VBE multipliers.

Since, BJTs input junctions conduct an electronic current exponentially, it needs some voltage for this little current to reach levels that are encountered in amplifiers. This is usually about 0.6V to 0.7V. However, this little voltage causes crossover problems if it is not addressed. The latter function, is usually done by VBE multipliers. This thread is intended for users to post those VBE multiplier topologies and even actual circuits which they found to perform best.

I am still hesitant about changing the VBE multiplier circuitry of my amplifier. It seems, it is impossible to implement a VBE multiplier or equivalent with a sharp step characteristic and with full temperature compensation. There are various variables involved often requiring approximations to the level of regulation needed.
 
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Why do you want to change the Vbe multiplier circuitry of your amplifier? Does it have problems over temperature?

The 'sharpness of the step characteristic' is not usually significant since the thermal inertia of the heat sink slows down everything.

Regards
 
jan.didden said:
Rick do you know what this 'sharpness of the step characteristic' means?

The voltage across a VBE multiplier is m*Vbe, where m is the multiplying factor and Vbe is the voltage across the base-emitter junction. As Ib, the base current, increases exponentially with Vbe, the behaviour of m*Vbe is logarithmic, i.e. the inverse of exponential.

A step-characteristic would be like that of an idealised Zener diode, whose current can reach infinity at the characteristic voltage. Once that voltage develops across such a multiplier, increases in Ic occur at no additional increases in Vce.
 
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I would have thought the more important consideration is finding the Vbe temperature monitoring junction (diode or transistor) having the closest match to the output transistor's temp-co. None of the circuits will work well if the temp-co's do not track, although they may be adequate.
 
The output stage is of the complementary feedback pair type with the drivers connected in the usual emitter follower configuration. These feed their collector currents into 22R resistors connected to each voltage rail. The resistors drive the output stage's transistors which have 0.22R resistors as their emitter resistors.

The amplifier is exhibiting quiescent current (class AB bias) drift which is very significant. The temperature of the room also makes a significant difference. One channel also has about 1V of DC at the output at all times. This makes it unuseable.
 
The voltage across a VBE multiplier is m*Vbe, where m is the multiplying factor and Vbe is the voltage across the base-emitter junction. As Ib, the base current, increases exponentially with Vbe, the behaviour of m*Vbe is logarithmic, i.e. the inverse of exponential.

A step-characteristic would be like that of an idealised Zener diode, whose current can reach infinity at the characteristic voltage. Once that voltage develops across such a multiplier, increases in Ic occur at no additional increases in Vce.
Yes I know that, of course, common knowledge. But in an amp there is a certain current through the Vbe mux, that does vary some, but it will not cut off.
There's no step in any shape or form.

jan
 
The amplifier is exhibiting quiescent current (class AB bias) drift which is very significant. The temperature of the room also makes a significant difference. One channel also has about 1V of DC at the output at all times. This makes it unuseable.
I'm not disputing anything said about the VBE multipliers or temperature tracking, but 1V DC at speaker terminals sounds more like a global biasing issue than VBE.
 
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The VBE stage is to adjust the idle current of the output stage as the temperature changes. It needs to keep the bias current constant.

Your comment that the amplifier output has a 1 volt DC offset is not caused by the VBE stage. The offset is caused by the DC feedback not working correctly. Typically the AC gain is 20x and DC gain is 1x. The CAP maybe leaking or bad.

Duke