Hello all,
apologies for the probably often asked question regarding the DoZ amplifier.
I hope someone would be able to help me.
I have a working power supply delivering 60V to the amplifier, but i'm unable to get the full voltage across the 10ohm safety resistor and ground. And when I measure the voltage of the output pin(speaker out), I'm unable to get the 1/2 supple voltage despite tinkering with VR1 and VR2.
I would appreciate any help rendered and apologies again for this often asked query.
apologies for the probably often asked question regarding the DoZ amplifier.
I hope someone would be able to help me.
I have a working power supply delivering 60V to the amplifier, but i'm unable to get the full voltage across the 10ohm safety resistor and ground. And when I measure the voltage of the output pin(speaker out), I'm unable to get the 1/2 supple voltage despite tinkering with VR1 and VR2.
I would appreciate any help rendered and apologies again for this often asked query.
There is a excelent forum of Rod Elliot at the website www.sound.au.com. Maybe you can try to get an awnser there too?
I think 60V is a very high supply. I wonder if the output transistor can handle that kind of power .. you need about 3.75A quensent current so the transistor dissipate 112Watt argggg that a lot!
If you can't get it to work, post again and I'll five you some DC-values to comparein your circuit. I'm sure we can get it to work,
Greetings,
Thijs
I think 60V is a very high supply. I wonder if the output transistor can handle that kind of power .. you need about 3.75A quensent current so the transistor dissipate 112Watt argggg that a lot!
If you can't get it to work, post again and I'll five you some DC-values to comparein your circuit. I'm sure we can get it to work,
Greetings,
Thijs
Hi,
Good to hear you got it to work! Can you tell us when was the problem?
Now tp calculate the the Iq
You will have about 28V peak in 4 Ohm, that's 7 ampere. But being a push-pull amplifier, Rod Elliot advises 0.75x the peak current: 0.75 x 7 = 5.25 Ampere..
Now beware: you are dissipation 315 Watts by now.. are you sure have you designed your amp to copewith that?
Goodluck,
Thijs
Good to hear you got it to work! Can you tell us when was the problem?
Now tp calculate the the Iq
You will have about 28V peak in 4 Ohm, that's 7 ampere. But being a push-pull amplifier, Rod Elliot advises 0.75x the peak current: 0.75 x 7 = 5.25 Ampere..
Now beware: you are dissipation 315 Watts by now.. are you sure have you designed your amp to copewith that?
Goodluck,
Thijs
You can use a capacitance multiplier as shown at http://sound.westhost.com/project15.htm
but add zener diodes to cut down the voltage.
This just spreads the heat across more transistors, as the heat you reduce from the amplifier transistors just goes to the power supply transistors.
Your power supply voltage is too high to run 4-ohm speakers with this amp. Not that it won't work, but the bias current will have to be set low enough so that your transistors don't get too hot. With a 60V supply, and say a 30C temperature rise, you will need to limit the bias current to about 1 amp given the heatsink you're using: 60V * 1A = 60W; 60W * 0.52C/W ~ 30C.
but add zener diodes to cut down the voltage.
This just spreads the heat across more transistors, as the heat you reduce from the amplifier transistors just goes to the power supply transistors.
Your power supply voltage is too high to run 4-ohm speakers with this amp. Not that it won't work, but the bias current will have to be set low enough so that your transistors don't get too hot. With a 60V supply, and say a 30C temperature rise, you will need to limit the bias current to about 1 amp given the heatsink you're using: 60V * 1A = 60W; 60W * 0.52C/W ~ 30C.
A thousand apologies to all! My speaker load impedance is 8ohms. Duh silly silly me! I apologise for the mistake.
With a supply of 58.9V, and having it halved to 29.5V by VR1 and 8ohm load. What kind of Iq should I look at to produce 20watts or somewhere around there?
Would the increased load impedance make it easier to have a lower Iq with 58.9V? Whats the optimal Iq to bias the quiescent current for 58.9V considering my two pieces of 0.52deg C/W heatsinks?
The transistors i'm using are MJ15003 if it helps any. I would like to get a smaller transformer but my budget overstretched and this transformer was from a salvaged project.
With a supply of 58.9V, and having it halved to 29.5V by VR1 and 8ohm load. What kind of Iq should I look at to produce 20watts or somewhere around there?
Would the increased load impedance make it easier to have a lower Iq with 58.9V? Whats the optimal Iq to bias the quiescent current for 58.9V considering my two pieces of 0.52deg C/W heatsinks?
The transistors i'm using are MJ15003 if it helps any. I would like to get a smaller transformer but my budget overstretched and this transformer was from a salvaged project.
Ok....heres what I did. I had a DOZ w/ 8! output transistors in parellel PER channel. PS voltage was 52-53v. I had heatsinks that dissipated 150 watts of heat PER CHANNEL. quiescent current was about 3A. So, I had about 150wpc of heat ...and 50wpc into 8 ohms ...or thereabouts. If you are going to run that high of voltage...I suggest you get more transistors than just the pair you have there....and BIG heatsinks. You might also want to have higher voltage driver transistors. Actually, for the driver transistors I used the same output transistors...I was using 2sc5200's I think. Worked pretty good. Make sure the driver transistors are heatsinked. they'll get hot. Later!
-Matthew K. Olson
-Matthew K. Olson
I would say that you will not get this thing to work long term with such a high voltage. For one thing my understanding is that the quality of the result increases with the quiesent current, so by reducing the current to at least half of its optimal value, it will decrease the quality of the output by half. If you are doing this type of amp for its audio benefits, then that got to be an unacceptable compromise. I run my Zen amp off a 12V AC transformer, and these are very similar amps. 24V would be a much better bet for you.
I would also guess that unless these 60V transformers are massive they might not have the VA rating for the job.
I got my 12V AC transfomer off ebay for about £10, so cost shouldn't be a real issue.
I hope this helps.
Shoog
I would also guess that unless these 60V transformers are massive they might not have the VA rating for the job.
I got my 12V AC transfomer off ebay for about £10, so cost shouldn't be a real issue.
I hope this helps.
Shoog
I gotta stop making silly mistakes like checking up the wrong specifications!
My heatsinks are rated 0.29 degC/W.
Would this work then?
29.5V output into 8ohms = 3.7A
giving 0.75 of 3.7A, Iq = 2.8A
Ambient temp = 60V x 2.8A x 0.29degC/w = 49degC
Could anyone double check and see if my calculations are right? Any corrections would be much appreciated.
What would be the power dissipation and power output be?
My heatsinks are rated 0.29 degC/W.
Would this work then?
29.5V output into 8ohms = 3.7A
giving 0.75 of 3.7A, Iq = 2.8A
Ambient temp = 60V x 2.8A x 0.29degC/w = 49degC
Could anyone double check and see if my calculations are right? Any corrections would be much appreciated.
What would be the power dissipation and power output be?
Hi
I calculate the final temperature like this: x=(P*(a+b+c+d+e))+amb.
Where:
x= junction temperature
p=power in watt
a=junction to case(C/W)
b=case to insulator(C/W)
c=insulator(C/W)
d=insulator to sink(C/W)
e=sink to air(C/W)
amb=room temperature
example: mj15003 with good insulator, 90W and the sink as in the above post I end up with this figures:
x=90(0.7+0.07+0.3+0.07+0.29)+25
Final temperature X=153 C
the above the transistor would fail because when the temperature rises the transistor get "weaker" in the case with MJ15003 the power handling goes down with 1.43 Watts per degree above 25C
250W-(1.43*128(rise)=67W MAX power handling
Keld
(I hope I got my numbers right, anyone feel free to correct them)
I calculate the final temperature like this: x=(P*(a+b+c+d+e))+amb.
Where:
x= junction temperature
p=power in watt
a=junction to case(C/W)
b=case to insulator(C/W)
c=insulator(C/W)
d=insulator to sink(C/W)
e=sink to air(C/W)
amb=room temperature
example: mj15003 with good insulator, 90W and the sink as in the above post I end up with this figures:
x=90(0.7+0.07+0.3+0.07+0.29)+25
Final temperature X=153 C
the above the transistor would fail because when the temperature rises the transistor get "weaker" in the case with MJ15003 the power handling goes down with 1.43 Watts per degree above 25C
250W-(1.43*128(rise)=67W MAX power handling
Keld
(I hope I got my numbers right, anyone feel free to correct them)
Thanks for the help all your guys kindly gave.
I think I will have to add more transistors in parallel in order to reduce the junction temperature. I was thinking of using 6 transistors per channel.
Can anyone help me with the calculations to determine if my 0.29degC/W heatsink is suitable for the voltage rail of 58.9V and Iq of 2.8A with a load impedance of 8 ohms?
Also how to calculate the output power would be useful.
Regards
I think I will have to add more transistors in parallel in order to reduce the junction temperature. I was thinking of using 6 transistors per channel.
Can anyone help me with the calculations to determine if my 0.29degC/W heatsink is suitable for the voltage rail of 58.9V and Iq of 2.8A with a load impedance of 8 ohms?
Also how to calculate the output power would be useful.
Regards
3055
Thanks for the help all your guys kindly gave.
I think I will have to add more transistors in parallel in order to reduce the junction temperature. I was thinking of using 6 transistors per channel.
Can anyone help me with the calculations to determine if my 0.29degC/W heatsink is suitable for the voltage rail of 58.9V and Iq of 2.8A with a load impedance of 8 ohms?
Also how to calculate the output power would be useful.
Regards
Thanks for the help all your guys kindly gave.
I think I will have to add more transistors in parallel in order to reduce the junction temperature. I was thinking of using 6 transistors per channel.
Can anyone help me with the calculations to determine if my 0.29degC/W heatsink is suitable for the voltage rail of 58.9V and Iq of 2.8A with a load impedance of 8 ohms?
Also how to calculate the output power would be useful.
Regards
Parallel Operation of Transistors
I hope this would reduce the junction temperature of the BJT. I intend to use 8 BJTs per channel. Hopefully, that'll decrease the dissipation by 4.
Does anyone know the type and power values of the resistors used for parallel operation of the transistors?
eg 1/4, 1/2 watt metal film/carbon film or 5watt/10watt wire wound?
Would wirewound resistors do? I have trouble finding metal film resistors of 0.1ohm and 0.22ohm.
Regards
Thanks again to all the help. Much appreciated!!
I hope this would reduce the junction temperature of the BJT. I intend to use 8 BJTs per channel. Hopefully, that'll decrease the dissipation by 4.
Does anyone know the type and power values of the resistors used for parallel operation of the transistors?
eg 1/4, 1/2 watt metal film/carbon film or 5watt/10watt wire wound?
Would wirewound resistors do? I have trouble finding metal film resistors of 0.1ohm and 0.22ohm.
Regards
Thanks again to all the help. Much appreciated!!
Attachments
just add them lar... 8 devices. with Imax 3.7amps l, 3.7/8 = 0.46amps per device.
with the emitter resistors being 0R1,
P= Isq*R
= 0.46sq*0.1
= 0.2 watts
having 0.2 watts dissipating on each resistor, it would only be right to double the power rating. so a minium of 0.5w resistors.
if you don`t feel safe, just double it again. get a couple of 1w 0R1 resistors. these are easily avaible from Sim Lim Towers if not RS components... infact, i might have a few lying around...
with the emitter resistors being 0R1,
P= Isq*R
= 0.46sq*0.1
= 0.2 watts
having 0.2 watts dissipating on each resistor, it would only be right to double the power rating. so a minium of 0.5w resistors.
if you don`t feel safe, just double it again. get a couple of 1w 0R1 resistors. these are easily avaible from Sim Lim Towers if not RS components... infact, i might have a few lying around...
hi guys
would this be a good circuit for a DC protection and speaker delay circuit for the DoZ? I think, it isn't necessary since there's a Zobel network, correct me if i'm wrong.
http://www.astro.uu.se/~marcus/private/dcprot.jpg
I have a spare 12V terminal from my soft start circuit, so this might seem easier to implement than Rod's speaker delay circuit.
Regards
would this be a good circuit for a DC protection and speaker delay circuit for the DoZ? I think, it isn't necessary since there's a Zobel network, correct me if i'm wrong.
http://www.astro.uu.se/~marcus/private/dcprot.jpg
I have a spare 12V terminal from my soft start circuit, so this might seem easier to implement than Rod's speaker delay circuit.
Regards
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.