Running QE AM-10 with +-15V

[omeiko]

Hi,

I've build some Quad-Eight op-amps and I've been wondering if I could run this op-amp with only +-15V.
It's ratet for +-28v originally but it would be more convenient if it could be used with a more common +-15v supply rails.
I rarely need the extra headroom provided by the high supply rails. So
is there some changes to be made to this circuit to accomplish this?

The circuit is here:

http://www.avensonaudio.com/tech/Quadeight/quad8am10op.jpg

Thanks,

Okko

 

[BuzAllen]

Should work just fine. The biasing in the circuit is supply independant so the different rails will not effect the idle currrent.

Brian

 

[omeiko]

Thanks Brian.

Just out of curiosity, what is the part that determines the idle current?
This circuit looks surprisingly similar to JE 990
( http://www.johnhardyco.com/pdf/990.pdf ). However, 990 needs some resistors to be adjusted to accept different supply voltages.
So what's the difference, for example, between these?

Cheers,

Okko

 

[BuzAllen]

I believe those 3 values for resistance are used for the 3 voltages to keep the current throught the biasing diodes for the current source the same whether at 12 or 24 volts. This will have much less effect on the overall following idle currents out of the transistor i source than if the current source was purely a resistive one where doubling the supply would essentially double the current through the resistor

In other words doubling the current through a resistor will double the voltage, in this circuit with the current range we're looking a, doubling the current through the diodes will not vary the voltage across them that much.

Basically the idle curent in the input pair and VAS stage is determined by the voltage left across either r3 or r8.

Lets look at the AM10 schematic and see what happens. At +/- 28 volts, there would be approx. 54.8 volts across r4, giving about 880uA throught the diodes. Simulation shows that they'll be on and at about 580mV each. With +/- 15V we get about 460uA, in this case the drop across the diode would be approx. 549mV.

Lets say that 1 diode exactly cancels Q8's Vbe and we then have one diode at either 580mV or 549mV across the 330 Ohm resistor. We'd get roughly 1.75mA or 1.6mA out of the source. Much more supply indepedant than a resistor only circuit but not completely so. If you wanted you could change r4 to account for this but I don't think it would matter all that much. This isn't exact as we assumed one diode and Q8 would cancel exactly but in reality that diode would also vary in voltage a bit and so would Q8's Vbe with the current change.

This is a bit idealized and real world/component choice would vary it.


Brian

 

[bcarso]

Concur (with the one exception below) with Brian. If you decide to change the resistors anyway the nearest 5% value to get the same diode current would be 33k (R4).

R6 as is may be a problem. It is apparently for current limiting and looks like it needs to be reduced for the much lower rails (the quiescent drop across it is of order 40V based on some assumptions about diode drops and Q Vbe's etc.). It also tends to spoil bandwidth a bit because of Miller effect. Better than blowing Q7 up from excess base current though. Notice that the 10:1 ratio of R5/R7 does not equate to a 10:1 ratio of collector currents in Q7 and Q6---I get about 2 mA Ic for Q7 and 670uA for Q6. I'd take R6 to 22k. I'd say hang a C across it as well but we are monkeying with a more-or-less tried-and-true design here at our peril.

Brad

 

[omeiko]

Thanks guys,

I have to read your answers through a few times to understand what's goining on in the circuit. I'm still such a novice but I'm slowly learning.

However, I changed the r4 to 32k and it seems to work fine.

Thanks,

Okko