I'm working on designing a class B amp from scratch based on the typical three stage design. The output is Darlington with 2-3 pairs of output transistors. I'm going for around 100W into 8 ohms.
The best explanation of bias circuit design I've found is from Bob Cordell's book, where he explains Vbe multipliers. I'm working it out in SPICE but when the tempco is correct the outputs have impossibly high bias current.
So my question is: how to set the overall quiescent current level separately from adjusting the tempco? Most circuits I've seen just have a pot in the lower half of the voltage divider around the bias transistor. Adjusting this would just change the multiplication factor, but I would prefer to have a fixed tempco and then adjust the offset to set the quiescent current in the outputs instead.
Can someone explain the reason behind this (very common) approach and how best to set the overall bias current while also having the correct tempco?
The best explanation of bias circuit design I've found is from Bob Cordell's book, where he explains Vbe multipliers. I'm working it out in SPICE but when the tempco is correct the outputs have impossibly high bias current.
So my question is: how to set the overall quiescent current level separately from adjusting the tempco? Most circuits I've seen just have a pot in the lower half of the voltage divider around the bias transistor. Adjusting this would just change the multiplication factor, but I would prefer to have a fixed tempco and then adjust the offset to set the quiescent current in the outputs instead.
Can someone explain the reason behind this (very common) approach and how best to set the overall bias current while also having the correct tempco?
It is difficult to get right tempco. If you want high bias current, i think you should make slightly over compensated.
Hi bimo,
Thanks for your reply but it doesn't really answer my question. I understand the overcompensating part, but setting the offset for the biasing to start with seems quite tricky. I can't figure out how to set the quiescent current separately from the compensation tempco.
None of the texts I've read really explain this part. Most designs seem to use very low values for the resistors which steer quite a lot of current away from the actual bias transistor which I think would introduce a lot of error.
Thanks for your reply but it doesn't really answer my question. I understand the overcompensating part, but setting the offset for the biasing to start with seems quite tricky. I can't figure out how to set the quiescent current separately from the compensation tempco.
None of the texts I've read really explain this part. Most designs seem to use very low values for the resistors which steer quite a lot of current away from the actual bias transistor which I think would introduce a lot of error.
Tempco will be vary, but it is OK. Unleash you use auto bias like LT1166 or something that Edmond Stuart design.
Changing tempco = changing multiplication factor.
So, you need to change the Vbe (or Vgs). You can do this with testing other bias spreader devices. Lowering multiplication factor = lowering tempco, this you can also do with adding extra voltage drop at the emmitter, by adding For example diodes.
Have a look at the first page of the document attached via the link.
Adding a base stopper at the bias spreader will as well lowering the tempco.
https://www.diyaudio.com/forums/sol...n-subwoofer-wideband-duty-13.html#post3382545
So, you need to change the Vbe (or Vgs). You can do this with testing other bias spreader devices. Lowering multiplication factor = lowering tempco, this you can also do with adding extra voltage drop at the emmitter, by adding For example diodes.
Have a look at the first page of the document attached via the link.
Adding a base stopper at the bias spreader will as well lowering the tempco.
https://www.diyaudio.com/forums/sol...n-subwoofer-wideband-duty-13.html#post3382545
The current in the resistors has to be much larger than the transistor base current in order to be accurate, which is why they are low in value. The properties of the bias transistor are then entirely in charge, multiplied by the resistor ratio.
Typical bias circuit might be seeing 6mA of current, and perhaps 2mA goes through the resistor chain, 4mA through the transistor, but the transistor has complete control of the bias voltage since its a negative feedback system,
assuming its got some current through it to work with.
There is an assumption that the Vbe get multiplied roughly by the number of junctions that have to be biased, ie x4 for drivers+outputs, x6 for drivers+darlington outputs. This can then tweaked with resistors in the collector/emitter path.
Its important the only pot is in the base-emitter leg of the circuit, as pots can fail open-circuit over time, and this is the only position where a pot failing doesn't lead to sudden and drastic over-bias (which destroys the output devices).