How do I choose a transistor for a VBE multiplier?

Question

I am currently designing a 150 W audio power amplifier for 12 ohms, so 60 V peak to peak and 5 A. I'm using a C5198/A1941 pair for AB amplification with D669/B649 in a Darlington configuration so I have on the top D669 with C5198 and at the bottom A1941 with B649.

I would like to bias all that using a VBE mutliplier. I think that the transistor that should be used is an NPN the same type as C5198, but I don't know if I need to use only C5198 for the VBE multiplier on the same heatsink as the Darlingtons or if I need to use a C5198/D669 pair to match. I also don't know how to choose R1/R2 as I don't know the impact of collector current of the yellow transistor.

I circled the one in yellow. Also if you have any improvements with explanations (I am an engineering student) it would be awesome.

PS: With my calculations I have Beta = 22 000 and a bias voltage of 231 µA for the upper Darlington.

Answer 1

Demystifying the "collector resistor"

Introduction

The so-called "Vbe multiplier" (aka "rubber diode") has always piqued my interest because it is one of the most basic transistor circuits with parallel (voltage-type) negative feedback. Its idea is: part of the output collector-emitter voltage is applied in parallel to the base-emitter input of the transistor. Figuratively speaking, the threshold base-emitter voltage Vbe (~0.7 V) of the transistor acts as an "input" voltage that is subtracted from the feedback voltage, and the difference is amplified by the transistor. The result of this is that the transistor "raises" its output voltage inversely proportional to the divider gain, and the input base-emitter voltage appears amplified ("multiplied") at the output.

The simpler version of this circuit is the legendary "active diode" (aka "transistor diode") where the entire output voltage is applied to the input (ie, there is no divider and the collector is connected directly to the base). As a result, the output voltage follows Vbe, and this "circuit" can be considered as a parallel equivalent of the emitter follower. Another famous circuit - current mirror, was built on it.

My amazement

As I was looking into the question and wondering why there was still no answer, a sentence in the @perilepsis's comment caught my attention:

At a minimum you will want to include a collector resistor in your VBE multiplier.

I had not seen this trick before, but it looked awfully familiar to me. Some distant notions about the transistor amplifier stage with an emitter resistor like "Rc-disturbed emitter follower" began to surface in my memory. I also remembered one of my explanations of the op-amp's role as a "negative resistor"... and the puzzle came together.

Figuring out Early effect

To understand how a simple resistor eliminates the Early effect, we must first have an intuitive understanding of this phenomenon. For the purposes of this curcuit, it is sufficient to explain it functionally without going into details. We can do it as follows.

"Ideal" current source

A transistor behaves as a constant current source; so we can explain Early effect by an equivalent but simpler and more understandable device - current source. Then let's measure and plot the IV curve (or a family of IV curves) first of an "ideal" current source. For this purpose, we connect a variable voltage source in parallel (run the DC Sweep Simulation).

^{simulate this circuit – Schematic created using CircuitLab}

The IV curves are absolutely horizontal.

Real current source

Then, if we connect a resistor in parallel to an "ideal" voltage source, it will divert a part of the current through itself, and the current source become imperfect, real, bad.

^{simulate this circuit}

The IV curves have a slope determined by the resistance.

Transistor

Now we can replace the humble current source with the more sophisticated transistor current source. We only need to add an input constant voltage source Vref.

^{simulate this circuit}

We get the same slant IV curves as above but we must be aware that there is no resistor or resistance here; the slope is due to phenomena in the semiconductor structure. Using the resistor above, we only modeled the Early effect. Similarly, we model the internal resistance of a real voltage source (eg battery) through a resistor, but there is no real resistance inside the voltage source either. The slope of the IV curve is (mainly) a result of chemical phenomena inside the source.

Vbe multiplier...

As we can see from the above graph, the transistor is (behaves as) a current source. How exactly it does this magic is its business. If we want to understand it, we can imagine it as a dynamic resistor that changes its resistance R in the same direction and to the same extent as the voltage V changes. As a result, the ratio V/R = I is constant.

... undisturbed...

But with the help of the negative feedback phenomenon we can reverse its behavior by making it change the current so that it keeps the voltage constant... to make it a voltage source. To do this, we just need to connect its collector to the base.

^{simulate this circuit}

We see that the transistor maintains its collector-emitter output voltage equal to its threshold base-emitter voltage (about 660 mV).

... R1-R2 disturbed...

To make it amplify (multiply) the "input voltage", we apply a well-known trick from our life - disturb it with the help of the R1-R2 voltage divider and take the voltage before this disturbance (from the collector). The transistor reacts to the disturbance by increasing its voltage (R1 + R2)/R2 times.

^{simulate this circuit}

However, we can see that the IV curve is not exactly horizontal. The reasons can be both the Early effect and the insufficiently high transistor gain. See for example how it depends on β:

β = 140

β = 1000: The curve is more horizontal.

... R1-R2 + Rc double disturbed...

So we decide to apply an even more clever trick - disturb the transistor once more by inserting an Rc resistor into the collector and take the voltage before this disturbance (from the collector). Thus the transistor is "double disturbed" and must "strain" (decrease its collector voltage) more than necessary to achieve the same result.

^{simulate this circuit}

As you can see from the IV curve family below, when for example, the "disturbing" resistor Rc increases its resistance as a parameter (0, 200, 400 and 600 Ω), the transistor always achieves its goal; so the four IV curves of the top voltage (above Rc) overlap each other.

This is because the transistor is forced to decrease its collector voltage more and more; the IV curve of its collector voltage slopes more and more to the right (clockwise), and finally becomes negatively sloped.

... "ideal" Vbe multiplier

At Rc = 200 Ω...

^{simulate this circuit}

... the curve (in brown) becomes completely horizontal, and this is our case; we use this voltage as a constant bias voltage. Besides the two curves (above and below Rc), I have also shown their difference (the voltage drop across Rc) at the bottom of the picture (in yellow brown).

Negative resistance viewpoint

The phenomenon of negative resistance has always been of interest because there is some mystery surrounding it. But it disappears once the idea behind it is figured out because the idea is very simple. So let's clarify what is positive and what negative here, and what this "negative" means.

Imagine a conceptual circuit where a resistor Rc dissipates power and a voltage drop I.Rc is lost across it. Conversely, a negative resistor Rce "produces" power (it is a source) and add a voltage I.Rce to the circuit. So if we connect them in series, the total voltage across them would be 0 V.

This would be true if the transistor was a true negative resistor (source). But it is a differential negative resistor that can only regulate the power given by an external source. Let's explore how this happens by replacing the transistor by a variable resistor Rce.

Iin = 1 mA

In fact, the circuit is a voltage divider supplied by a 1 mA constant current source Iin.

^{simulate this circuit}

But this is a more special divider - dynamic voltage divider.

Iin = 2 mA

If, for example, we double the current, the voltage drop VRc will increase by 200 mV. At the same time, Rce will reduce its resistance by such that the voltage drop across it will decrease by 200 mV. As a result, the total voltage Vtop across the network will not change. It is as if the variable positive resistor Rce is a constant -200 Ω negative resistor that neutralizes the constant 200 Ω positive resistor Rc.

^{simulate this circuit}

Iin = 0.5 mA

Conversely, if we decrease twice the current, the voltage drop VRc will decrease by 100 mV. Rce will increase its resistance by such that the voltage drop across it will increase by 100 mV so the total voltage Vtop across the network will not change. As above, the variable positive resistor Rce acts as a constant -200 Ω negative resistor that neutralizes the constant 200 Ω positive resistor Rc.

^{simulate this circuit}

Conclusions

• A constant negative differential resistor is a variable positive resistor.

• The transistor Q in the circuit of a Vbe multiplier with collector resistor Rc can be considered as a negative differential resistor Rce that neutralizes the equivalent positive resistor Rc.

Answer 2

With \$\beta=20,000\$, and output transistor collector currents up to 5A, you are expecting base current to never exceed \$\frac{5A}{20000} = 250\mu A\$, which must be sunk/sourced by your yellow transistor.

I don't think there's a transistor in existence that can't handle that tiny current, or even 100 times that. Given that yellow's \$V_{CE}\$ will be only 4V or so, even passing tens of milliamps it would only dissipate tens of milliwatts, so you could use pretty much any signal transistor in this role.

It's hard to explain with no component labels, so I'll draw a simplified equivalent here:

^{simulate this circuit – Schematic created using CircuitLab}

This is just the upper half of the push-pull pair, with a load that draws about 5A at maximum voltage.

I think your 4.7Ω resistors are not useful. Did you misplace resistor R1? R1's role here is to provide a discharge path for Q3's stored base charge, so it can switch off quicker.

As output current (shown by ammeter AM4) rises, Q2's base current (ammeter AM1) does too, always a factor of about 20000 lower. To be safe, let's assume base current at maximum load won't exceed 300μA.

All of that current must be sourced from whatever is driving the base of Q2. If you are not driving IN directly from some voltage source, as is your case when using the V_BE multiplier, that base current can only come via the top-left resistor in your schematic. that's R3 here:

^{simulate this circuit}

As you can see, base current is causing R3 to develop a significant voltage, 2.7V shown on VM2. This limits the maximum possible base potential to be significantly lower than the 35V supply, and consequently maximum output voltage is reduced by that amount too. Therefore, before you begin considering the resistances for your V_BE multiplier, you must first establish a reasonable value for R3.

You must consider that the voltage across R3 can rise to 70V, as input (Q2 base) potential is pulled down towards -35V. Even though Q2's base current will never exceed 300μA or so, R3 current will be significantly more than this.

On top of that, you will require significantly more than 300μA through R3 (with Q2 base potential at maximum), because you must have current "left over" to bias your V_BE multiplier.

With this design you are forced to have huge variation in current through R3, as the input swings between +35V and -35V, and this will affect the V_BE multiplier's own voltage drop. To keep the voltage across the V_BE multiplier reasonably constant, at about 4V, you want to keep that current swing to a minimum.

Your choice of R3 is at least as important as your choice of resistances for the V_BE multiplier, so let's examine how you might choose R3. I'd start with a guess as to what a reasonable bias current for the V_BE multiplier would be, add that to base current, and see what value for R3 emerges.

Since you require 30V maximum at the output, you require \$30V + 2V_{BE} \approx 32V\$ at Q2's base, meaning that R3 cannot ever be permitted to drop more than \$35V - 32V = 3V\$ when the input is near maximum.

Q2 base current and V_BE multiplier current is all sourced via R3 (from the +35V supply rail), and must not cause R3 to develop more than 3V. Assuming that the V_BE multiplier requires 1mA, then total current through R3 will be \$1mA + 300\mu A = 1.3mA\$, and R3 must be at most:

$$ R_3 = \frac{3V}{1.3mA} \approx 2.2k\Omega $$

When the input is -35V, the potential at Q2's base is pulled down to around -31V (accounting for the V_BE multiplier's voltage of 4V or so), and the voltage across R3 rises to \$(+35V) - (-31V) = 66V\$.

With 66V across R3, current in R3 will be:

$$ I_{R3} = \frac{66V}{2.2k\Omega} = 30mA $$

Power in that resistor, with 66V across it, would be:

$$ P_{R3} = \frac{V^2}{R_3} = \frac {(66V)^2}{2.2k\Omega} = 2W $$

R3 will have to be quite beefy, but it only spends very little time during an input cycle (when the input is near -35V) dissipating 2W, so probably a smaller device would be fine.

This means that the current drawn through the V_BE multiplier will vary between 1mA and 30mA! Let's hope that the V_BE multiplier voltage will remain close to 4V throughout this large variation in current. It won't, but there is hope:

^{simulate this circuit}

Here I've calculated R4 and R5 as follows. First, when current is at a minimum of 1mA, I want Q1 to be passing at least 20% of that (as a rough starting figure), and as current increases the transistor passes an ever greater fraction. I also want 4V across this V_BE multiplier, enough to bias all output transistors slightly "on".

R4 and R5 form a potential divider, and their junction will have potential \$V_{BE} = 0.65V\$:

$$ 4V \times \frac{R_5}{R_4+R_5} = 0.65V $$

At a guess, current through R4 and R5 should be roughly 80% of the minimum bias current of 1mA (that's 800μA), leaving 20% (200μA) for the transistor:

$$ R_4 + R_5 = \frac{4V}{0.8 \times 1mA} = 5k\Omega $$

Solving those two simultaneous equations gives me:

$$ R_4 \approx 810\Omega $$ $$ R_5 \approx 4200\Omega $$

This is what happens when I sweep current (x on the horizontal axis) between 0 and 30mA. The blue trace below is OUT1 from the circuit above left, and orange is OUT2 on the right:

The classic V_BE multiplier doesn't work very well, but behaviour can be improved by adding R6. The value of R6 is calculated as the slope of the blue trace in the straight region, inside the green markers. This slope is the ratio change in voltage divided by change in current:

$$ \frac{\Delta V}{\Delta I} = \frac{880mV}{21mA} = 42\Omega $$

That's the dynamic resistance of the V_BE multiplier, causing it to add a certain voltage per ampere of current, rather than staying fixed. By inserting a resistor R6 of this same value, we effectively subtract the same voltage per ampere of current, that the entire V_BE multiplier adds, which flattens out the curve.

You should try your circuit with all these resistor values in a simulator, under various load and input conditions. You'll no doubt need to tweak things a bit, but at least you have an idea of how I would approach the problem.

I don't think the design is a good one; you could achieve much better results using a constant current source instead of R3. Take a look at my answer about the diamond buffer to see a current source used in this very scenario.