# Differential Amplifier Biasing Problems

This is the circuit that I built: (desired voltage levels are described next to the circuit) simulate this circuit – Schematic created using CircuitLab

I tried to make proper bias conditions for this circuit, but failed (I tried many different similar examples before). The problem is that there is too high Vbe on Q2 and Q3, but the Vbe of Q1 is actually negative; Q2 and Q3 are saturated; Q1 and Q2 are unbalanced, of course.

You may be wondering why Rc* and Rc** are not the same - I tried to make a voltage divider for Q3 (to properly set Q-point) with Rc*, so I would be making another divider for Q3 (one resistor less).

R4 and R5 were meant for a negative feedback for later on AC experimentation, but already failed at biasing point.

• Something is very wrong in this circuit but I cannot find what would that be. Can anyone spot and describe mistakes I made?
• You a lucky man. You just have discovered the positive feedback. And increases in Q3 emitter voltage will "open" Q2 more and at the same time, Q1 will conduct less current. Hence the Rc* voltage drop is smaller ( Q1 collector voltage will rice ). Or we can say that the "more part" of a Rc* current is now flowing into the Q3 base (Q1 conduct less current, Q1 steals lees current from Rc*). And more current into Q3 base means Q3 will conduct even more current. And the voltage at Q3 emitter will rise even further. Hence the Positive Feedback. Try to think about it little more. – G36 Dec 7 '17 at 18:30
• @G36 Hmmm... I only heard about negative feedback, and now I did something reverse to that. Thanks for noticing that mistake – Keno Dec 7 '17 at 18:59
• @G36 On the other hand, I was kind of copying your design from my last question you answered to. (Here: electronics.stackexchange.com/questions/343271/… ) – Keno Dec 7 '17 at 19:02
• But in this circuit (electronics.stackexchange.com/questions/343271/…), we have a negative feedback. Notice that any rise in Vout voltage will cause Q2 to conduct more current. So Q2 will "steal more" current from Q3 base. So Q3 will conduct less current and Vout will drop. Negative feedback tends to promote a settling to equilibrium and reduces the effects of disturbance. Do you see the difference? – G36 Dec 7 '17 at 19:12

I like to think of the long-tailed pair as more like a teeter-totter (to give it a kind of "dynamics.") To give some clarity, imagine that you built this teeter-totter out of two lengths of gutter, joined together with a kink so that when balanced equally where the joint itself rests on a fulcrum, each side has a slight but equal downward slope. If you start a water flow from a raised garden hose or water source with the opening right above this joint, and do it perfectly, then there will be an equal amount of water flowing away down the left and right side gutters.

Now, imagine you take your hand and tilt this teeter-totter slightly. More of the water will be directed down one side than the other. And if you tilt it enough, you might cause one of the gutters to become completely horizontal (or more) and so all the water goes just in one direction, now.

That's the long-tailed pair shown here: simulate this circuit – Schematic created using CircuitLab

The current source is that garden hose. It sets the flow of water. This could be a resistor, too (if there is some attempt made to stabilize the voltage across it, so that the current through it is usefully stable.) The collectors are the left and right gutters where the flow goes. If $V_L=V_R$ then the teeter-totter is balanced and the current in $I_1$ is divided exactly in half, so that both collector currents are equal to each other and each of them will be one-half of $I_1$. But just the slightest change in the relative voltage values of $V_L$ and $V_R$ will mean that the current "diverts" somewhat and the two collector currents will no longer be equal to each other.

It only takes a very slight difference in voltage between the two inputs to make quite a difference between the flows. If you arrange things so that $\mid\:V_L-V_R\:\mid=60\:\textrm{mV}$, then there will be about a 10:1 difference in the collector currents. Looking at the above schematic, if $V_L=V_R+60\:\textrm{mV}$ then $Q_2$'s collector current will be about 10 times the collector current of $Q_1$.

Note: You usually don't drive a long-tailed pair like this by even that much, though, because it's far too easy to wind up saturating one of the BJTs and to start drawing significant base current. (You can test this easily enough with a BJT type opamp, where the "rule" that the inputs draw negligible current quickly becomes violated if you don't allow the opamp to keep the two input voltages very near each other's voltage value.)

You can insert resistors in either (or both) collector legs if you want to turn these directed currents into resulting voltages. (So long as you have the compliance voltages required, of course.) But it isn't necessary.

That's mostly it. Now you have the visualization needed to see what went wrong in your schematic. I've disconnected a few things in your schematic and highlighted the long-tailed pair by surrounding it with a box: simulate this circuit

(I'm not even sure what $R_3$ is supposed to do, so I'm not inclined to even re-connect it later.)

At this point you can see that the left side has $Q_1$'s base connected to a voltage divider. The voltage there is easy to compute as being $10\:\textrm{V}$. So the only remaining question is "What happens as $Q_2$'s base moves slightly above or slightly below $10\:\textrm{V}$?" If above, then $Q_2$ will pull more current towards its collector, leaving less for $Q_1$'s collector. So the voltage drop across $Q_2$'s collector resistor increases (lowering its collector voltage) and the voltage drop across $Q_1$'s collector resistor decreases (raising its collector voltage.)

So, looking at the above schematic, mentally imagine connecting the base of $Q_3$ back up to the collector of $Q_1$. Doing so means that the emitter of $Q_3$ will "follow" the voltage at the collector of $Q_1$. You apply a divider here and feedback this divided voltage to the base of $Q_2$. (Keep in mind that you have a voltage divider at the base of $Q_1$ already that provides a "reference voltage" that [hopefully] doesn't change while we consider the rest here.) If for some reason the base of $Q_2$ rose upward so that it is just slightly above $10\:\textrm{V}$, then this would increase $Q_2$'s collector current and decrease $Q_1$'s collector current. A decrease in $Q_1$'s collector current means an increase in $Q_1$'s collector voltage. An increase in $Q_1$'s collector voltage, raises $Q_3$'s base, also raising its emitter. But in increase in $Q_3$'s emitter voltage means an increase in the divided voltage at the base of $Q_2$. Since this whole process started out with a hypothetical, but very slight increase in the voltage at the base of $Q_2$, and since we've worked out that the circuit then follows this "event" up by responding with an increase in the voltage at the base of $Q_2$, I think you can easily realize now that what you've done is to create "positive feedback" for "voltage disturbances." The result is that the circuit rapidly moves to a quiescent point where it runs out of voltage headroom (and the negative feedback from that new effect [and unanalyzed at this point] stops the process.)

The obvious thing to do here would be to connect the base of $Q_3$ to the collector of $Q_2$, instead. Since the collector voltage at $Q_2$ declines as more current is directed to its collector (when its base voltage increases due to a disturbance), this will cause the voltage at the emitter of $Q_3$ to decrease as well. So this will respond by countering (lowering) the disturbance. Without any calculations, you can see that there is negative feedback to counter disturbances. By itself, that's not enough to calculate where things will go. But qualitatively, at least, the idea is sound.

Now, let's reconnect things back up and get rid of $R_3$: simulate this circuit

A question here might be: "What's the stable operating point of this circuit?" The approach isn't hard, but the closed solution requires the LambertW function (which I'll avoid.)

Here's the approach. Starting at the collector of $Q_2$, we can say at the voltage there is $20\:\textrm{V}-I_{C_2}\cdot R_{C_2}$. We now subtract from this $V_{BE_3}\approx 700\:\textrm{mV}$ to get to the emitter of $Q_3$. Then apply the voltage divider there to get the base voltage of $Q_2$ as:

$$V_{B_2}=\left(20\:\textrm{V}-I_{C_2}\cdot R_{C_2}-V_{BE_3}\right)\cdot\frac{R_5}{R_4+R_5}$$

We also know that the ratio of the two currents, $I_{C_1}$ and $I_{C_2}$, is: $\frac{I_{C_2}}{I_{C_1}}\approx e^\frac{V_{B_2}-10\:\textrm{V}}{V_T}$. If we now figure that the current sink represented by $R_E$ is $I_{R_E}=\frac{10\:\textrm{V}-V_{BE_1}}{R_E}$ then we can instead write:

$$\frac{I_{C_2}}{I_{R_E}-I_{C_2}}\approx e^\frac{V_{B_2}-10\:\textrm{V}}{V_T}$$

And solve for $I_{C_2}$:

$$I_{C_2}=\frac{I_{R_E}}{1+e^\frac{10\:\textrm{V}-V_{B_2}}{V_T}}$$

We can now substitute, to get:

$$V_{B_2}=\left(20\:\textrm{V}-\frac{I_{R_E}\cdot R_{C_2}}{1+e^\frac{10\:\textrm{V}-V_{B_2}}{V_T}}-V_{BE_3}\right)\cdot\frac{R_5}{R_4+R_5}$$

With $V_{B_2}$ on both sides of the equation and involving an exponent, the LambertW function is required for an closed solution. Also, if you use iteration with the above form, it's likely to diverge and not converge. So it needs to be solved for the other $V_{B_2}$:

$$V_{B_2}=10\:\textrm{V}-V_T\cdot\operatorname{ln}\left(\frac{I_{R_E} R_{C_2}-\left[20\:\textrm{V}-700\:\textrm{mV}-V_{B_2}\left(1+\frac{R_4}{R_5}\right)\right]}{20\:\textrm{V}-700\:\textrm{mV}-V_{B_2}\left(1+\frac{R_4}{R_5}\right)}\right)$$

But you can plug in some assumed values and see where it gets you after a couple of iterations. Start with $V_{B_2}=10\:\textrm{V}$ and an estimate for $I_{R_E}$ and you will get $V_{B_2}\approx 10.028431\:\textrm{V}$. Plug that in and get $V_{B_2}\approx 10.028042\:\textrm{V}$. Another time and you get $V_{B_2}\approx 10.028047\:\textrm{V}$. You can see that it settles down pretty quickly.

Note: Of course, I've taken the standard value for $V_{BE}$. The actual value depends on the collector currents for each BJT and will be somewhat different. Also, the two BJTs used in the long-tailed pair are assumed to be identical devices. If you buy discrete BJTs (not in matched pairs), then a real circuit will settle somewhere else, as well. Resistor values vary. Etc. Just keep in mind this has been a theoretical discussion making broad assumptions. But it shows the general approach one might start out with, to get an idea about how to approach a rough prediction.

And now you know that there is about a $28\:\textrm{mV}$ difference in the base voltages when your modified circuit (no $R_3$ and a different collector connection) sits quiescently.

• I think I will have to read and rethink you answer a couple of times to fully understand this thing. – Keno Dec 8 '17 at 16:22
• On your modified (last) circuit; shouldn't there always be the output of diff-pair from the collector of Q1 like I did, an not from Q2? Like here oi57.tinypic.com/mcxdw6.jpg – Keno Dec 8 '17 at 16:37
• @Keno That's your misunderstanding of that circuit, not actual fact, and also misunderstanding of the long-tailed pair (I think) that are both conspiring towards your confusion here. There is only one output from that long-tailed pair -- the one that moves towards $C_9$ in that circuit -- and the choice of which of the two to use there depends on the "desired meaning" of the inputs and the rest of the circuit. The long-tailed pair, itself, is purely symmetrical. There is no "preferred" side. Somehow you imagine human visual left of a schematic is magically known by a circuit? – jonk Dec 8 '17 at 18:10
• No. But is there a huge difference, whether the output of diff-pair is taken from Q1 or Q2? – Keno Dec 8 '17 at 18:31
• @Keno It's symmetrical. So, if you "define" one BJT as taking the (-) input and the other BJT as taking the (+) input, then you have made a choice about which leg is which leg. You could choose to swap that arrangement mentally. The circuit actually doesn't care, since it is symmetrical. But flipping the meaning of the inputs also flips the meaning of the collector legs. You make an input assignment choice and the output collector leg choice follows inescapably. But you are free to make that input assignment choice. – jonk Dec 8 '17 at 20:39 There are a number of things wrong here, but the most fundamental is that you have positive feedback. That's why the output is slamming to one rail.

You didn't say, but I'll take the base of Q1 to be the input to this circuit, and the output the emitter of Q3. The signal is inverted from B of Q1 to C of Q1. Q3 provides more drive current, but doesn't invert. The output is therefore inverted from the input. This means the base of Q1 is the negative input, which makes the base of Q2 the positive input.

You then feed back part of the output to the positive input. DC positive feedback causes latching behavior, which is exactly what you see.

## Added in response to comment

Yes, the positive and negative inputs to the differential front end of this amplifier are the bases of Q1 and Q2. Each effects the final single-ended output with opposite polarity.

There can be multiple inversions between the diff amp input and the final output. That only effects which one of the two inputs is considered the positive and which the negative. If done right, they are essentially symmetrical, so you simply flip their usages to accommodate a net inversion. Note that changing which collector the single-ended signal is take from also flips the polarity.

• But isn't it normal for differential amplifier (in such configuration) to have such inputs at the bases of Q1 and Q2? I mean, if negative feedback is desired, then the output of amplifier should be connected to the base of other transistor in diff-pair and there should also be a 180 degree shift between input and output signal for proper operation of amplifier – Keno Dec 7 '17 at 19:10

Try this circuit (with negative feedback instead of a positive feedback). simulate this circuit – Schematic created using CircuitLab

This is all you need to properly DC biased this simple circuit.

And for DC experiments RF2 and C1 are not needed.

Additional notice how easily we can set the Q-point for Q1 and Q4.

Because of the fact that the Q1 base volate is around 10V the Iee current is:

$$Iee = \frac{10V - Vbe}{Re1} \approx 1.9mA$$

And Q1 collector current is

$$I_{C1}\approx \frac{Vbe4}{R_{C1}} \approx 1mA$$

And because we have the negative feedback via RF1 resistor the voltage at Q4 collector will also be around 10V. Hence Q4 collector current is equal to:

$$I_{C4} = \frac{10V}{Rc2}\approx 10mA$$

I hope you see how easy it is.

To help you recognize which type of a feedback we are dealing with this diagram might help. 1 - Any rise in the voltage at point X will cause that the voltage at point Z also will rise.So if we connect together these two points (X and Z) we will have a Positive feedback on the circuit.

2 - This time any rise in voltage at point A corresponds to the voltage drop at point D. So, if we connect this two points together (A with D) we will end with a Negative feedback circuit.

• Since you used PNP instead of NPN for Q4, you now have 360 degree phase shift, whereas negative feedback shouldn't operate properly (there should be only 180 degree phase shift). – Keno Dec 7 '17 at 19:16
• From where to where do you see the 360-degree phase shift? Assume for now that the voltage at the Q4 collector is rising. So, Q2 will conduct more current and Q1 will conduct less current (do you know why?). Less current in Q1 collectors means less current for Q4 base. Hence Q4 collector voltage will drop. So I see a negative feedback. Do this kind of "mental" analysis by yourself. – G36 Dec 7 '17 at 19:24
• First, base of Q4 is connected to collector of Q1; therefore 180 degree shift. Second, base of Q2 is connected to the collector of Q4 via resistor; therefore another 180 degree shift - all together 360 degree shift. Did I miss something? – Keno Dec 8 '17 at 13:20
• I really looks like you made a circuit with positive feedback, not me. – Keno Dec 9 '17 at 20:42
• @Keno Why do you think so? – G36 Dec 10 '17 at 9:53