LM13600 - why is it getting hot?

[blackieNYC]

A gator on breadboard.  audio passes thru and everything seems to work just fine.  And after a few seconds, then the 13600 starts getting hot and the supply voltage drops a little from 9.6 to 9.45.
I've wired the 13600 like this - http://www.diystompboxes.com/smfforum/index.php?topic=105341.0
pins 9,10,16 are grounded, and 12 is tied to 13 as in Merlin's Engineer's thumb.
13600 voltages, quickly taken, are - pin 1=.5volts(and the output signal doesn't seem to respond to the setting of the 250k Level control, but it is connected to Vref via the 10k)
pins 3&4 = 4.7v
pin 5 is 3.6v, and may be dropping slowly
pin6 = 2v.
other pins are n/c

[Fender3D]

Are you sure about this schematic?
There's no negative supply for 3080...

[R.G.]

Mother Nature is whispering in your ear.

If your [X] is getting too hot, then it's either being externally heated, or it's generating more heat inside than it can get rid of through its surface. There are no other answers.

But you knew that, and you want me to get on with why it's generating too much heat internally, right? 

OTAs are purely current-differencing devices, with no resistors or caps inside. They're purely transistor pairs in mirrors and difference amplifiers. When you put a voltage across the power supply pins and put NO current in the Iabc pin, there is no current flowing inside, and so it does not self heat at all.

When you put current into Iabc, that same current flows to all of the mirrors and differential pairs inside. Most importantly, the output pin is driven by one set of transistors between the V+ supply and the output pin, and one set of transistors between the V- supply and the output pin. When you set Iabc, that same current flows from V+, through the + side output transistors, through the - side output transistors, and back into V-. It's only a difference between the voltage on Vin+ and Vin- that makes some of the flow-through current go in/out the output pin.

So you're dissipating the total power supply times Iabc in the output stage 100% of the time in an OTA.

The spec sheet on the LM3080 said that Iabc was a maximum of 1ma. They were's kidding. Exceed 1ma, and it changes into a Darkness Emitting Diode (DED).

The LM13600 is iimproved. It will live up to 2ma of Iabc.

It's getting too hot because Iabc is too big.

[PRR]

> There's no negative supply for 3080...

IC4B Out pulls OTA V- down. This sucks more current in 10K+250k.

Yes, it is a mind-warp to be pulling power supply pins around. But the OTA is not fussy about its supplies, they have very little effect, except here the resistors from Iabc to Vref vary Iabc current without adding a current generator.

[blackieNYC]

Thanks guys. I'll look closer at that.
Is it time we redesigned the gator for the 13600, using its pin1 as (what seems to me to be) intended, instead of wiggling a negative rail around?  The nice long attack and quick release of the side chain is fine, if maybe a little inflexible. (Not very useful as a noise gate IMO. Just a sweller and that's great.)

[R.G.]

When you're working with OTAs, it's good to remember that the Iabc pin and the V- pin really can't wiggle about independent of each other, at least not very much. In the 3080, they were effectively connected by a silicon diode. In the LM13600, they're effectively connected by two diodes in series.

Iabc is a critical thing for everything going on inside. Since Iabc is effectively a diode's (or two diodes') forward voltage(s), the voltage between V- and the Iabc pin is the equation for diode voltage as a function of diode current, with sudden death being over 1ma (2ma for the LM3600). The things work at low Iabc, so the voltage between Iabc and V- in the positive direction is between maybe 300mV and 600mV. Any more and the chip dies.

Reverse break the current mirror that's what's really there and the chip is dead.

One really good way to deal with the oddities of Iabc is to use a PNP current mirror to "drizzle" bias current into Iabc from a more-positive supply so you don't have to worry about either the odd and changing offset between Iabc and V- or put in a big resistor to limit the bias current and hope that the inaccuracies of the voltage between Iabc and V- don't change the current you're trying to put in there too much.

But that's just me. I think current mirrors are kewl.   

[Kipper4]

One really good way to deal with the oddities of Iabc is to use a PNP current mirror to "drizzle" bias current into Iabc from a more-positive supply so you don't have to worry about either the odd and changing offset between Iabc and V- or put in a big resistor to limit the bias current and hope that the inaccuracies of the voltage between Iabc and V- don't change the current you're trying to put in there too much.

But that's just me. I think current mirrors are kewl.   

Somehow I can't picture this in my head. So if someone would care to draw up a scheme for it I'm sure it would help.
Thanks
Rich

[R.G.]

Somehow I can't picture this in my head. So if someone would care to draw up a scheme for it I'm sure it would help.

I'm doing some recuperation and it's hard for me to get to my drawing machine right now, so let me try one more version with words.

Imagine two NPNs, both with their emitters tied to ground. The bases are tied in parallel, and one NPN's collector is also tied to the parallel bases. The remaining collector is pulled up to some V+ voltage.

The parallel bases and one collector is the input. The other collector is the output.

When you let a little current flow into the input to ground, the input transistor lets some of the current through its base-emitter junction, and hfe times as much current through the collector-emitter junction. The whole thing acts like a simple forward biased diode, following the diode equation for the current being an exponential function of the base voltage. But thSo ere's that second NPN there. It's base emitter is paralleled to the first NPN's base emitter, so its base-emitter voltage must be equal to the input NPN's base emitter voltage. If the two transistors are close to identical, then for identical base-emitter voltages, they must have identical collector emitter voltages.

In practice, they do, very closely, and there are ways to make them come even closer to "perfect.

So we have two NPNs, both emitters grounded, one base/collector as an input, and one collector only as an output.

We can allow, say, 100uA, to flow in the input side to ground and then measure how much current flows to ground through the output collector (which was pulled up to some V+, remember, so there is current available for it). We'll find that 100ua +/- some percentage of accuracy flows. Generally we can consider them to be identical.

A current drizzled from a positive voltage source into the input side is "mirrored" by an identical current sucked down into the output side.

This is very close to what is happening on the Iabc pin of the LM3080. It's Iabc pin ***is*** the input to an NPN current mirror.

This works great if you only want to drop current into Iabc from a more positive DC voltage. It is a PITA if you are worried about DC accuracy down around your V- power supply because of that one-diode-drop offset.

Enter the PNP current mirror. Make a PNP circuit with both emitters tied to V+, both bases and one collector tied in parallel, and one collector pullled down to V-. Works same as the NPN mirror, but now any current you pull down from V+ through the mirror input is mirrored as a current flowing down from the PNP mirror output.

My obvious next step is to tie the output of the PNP mirror to the input of the Iabc NPN mirror. Now I can pull down on the PNP mirror's input, and the same current flows into the NPN mirror's input.

This probably seems like a silly waste of more parts, but it turns out that with today's ICs, it's easier to pull a controlled current from a higher voltage than to generate a controlled current into V+. And the insertion of those two PNP transistors lets me save more parts and circuit complexity than they cost me.

[Kipper4]

Bump. It's taken me ages to find this in search.
I need to reread.
Carry on fellas
Ignore me.

[Kipper4]

Like this?

[R.G.]

That's the basic idea. I'd put 10 to 47 ohm resistors in series with the emitters of the transistors, use PNPs and connect your "ground" to some +V, as well as eliminating that 1M resistor.

[Kipper4]

Thanks RG.
I went with the first example.
I'll go back to the drawing board.
Get well soon buddy.

[Kipper4]