I'm hoping someone can help me figure out why a component keeps burning on the circuit I'm testing.
In the attached schematic, the part of the circuit in the upper right is a current source that is supplying bias currents to the OTA's. It's designed to supply up to 600μA to each OTA, the upper end of which is limited by R98. In simulation it is limited to 4.84mA total out of Q4. I simulated each of the eight OTA's as two diodes in series, because the Iabc input to the OTA (LM13700) is two diode drops above the negative supply. Even simulating a short from the collector of Q4 to the negative supply doesn't cause the output current to rise above the limit set by R98.
There's a little more to the circuit than what you see in the schematic, but I cut it out for simplicity. I don't think anything I cut out would be part of the problem.
The problem is that unless I put a resistor between the collector of Q4 and the OTA's, Q4 burns. Not always immediately, sometimes it takes a few seconds after turning on power, but it's always Q4, and I'm pretty sure it's always the emitter of Q4 that has a visible burn mark on the component. On my breadboard test circuit Q3 and Q4 come from a FPQ2907, which is four 2N2907's in a PDIP package. I've burned up a bunch of these things now. I have used a resistor down to 470Ω without anything burning up.
- The transistor is rated for 150mA, I'm only asking it to supply 5mA.
- The transistor is rated for Vce of 40V, it should only be getting Vce of 22-23V.
- The OTA's aren't getting damaged, only Q4. I know from past experience how easy it is to burn up a LM13700 by supplying too much current to the Iabc input, but that's not happening here.
- I've used the same current source, with different resistor values to provide different current limits, in OTA circuits using two and four OTA's, without adding another resistor and without Q4 burning. Those circuits worked as designed and simulated.
- I've tested the output of Q4 by disconnecting it from the OTA's and putting the current into a resistor to ground. Testing this way, it's pretty darned close to what it should be.
I've also attached a picture of my breadboard. The PDIP containing Q3 and Q4 is the first black one on the left (not counting the SOIC on the green adapter board); the four to the right of it are the OTA's. It's kind of hard to see but there's a resistor from the upper right pin of the transistor array to a point between the two DIPs, and from there it goes to the four OTA packages through a short red wire, two long loopy green ones and a long loopy black one. Each of those four wires goes to one Iabc input of the OTA chip, and then a short jumper wire goes to the second Iabc input in the package (two OTA's per package). Without that resistor, I had those four wires going directly to the upper right pin of the transistor array (collector of Q4).
Sorry for the length of this post, I just wanted to give you all the info. Any idea why Q4 is burning without a resistor?
In the attached schematic, the part of the circuit in the upper right is a current source that is supplying bias currents to the OTA's. It's designed to supply up to 600μA to each OTA, the upper end of which is limited by R98. In simulation it is limited to 4.84mA total out of Q4. I simulated each of the eight OTA's as two diodes in series, because the Iabc input to the OTA (LM13700) is two diode drops above the negative supply. Even simulating a short from the collector of Q4 to the negative supply doesn't cause the output current to rise above the limit set by R98.
There's a little more to the circuit than what you see in the schematic, but I cut it out for simplicity. I don't think anything I cut out would be part of the problem.
The problem is that unless I put a resistor between the collector of Q4 and the OTA's, Q4 burns. Not always immediately, sometimes it takes a few seconds after turning on power, but it's always Q4, and I'm pretty sure it's always the emitter of Q4 that has a visible burn mark on the component. On my breadboard test circuit Q3 and Q4 come from a FPQ2907, which is four 2N2907's in a PDIP package. I've burned up a bunch of these things now. I have used a resistor down to 470Ω without anything burning up.
- The transistor is rated for 150mA, I'm only asking it to supply 5mA.
- The transistor is rated for Vce of 40V, it should only be getting Vce of 22-23V.
- The OTA's aren't getting damaged, only Q4. I know from past experience how easy it is to burn up a LM13700 by supplying too much current to the Iabc input, but that's not happening here.
- I've used the same current source, with different resistor values to provide different current limits, in OTA circuits using two and four OTA's, without adding another resistor and without Q4 burning. Those circuits worked as designed and simulated.
- I've tested the output of Q4 by disconnecting it from the OTA's and putting the current into a resistor to ground. Testing this way, it's pretty darned close to what it should be.
I've also attached a picture of my breadboard. The PDIP containing Q3 and Q4 is the first black one on the left (not counting the SOIC on the green adapter board); the four to the right of it are the OTA's. It's kind of hard to see but there's a resistor from the upper right pin of the transistor array to a point between the two DIPs, and from there it goes to the four OTA packages through a short red wire, two long loopy green ones and a long loopy black one. Each of those four wires goes to one Iabc input of the OTA chip, and then a short jumper wire goes to the second Iabc input in the package (two OTA's per package). Without that resistor, I had those four wires going directly to the upper right pin of the transistor array (collector of Q4).
Sorry for the length of this post, I just wanted to give you all the info. Any idea why Q4 is burning without a resistor?
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