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fglabach 17th July 2012 09:54 PM

Optimizing the VBE Multiplier
1. Bob Coredell says that you can determine the current compenstaion resistor of the VBE multipler by multiplying the intrinsic base-collector resistance of 2.9 ohms by the multiplication factor, which gives you the dynamic impedance. You then pick a resistor of the same value to compenstate. So if my mulitplication factor is 6, I would use an 18 ohm resistor.

Douglas self uses an 18 ohm resistor in one of his amps with only a mutiplication factor of 3, which should dicate a 10 ohm resistor. He also recommends attaching a variac to the amp then changing the resistance until you get the least variation in current over voltage changes. This is difficult to do.

What I glean from this is that the higher the mulptiplication factor the more the circuit would benefit from a current compensation resistor. So is there any easy way to determine the value?

2. I have read in the web to put a capacitor across the VBE mutiplier output to stabalize it, particularly from high frequencies. I have seen advice on the value and type of capacitor so varied that there is no rule of thumb here.

My VBE muliplier drives a predriver, that drives a driver transitor that drives 12 output transistors. Does the fact that the prediriver is an easier load on the VBE mutiplier mean I can get away with a smaller cap. I propose to use a small 1 uf polyprop cap, but then others say use larger electrolytic caps here, life Self and Cordell. Is there any science here at all or just opinions?

keantoken 13th January 2013 03:58 PM

I'm surprised no one responded to this.

I don't know what this "intrinsic base-collector resistance" is. I think it has very little to do with Vbe multiplier regulation. Diode ( = transistor) resistance is inversely proportional to forward current. The proportion is .033. So, to calculate the resistance of a diode, solve .033/If. The same is true for a transistor connected like a diode.

Say we have a transistor with 6mA Ic. If we short the base and collector, the transistor will show a small-signal AC resistance of 5.5R. The compensation resistor is inserted in series with the collector, and the output taken from between the collector and resistor. If the resistor equals the dynamic resistance, it will cancel Vf variations caused by Ic changes.

However since the resistor has constant resistance and the transistor's resistance is dynamic, they only cancel at the given Ic. The dynamic mismatch is such that if Ic goes over or under the nominal value, the voltage drop decreases. This works to our advantage for a Vbe multiplier, since in both fault conditions output stage bias will decrease rather than increase.

For a Vbe multiplier, the resistive divider affects things. It is basically true that the Vbe multiplying factor also multiplies the dynamic resistance. So for a x2 Vbe multiplier at 6mA, the compensation resistor is roughly 11R.

However transistors commonly used for the Vbe multiplier typically have low Hfe of 100 or so. This has to be accounted for. The source impedance seen by the transistor base is divided by Hfe. So say we have a x2 multiplier at 6mA with Hfe=100 and 2 1K resistors as the voltage divider. The total divider resistance seen by the base is 500R. 500R/100Hfe=5R. This alters our compensation resistor significantly, from 11R to 16R.

The .033 proportion can change between transistors and diodes as a result of doping differences. I'm not sure but I think .033 is usually right for small-signal silicon transistors. Schottkey diodes are more like .045.

Theoretically, if we replace the compensation resistor with a diode, the dynamic resistances would track and null perfectly. Unfortunately this cancels the voltage drop we need as well as forcing the transistor into Vcesat, so it won't track right. Furthermore any significant current draw from the output node disrupts tracking. If there is a schottkey diode with the same dynamic resistance factor as a silicon transistor, then we could use that and have a voltage drop of .3V. We would have to use something like the BC5xx or BC3x7 so Vcesat would not be an issue. Simulation models suggest some schottkeys will actually work for this, but I don't know whether to trust them. In any case using schottkeys this way you'd need to double your multiplier in order to have the same voltage drop. You would need as many schottkeys as your multiplier. This idea needs to be proven

The Vbe multiplier bypass cap is a subject of mostly opinion and individual experiences. It is important to realize this cap is a big piece of surface area connected to the VAS, and radiates to nearby circuitry. So the size and dimensions of the cap used here may alter stability, and this may be just as audible as any effect on bias voltage. I suspect this may account for some of the variability of the chosen cap. If you want to eliminate this mechanism, you can use a strip of grounded foil around the cap to short its static radiation to ground. Or several turns of a grounded wire may be enough.

As far as science and ruggedness, if significant current will be drawn through the Vbe multipler, which is the case for a double EF output stage, a Vbe multiplier bypass of up to several hundred uF is justified. For a triple EF or otherwise heavily buffered output stage, the possibility if dynamic bias pumping is diminished. In this case small RF decoupling is still justified, because a Vbe multipler can be slow and present a significant impedance at RF and this can cause stability quirks. For this purpose I think no more than 10nF is required, and 1nF is likely to pass with flying colors.

If we were to attempt to consider the cap audiometrically, we would be thinking about the RC time constant between it and the Vbe multiplier resistance. At a resistance of 16R, it's unlikely a cap here will be large enough to have a corner in the audio band. Even so I've heard changes in the sound when changing this cap so it's something I'd try at least once.

sreten 13th January 2013 04:19 PM


The capacitor across the Vbe multiplier is called by Self a "switch-off"
capacitor and has not much to do with the Vbe multiplier. It provides
a low impedance path for turn off currents in the drivers and outputs.

rgds, sreten.

naf 13th January 2013 04:36 PM

thanks kean, i learn something new on vbe multiplier... i see this on Honey Badger Amp but don't know how to calculate.

Struth 13th January 2013 06:19 PM

Hi Guys

Self also showed that adding one more BJT in CFP fashion provides a much lower output impedance than the compensated single-BJT can (fig.15-34, in Audio Power Amplifier Design Handbook 5th ed). This is really the way to go. The BJT used for thermal tracking can be a large case to accommodate mounting to drivers or outputs, with the other just a high-gain TO92.

The Vbe bypass cap can be left out and most circuits will function fine. However, for best high frequency response 100nF to 10uF is typically added.

Have fun
Kevin O'Connor

keantoken 13th January 2013 06:39 PM

A CFP will mostly get rid of this problem, but has its own complex behavior. It can go unstable, which is seriously bad. And tempco may be off. Ultimately one needs to determine what kind of bias generator to use based on the tempco requirements, as this will limit the topology to the compatible types.

Bigun 13th January 2013 06:46 PM

I like the Hagerman, more stable wrt Vas current variations. At least I thought so when I used it here:

post 22 has the Hagerman notes attached.

keantoken 13th January 2013 07:23 PM

I like the Hagerman too. It seems you would have to mount both transistors to the outputs though to get proper tempco. Q13 can be replaced with a diode string if you need more voltage. Reference:

Bigun 13th January 2013 07:28 PM

I used the version with one transition and a diode, mounting only the one transistor against the body of the output device package. I believe the Vbe does better when it's not mounted on a large heatsink - too much thermal lag, better to be mounted on the power device itself.

Dave Zan 14th January 2013 12:34 AM

Thanks for the link to the PDF but the comparison there does not look square.
The Vbe multiplier circuit (2b) has lower base-emitter resistor than 2c and 2d and starts to drop out at the lower current levels of the chart.
Also there is no comparison of a Vbe multiplier + resistor with the proposed "improved" circuit (2d).
If a resistor is used for current correction then I am not sure the extra diode-connected transistor offers much improvement.

Best wishes

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