I'm working on a new board and configured the output stage like the attached image; it resembles the internal layout of a darlington transistor and does'nt use the switchoff network.
Should it work well or there could be problems ?
Simulation seems to suggest a good behavior
Should it work well or there could be problems ?
Simulation seems to suggest a good behavior
Attachments
Hi,
I think you might have stability issues if one transistor of three cundcuts more current then the others, it will apply more local feedback for itself and then inturn for the others making it do all the work.
You might want to put your ballast resistors on the collector side and tie all three output emmiters to one spot and then if you need short circuit resistors on the emitters then add them after you tie all three's together. Then use only one feedback resistor per rail. I'm not sure that even this would help, you might need to just remove the 100's all together but then you just a regular old ouput stage.
Cheers
I think you might have stability issues if one transistor of three cundcuts more current then the others, it will apply more local feedback for itself and then inturn for the others making it do all the work.
You might want to put your ballast resistors on the collector side and tie all three output emmiters to one spot and then if you need short circuit resistors on the emitters then add them after you tie all three's together. Then use only one feedback resistor per rail. I'm not sure that even this would help, you might need to just remove the 100's all together but then you just a regular old ouput stage.
Cheers
darlington connection
A true darlington connection like this is inherently unstable, mostly because of temp. coeffecient affecting the DC bias. Yes it is true that some amp circuits use this type of connection, but the darlington devices are integrated. A "darlinton" transistor consists of two integrated transistors that are perfectly matched with each other. If you try to use descrete transistors even of the same type, in a darlington connection you will get nonlinear operation and unstability.
If you are going to use descrete transistors for the output, then increase the power of your driver stage to handle the lower beta.
A true darlington connection like this is inherently unstable, mostly because of temp. coeffecient affecting the DC bias. Yes it is true that some amp circuits use this type of connection, but the darlington devices are integrated. A "darlinton" transistor consists of two integrated transistors that are perfectly matched with each other. If you try to use descrete transistors even of the same type, in a darlington connection you will get nonlinear operation and unstability.
If you are going to use descrete transistors for the output, then increase the power of your driver stage to handle the lower beta.
@easy amp:
...cannot follow your points...
1.
Especially I think that direct parallel connection of basis and emitter, will result in poor current sharing!
2.
Stability vs oscillation is a concern all configurations, cannot answer this in a short simple sentence...
@ Cunningham:
Yes, thermal runaway has to be taken into account.
But lots of amplifiers use such output stages and with a
proper thermal compensation and resistors at the emitters, they usually work thermally stable.
...cannot follow your points...
1.
Especially I think that direct parallel connection of basis and emitter, will result in poor current sharing!
2.
Stability vs oscillation is a concern all configurations, cannot answer this in a short simple sentence...
@ Cunningham:
Yes, thermal runaway has to be taken into account.
But lots of amplifiers use such output stages and with a
proper thermal compensation and resistors at the emitters, they usually work thermally stable.
By the way: Integrated darlingtons have the same issues with thermal runaway.
Any BJT goes to higher current if temperature is increased.
Rule of thumb:
For compensation you have to decrease the basis emitter voltage by 2mV/°C. For a darlington you need to compensate two basis emitter paths ==> 4mV/°C.
If you have a complementary darlington output stage then your thermal compensation must reduce the bias voltage about 8mV/°C.
Any BJT goes to higher current if temperature is increased.
Rule of thumb:
For compensation you have to decrease the basis emitter voltage by 2mV/°C. For a darlington you need to compensate two basis emitter paths ==> 4mV/°C.
If you have a complementary darlington output stage then your thermal compensation must reduce the bias voltage about 8mV/°C.
@Mambo:
Some points to consider:
-Your emitter resistors are very low, this makes proper thermal compensation more difficult.
-Also if the current gain is different from the original darlingtons, then the thermal compensation may show an inadequate temp gradient (current which is drawn from the output stage does affect the amount of mV/°C of the compensator).
If your new assembly has higher current gain then you may get thermal undercompensation and with this thermal runaway.
-When you say:" ...does not use the switch off network..." Do you mean the reverse diodes which are usually integrated in darlingtons, or do you mean the RC-network at the output?
-General issues with oscillation:.... just wishing good luck!!
Bye
Markus
Some points to consider:
-Your emitter resistors are very low, this makes proper thermal compensation more difficult.
-Also if the current gain is different from the original darlingtons, then the thermal compensation may show an inadequate temp gradient (current which is drawn from the output stage does affect the amount of mV/°C of the compensator).
If your new assembly has higher current gain then you may get thermal undercompensation and with this thermal runaway.
-When you say:" ...does not use the switch off network..." Do you mean the reverse diodes which are usually integrated in darlingtons, or do you mean the RC-network at the output?
-General issues with oscillation:.... just wishing good luck!!
Bye
Markus
It makes little diff where you place the ballast resistor emitter or collector they provide similar functions of curent sharring between devices. Emiter ballast are optimum for use as short circuit networks but work similarly on the collectors for current sharing.
It also was only an option for multi power devices in a darlington output, wich you hit on as being bad from the start, but a possible solution to induce better current sharing my thought was to remove the multiple local fedback lines running to a single point controlling multiple devices which is addressed with my removale of multiple feedback lines, letting a sumation of the three outputs control the LNFB.
Mambo,
As I see it, it makes a HUGE difference where you put your 0.1 ohm resistors (collector or emitter). They should surely be on the emitter (where you have put them) when your bases are tied together at the same potential.
If they're on the collector side, you cannot compensate for Vbe differences in the parallelled power transistors, which will lead to the current flowing through only one of the tansistors, frying it.
May I ask... why not use the well tried, and reliable complelemtary output stage with a driver and n x power transistors for each supply rail (NPN and PNP push-pull stage).
With this layout, you put a rather varying load on your driver, and thus a varying load on the Vas stage of your amplifier. to me, this doesn't point in the direction of good linearity.
Jennice
As I see it, it makes a HUGE difference where you put your 0.1 ohm resistors (collector or emitter). They should surely be on the emitter (where you have put them) when your bases are tied together at the same potential.
If they're on the collector side, you cannot compensate for Vbe differences in the parallelled power transistors, which will lead to the current flowing through only one of the tansistors, frying it.
May I ask... why not use the well tried, and reliable complelemtary output stage with a driver and n x power transistors for each supply rail (NPN and PNP push-pull stage).
With this layout, you put a rather varying load on your driver, and thus a varying load on the Vas stage of your amplifier. to me, this doesn't point in the direction of good linearity.
Jennice
Mambo said:
I used almost same configuration in some of my amplifiers. It can work well, with proper thermal compensation. The bias transistor mounted to the case of one of the output devices (preferably the hottest one), and the driver on separated heatsink,- to keep it cool-, can be the good solution. If any thermal runaway appears, increase the emitter resistors to 0.22 - 0.33ohms will help. But my point of view: keep the emitter resistors as small as possible!
Oscillations: I found very good solution to avoid it. Just apply small series resistor with the base of the output devices 2.2 - 4.7 ohms is good enoug. With this solution, my amplifier can drive single 10uF load without any series or parallel resistor. No overshot, no oscillation.
sajti
I'm surprized at these remarks, I agree it makes a difference but a HUGE difference I don't agree with. Kirchoff's law current in current out, other then the drivers current passing through the b-e of the output transistor the exact same amount of electrons are moving on each side collector and emiter this is all the resistor see's. It doesn't know which side it's on, same as the transistor it doesn't know which side the resistors on being that the resistor effect on electrons flowing though the current loop (rail to load) will be the same wether it's on the collector or emiter side, Barring of course the drivers current passing through the outputs b-e. One of the resistors roles is to provide heat based regulation, decreasing the resistance on which ever transistor is doing more work and hence forces others in the network to due more work.
This function can be obtained from either side as current will be almost the same on either side so will resistor regulation. You can run into trouble though when your output transistors hfe falls to a level low enough to signigantly pass the driver transistors current through the outputs transistor b-e and not the output transistor doing the work through it's c-e which reduces the effective regulation or current sharing provided by the resistor as well as other matching characteristics which could cause loss of hfe and vbe matching and poooffff.
I agree this is not the best topology, and that the resitors work better on the emitter side but with the original design it's easy to see that with all three transistor giving there feedback to single source and then that single souce controlling all three output device's it will never work correctly. Anytime one transistor did a little more work then the one next to it which is going to happen, it would intern send more negative feedback to the same place all three were getting there signal from reducing the amount of current all three were sharing equally, sounds good, well the others were allready conducting less so they continue to stay below this transistor, which everone got out ahead first, and while this one is doing all the work it continues to heat increasing it's hfe and the others fall away doing less and less cascading into the one doing all the work failing.
If you increased the output resitors value to higher value and I'm not sure what it would be you could probably stop the single transistor doing all the work from blowing up but the amp would run lop-sided with one transistor still doing most of the work and the other 2 just hanging out coming on only when there was signifigant current heating up the one doing all the work's resistor.
And I have actually built amplifiers for testing this eaxct idea and it works as mentioned above failing to regualte under high load conditions. I'm building a triple output darlington similar to the original post just to see what it does, theory is nice but I find realality is always different. I love playing with transistor's.
Cheers
Cheers
Easy,
emiiter resistor has two purposes:
1, The DC voltage across is gives local feedback for bias regulating over wide temperature range.
2, The AC voltage (resulted by the sound) across it, reduce the Vbe on the transistor which takes more current. This local feedback helps to equal the current, and dissipation.
The current flows thru the output devices not depended by the collector voltage (if it high enough). So Your solution has some resistors which are not necessary to use, and there are some missing...
sajti
emiiter resistor has two purposes:
1, The DC voltage across is gives local feedback for bias regulating over wide temperature range.
2, The AC voltage (resulted by the sound) across it, reduce the Vbe on the transistor which takes more current. This local feedback helps to equal the current, and dissipation.
The current flows thru the output devices not depended by the collector voltage (if it high enough). So Your solution has some resistors which are not necessary to use, and there are some missing...
sajti
Easyamp said:
The original post looks very good for me. You can increase the performance, if You put 100ohms+100nF parallel between the emitter of the drivers. It will keep open the drivers, which means class A working for them, and the 100nF helps to close the output devices faster, if You drive the amp with fast signal.
sajti
Easyamp,
My claim:
The original post will work. Moving the location of transistors will not, if you load it more than one transistor is capable of. You will not have much current charing if the 0.1 ohm resistors are on the collector side.
My concern was, that if you move the resistors to the collector side, you will force all transistors to share voltage potentials at base and emitter, respectively.
Vbe is approximately 0.7V, but not completely identical. Therefore the transistor with the lowest Vbe will switch on, and the others remain off. This one transistor takes the current and heat (literally), which further lowers Vbe. The thermal runaway is almost guaranteed.
Please show me one reliable design which does have the resistors at emitter, and I will rest my case.
Jennice.
My claim:
The original post will work. Moving the location of transistors will not, if you load it more than one transistor is capable of. You will not have much current charing if the 0.1 ohm resistors are on the collector side.
My concern was, that if you move the resistors to the collector side, you will force all transistors to share voltage potentials at base and emitter, respectively.
Vbe is approximately 0.7V, but not completely identical. Therefore the transistor with the lowest Vbe will switch on, and the others remain off. This one transistor takes the current and heat (literally), which further lowers Vbe. The thermal runaway is almost guaranteed.
Please show me one reliable design which does have the resistors at emitter, and I will rest my case.
Jennice.
Thanks Jennice and Satji!
Hi easyamp,
let's check the following example.
-One power supply 10V.
-One control voltage source 1V
Both connected with their neg pole to ground.
Now pick a power BJT with B=100.
Connect the basis of the BJT to the 1V voltage source.
Connection A:
Collector of BJT connected to positive 10V rail directly.
Emitter connected through 0.1 Ohms towards ground.
Vbe drop will be around 0.6V....0.7V (Depending on temperature
and chosen BJT, let' say Vbe=0.65V). Volatge across 0.1 Ohms
will be 0.35V (Kirchhof).... resulting in a current of about 3.5A.
Base current will be little less than 35mA.
Collector current will be around 3.47A.
Connection B:
Emitter is connected to ground directly.
Collector is connected to 10V rail through 0.1 Ohms.
Now Vbe drop is brutally forced to 1V, resulting in excessive
base current. Let's assume that BJT would survive this for a certain time... This high base current will massively turn ON the BJT, means will pull down the collector. Collector current will be limited by
0.1 Ohm resitor.
I= (10V-Vce_saturation)/0.1Ohm ==> approx 90A...
It is definitely a difference were you put the resistor.
Bye
Markus
Hi easyamp,
let's check the following example.
-One power supply 10V.
-One control voltage source 1V
Both connected with their neg pole to ground.
Now pick a power BJT with B=100.
Connect the basis of the BJT to the 1V voltage source.
Connection A:
Collector of BJT connected to positive 10V rail directly.
Emitter connected through 0.1 Ohms towards ground.
Vbe drop will be around 0.6V....0.7V (Depending on temperature
and chosen BJT, let' say Vbe=0.65V). Volatge across 0.1 Ohms
will be 0.35V (Kirchhof).... resulting in a current of about 3.5A.
Base current will be little less than 35mA.
Collector current will be around 3.47A.
Connection B:
Emitter is connected to ground directly.
Collector is connected to 10V rail through 0.1 Ohms.
Now Vbe drop is brutally forced to 1V, resulting in excessive
base current. Let's assume that BJT would survive this for a certain time... This high base current will massively turn ON the BJT, means will pull down the collector. Collector current will be limited by
0.1 Ohm resitor.
I= (10V-Vce_saturation)/0.1Ohm ==> approx 90A...
It is definitely a difference were you put the resistor.
Bye
Markus
A BJT has an internal resistance between B & E and this is related to current capacity and hfe. (re) Adding an emitter resistor, maybe 5-10 times 're' puts it in series with 're' and reduces the percent difference effect of 're' in both transistors, making them act as if they are more matched. Yes gain is reduced, but equal operation of the devices may be more important.
Chris
Chris
ChocoHolic said:@Jennice:
...did you miss out a "not" in your last sentences???
...or wanted to say "collector" instead of emitter??
chokoholic,
You're right. I think the M$ devil was erasing my "not"... ;-)
It's probably used to the sentence "I'm not pleased with M$ products!", and is just used to remove my "not"'s.
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