# Classic transistor differential ampliﬁer - problem with a single-ended output

I have a differential amplifier like at the pic below: And I am not sure if I am doing everything right.

We know that the input signal is a triangle wave with $V_{pp}>100\text{mV}$, no offset and $f=1\text{kHz}$ and our output signal is a square wave that changes from $0$ to $5\text{V}$. And my task is to calculate:

-$V_{CC}$

-$R_C$

-$G_{diff}$

The first thing is to calculate $I_0$ - on both $10k\Omega$ resistors there is $7.5V$ voltage, thus $7.5V-0.7V-I_0\cdot 3.4k\Omega = 0\rightarrow \ I_0=2\text{mA}$

Next step: we know that the maximum output voltage is equal to 5V, so $V_{CC}=5\text{V}$

Now let's say, that we have such input voltage, that T1 is in cutoff mode, then $I_{C}=I_0=2\text{mA}$

This means, that $V_{CC}-I_CR_C$ is equal to the minimum output voltage, i.e. 0V. Thus: $R_C=\frac{V_{CC}}{I_0}=2.5k\Omega$

As for the gain: I think it is $G_{diff}=\frac{R_C\cdot I_0}{2\varphi_T}$

Is my solution correct? Thank you in advance.

OK for Vcc and Ic

As for the gain, each transistor behaves like it has an internal emitter resistor of value:

$r_e = \frac{1}{g_m} = \frac{V_T}{I_C}$ with Vt about 25mV.

So, with zero input voltage, Ic=1mA per transistor (2mA total), and their re are both 25 ohms.

To calculate the gain we need $\frac{dIc}{dVin}$ so we approximate both transistors as perfect transistors with emitter resistor re added. Since each transistor has its own re, they appear in series, so

$\frac{dIc}{dVin} = \frac{1}{2 r_e} = \frac{I_C}{2V_T}$

Multiply by Rc to get the gain...

$\frac{I_C Rc}{2V_T}$

I think you lost a "2" somewhere along the way, as Ic=Io/2 my result doesn't match yours...

Note that the gain depends on transistor gm which depends on collector current for each transistor. If the collector currents are not balanced then the transconductance $g_m$ of one transistor will increase, and the other will decrease. Same for their $r_e$ since $r_e = 1/g_m$

At low input voltages these variations compensate each other, and linearity is good. However when input voltage is high enough to produce a noticeable variation of current in the transistors, or even further, when one transistor gets close to turning off, then the gain will be lower. With one transistor off, the gain is zero. So keep in mind this circuit is only linear when input voltage is quite small. If you want to use it as a comparator, this is not an issue.

• Thanks, however there is 1 thing I don't see: as Ic=Io/2 - there is no Rc on the left side of the circuit, so why the current splits symmetrically? – SantaXL Jun 20 '18 at 11:12
• Rc does not set Ic, it does have a small effect due to Early effect on T2, but as a first approximation this can be neglected. When Vin=0, then both transistors' bases are at the same voltage, their Vbe are equal, and their collector currents are equal at Io/2. – peufeu Jun 20 '18 at 12:17

I think @peufeu may be giving this more thought than this problem needs, based on the info you gave us. This form of diff amp can be considered high enough gain that, without feedback, it's essentially a comparator. With 100mV in, except for switching at the zero crossings, your 2ma Io will be going either through T1 or T2. I agree with your answer.

As far as the symmetry goes, think of it as above--without a collector resistor, T1 still passes whatever portion of the 2mA isn't being passed by T2. A collector resistor would reduce the power dissipation, but since we're not using it as an output, there's no necessity for it.