You just set up three simultaneous equations (very easy to write them down) and solve them for the three unknowns. The answer is easy to remember and makes a lot of sense, too. I could just write out the equation for you, it's easy to do. But I want to allow you a chance to figure it out on your own. It's not at all complicated. \$\endgroup\$ – jonk Jun 27 '18 at 1:34
  • \$\begingroup\$ Oh. Hmm. I just noticed something I wish I'd noticed earlier. Are your bases grounded in the schematic? Both of them? I may have been talking at cross-purposes with you because I didn't read you well enough. So are the bases grounded as it now seems to me they are? \$\endgroup\$ – jonk Jun 27 '18 at 1:46
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Without more information to go on I would probably use the following approximation:

Looking at the voltages given, all BJT's are in the forward active region. This means that the collector will source/sink a more or less constant current, equal to \$\frac{I_0}{2} = 1mA\$. These current sources will shield the output voltage and current from everything happening elsewhere, so it does not influence \$\frac{I_{C1}}{I_{C2}}\$ in any meaningful way - as long as all transistors are in the active region.

The other way around does happen though. If a differential input voltage is applied, then \$I_{C1}\neq I_{C2}\$, and a non-zero current would flow through the resistor as you have stated (\$I_{out} \approx I_{C1} - I_{C2}\$). The difference between these currents will be proportional to Q1's and Q2's transconductance gain \$g_m\$, or

$$I_{C1} - I_{C2} \approx g_m\cdot (v_{B1} - v_{B2})$$

The output current will still be \$\approx I_{C1} - I_{C2}\$, and this is not influenced much by \$V_S\$ and \$R_C\$ because the output transistors still act like current sources.

The output voltage does change drastically though. If both base voltages are kept constant, then \$I_{out} \approx 0\$, and so \$V_{out} \approx V_s = 5V\$. Applying a differential voltage would mean that

$$\begin{align} V_{out} &\approx g_m\cdot (v_{B1} - v_{B2})\cdot R_C + V_S \\ &= g_m\cdot (v_{B1} - v_{B2})\cdot 1k\Omega + 5V \end{align}$$

If you would not connect \$R_C\$, then all current would be forced through the output impedance of the output transistors. These output impedances are generally very high for BJT's in the active region, so they can usually be neglected for \$R_C \ll Z_{out}\$. If this is not the case, you can replace \$R_C\$ with \$R_C || Z_{out}\$ to get a better approximation for the small-signal gain.

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