Hi all.
I have shown a few design ideas on this forum. One of the reasons I have been testing and simulating many different ideas is that I was trying to find an amplifier that I could build to give low distortion into an 8-ohm load loud enough to fill my room (which is quite small).
This amplifier sims good enough for me to try as a project, and I haven't run into any stability problems.
Simulation results:
THD @ 6W, 8ohms, 1KHz is about .0002%
THD @ 6W, 8ohms, 20KHz is about .0012%
I have enough parts (minus bias resistors and heatsinks) to build two of these for a stereo setup.
However, it still might take me some time to buy the few needed components.
What do you think?
- keantoken
I have shown a few design ideas on this forum. One of the reasons I have been testing and simulating many different ideas is that I was trying to find an amplifier that I could build to give low distortion into an 8-ohm load loud enough to fill my room (which is quite small).
This amplifier sims good enough for me to try as a project, and I haven't run into any stability problems.
Simulation results:
THD @ 6W, 8ohms, 1KHz is about .0002%
THD @ 6W, 8ohms, 20KHz is about .0012%
I have enough parts (minus bias resistors and heatsinks) to build two of these for a stereo setup.
However, it still might take me some time to buy the few needed components.
What do you think?
- keantoken
Attachments
Except those mentioned from the good friend Nelson, from my recent experiments on my project, i have the following remarks:
1. C3-R11 network it is placed usually between the collectors of LTP.
2. The C2 of 100pF needed in the case of using an input attenuation pot (before C5) to suppres the oscillations caused from the interaction between pot and C5, and it reduces high frequencies response. C2 and C6 are the components that affects the high frequency responce. You can play with the values of these two caps to obtain the better result which is small rise time and no Miller oscillations.
You can also decrease the value of C5 down to 2,2ìF without loss in low frequencies if you intend to use some good quality foil polypropylene cap, for having the smallest posible inductance.
Regs
Fotios
1. C3-R11 network it is placed usually between the collectors of LTP.
2. The C2 of 100pF needed in the case of using an input attenuation pot (before C5) to suppres the oscillations caused from the interaction between pot and C5, and it reduces high frequencies response. C2 and C6 are the components that affects the high frequency responce. You can play with the values of these two caps to obtain the better result which is small rise time and no Miller oscillations.
You can also decrease the value of C5 down to 2,2ìF without loss in low frequencies if you intend to use some good quality foil polypropylene cap, for having the smallest posible inductance.
Regs
Fotios
Oh, thanks Mr. Pass. 
I just tried it but didn't see any affect on distortion. I'll leave them in though.
Things I need to know before building this:
1: How large a heatsink I need.
2: What kind of transformer I need. Rail voltage is 18V each, max current draw is 2A.
Also, I solved a really weird problem with two diodes. The output would sometimes short to one rail because of current through the C-B junctions of Q11 and Q12.
See attachment.
Also, fotios, thanks for the advice, especially for C5.
R20 and C3 work well to keep the circuit from oscillating, as I have discovered recently. A great many of my old designs should be able to become more stable as a result of my recent learnings.
Thanks all,
- keantoken
I just tried it but didn't see any affect on distortion. I'll leave them in though.
Things I need to know before building this:
1: How large a heatsink I need.
2: What kind of transformer I need. Rail voltage is 18V each, max current draw is 2A.
Also, I solved a really weird problem with two diodes. The output would sometimes short to one rail because of current through the C-B junctions of Q11 and Q12.
See attachment.
Also, fotios, thanks for the advice, especially for C5.
R20 and C3 work well to keep the circuit from oscillating, as I have discovered recently. A great many of my old designs should be able to become more stable as a result of my recent learnings.
Thanks all,
- keantoken
Attachments
keantoken said:Oh, thanks Mr. Pass.
I just tried it but didn't see any affect on distortion. I'll leave them in though.
Things I need to know before building this:
1: How large a heatsink I need.
2: What kind of transformer I need. Rail voltage is 18V each, max current draw is 2A.
Also, I solved a really weird problem with two diodes. The output would sometimes short to one rail because of current through the C-E junctions of Q11 and Q12.
See attachment.
Also, fotios, thanks for the advice, especially for C5.
R20 and C3 work well to keep the circuit from oscillating, as I have discovered recently. A great many of my old designs should be able to become more stable as a result of my recent learnings.
Thanks all,
- keantoken
For Vcc=18Vdc obtained from full wave bridge rectifier, you can add 1,5 Volt drop from bridge diodes, as well yet 1,5 Volt from ripple (if you are not using a regulated supply), thus you need Vcc = 18V + 1,5V + 1,5V = 21Vp which translated in a transformer with secondary of 21V/1,41 = 14,89Vac thus 15Vac. For a split power supply you need a xformer with 2 X 15Vac secondaries @2A.
Now for the power, you said 2A... i don't know if are peak or rms. Let us to say rms. For Vsec = 2 X ( 2 X 15 ) = 60 Vac and current Isec = 2 X 2 Arms = 4 Arms, the core of transformer must be rated at: P = V X I X cosö (where ö is the phase angle between V and I due to inductive load of xformer and for 45 deg mean phase shift cosö = 0,707) = 60V X 4A X 0,707 = 169,68VA. To be safe, you must get a: 200VA xformer with 2 X ( 2 X 15 Vac ) secondaries made from wire rated for 2A. If you have understand, i talk about a xformer with individual supplies per each channel to avoid the gnd loop caused from the common gnd node. The price is the same. Or you can use two seperate xformers one per each channel. In this case, you need two 100VA rated xformers. All the rest are same.
Keane, i forgot to refer that. Have you seen for OnSemi D44H11-D45H11 for output? They are far better from MJE340-350 in this place. Take a look please on datasheet of OnSemi.
Regs
Fotios
Thanks for the info, fotios. I was meaning 2A peak, not RMS. My preference would be two toroids, one for each channel...
Are you sure? The D4x devices are rated for 10A while the MJE350s are .5A. Overkill! I only need 6W, and I'm afraid if I go higher my breadboard will explode.
- keantoken
fotios said:
Are you sure? The D4x devices are rated for 10A while the MJE350s are .5A.
Overkill! I only need 6W, and I'm afraid if I go higher my breadboard will explode. - keantoken
keantoken said:Thanks for the info, fotios. I was meaning 2A peak, not RMS. My preference would be two toroids, one for each channel...
Are you sure? The D4x devices are rated for 10A while the MJE350s are .5A.
- keantoken
My calculations for 2 Arms consumption of course results in bigger values, but the thing is that the ready offered toroids are rated usually in standard cores of 50VA - 100VA - 150VA - 200VA etc. You can't use 50VA xformer (per channel) it is small, thus you are going again in two 100VA toroids. Moreover the extra cost it is small. Here in Greece, it is only 5 euros extra.
As for D44, because are rated for 10A does not means that they consume 10A! Instead they sustain 10A. I use the dpak versions MJD44H8-45H8 as output devices, in my small operational amplifier implemented with smd discrete devices. It operates with +/-24Vdc and it consumes only few mA. No problem friend, don't be afraid. Simply those are modern and much better devices from the old good MJE340-50.
Regs
Fotios
fotios said:
The D4x devices aren't suited for this amplifier because of the high Cob. It is about 100 times higher than the MJE340/350. The gain is also lower. I don't believe the MJE340/350 are inferior.
MJL21193 said:
...Which will have lower Ft, higher Cob and will provide higher HF distortion. The current setup has high gain and high Ft. Plus, I don't want to be wasting 10A transistors on a 6W amplifier...
I will bredboard this first, to check that it works and then I will put it in a case. This is not something that I expect to throw together in 15 minutes.
- keantoken
In some of the old CRT's you said were in the "hood" ,they
use real transistors for the vertical output section , usually
rated 2-3A , and usually faster 2SB/2SD-XXX devices, too.
Just a couple pair would do.
Older compact stereo systems also use 2-5A PNP/NPN
to-220 pairs and have a decent heatsink attached as well.
Find some scrap.
I've built the "big boys" on a breadboard, just up to the
VAS.. put the outputs ,drivers and emitter resistors offboard,
as that much current will fry the contacts.
OS
ostripper said:
This is excellent advice. I have picked up receivers at Goodwill that have a ton of usable parts, some of these rare Japanese transistors. Transformers, heatsinks, screws, knobs, switches - all good usable stuff, enough out any old receiver to build a low power class A amp for next to nothing.
Well, I just got through ripping apart one of those monitors.
It had been sitting outside for a while, so the plastic was brittle and easy to rip off.
There were several loops of wire at the sides of the front of the CRT, which are being used for my FM aerial now.
I salvaged some of the CRT positioning coils to see if I could do something with them. Might do something with the flyback.
This was an Apple monitor, seemingly to be used with a Mac. Fairly easy and straightforward to take apart, unlike my last one. Some small heatsinks which I should be able to use to make a small headamp version of this. If only I had headphones. Wait though, why does anyone need a headphone amp? Don't they get loud enough?
Then there are a few trafos, some weird magnetoelectric thingies that should serve some purpose. Variable trafos, nice. Toroid RF chokes, which I have used to good effect on interfering appliances.
There were more goodies on my last monitor, though. I think it was a CTX. The inside was obviously sophisticated. Problem was that the RF transistors would melt before the solder. 
There are also HV caps and high wattage resistors.
some 2SC devices, but the main deflection transistor was an MJW something.
- keantoken
It had been sitting outside for a while, so the plastic was brittle and easy to rip off.
There were several loops of wire at the sides of the front of the CRT, which are being used for my FM aerial now.
I salvaged some of the CRT positioning coils to see if I could do something with them. Might do something with the flyback.
This was an Apple monitor, seemingly to be used with a Mac. Fairly easy and straightforward to take apart, unlike my last one. Some small heatsinks which I should be able to use to make a small headamp version of this. If only I had headphones. Wait though, why does anyone need a headphone amp? Don't they get loud enough?
Then there are a few trafos, some weird magnetoelectric thingies that should serve some purpose. Variable trafos, nice. Toroid RF chokes, which I have used to good effect on interfering appliances.
There were more goodies on my last monitor, though. I think it was a CTX. The inside was obviously sophisticated. Problem was that the RF transistors would melt before the solder.
There are also HV caps and high wattage resistors.
some 2SC devices, but the main deflection transistor was an MJW something.
- keantoken
Hi,
16Vdc rails should allow around 11Vpk output into your 8r0 load. I suspect you'll get a bit more.
11Vpk = 7.5W and = 1.4Apk.
The bias of a ClassA push pull amplifier must be set to around 700mA to achieve that 7.5W of classA.
Each output device will pass 700/4 = 175mA of bias current in the quiescent state.
Look at the voltage drop on that 4r7 collector resistor on the first output device pair.
822mV across the the Base Emitter junction OOPS
your protection scheme prevents you setting up the bias current. even if you reduce the bias current to get ClassAB in the output stage the protection will start to trigger @ ~ 85mA. That's just 340mA to the 8ohm load. That's the protection triggering at less than 1W output when in ClassAB and preventing operation in ClassA.
Learn to design on the back of a postage stamp before building a prototype.
Put in those emitter resistors.
Nelson told you, do it.
BTW,
an 8ohm severe reactance speaker can draw peak currents approaching 3.9Apk on fast transients from 11Vpk.
Your output stage must be designed to meet this demand or it's not going to sound nice. Yes, it goes into ClassAB on transient peaks, but it'll sound better if that capability is built in.
4pair of MJE can just about manage this if you keep them cold.
16Vdc rails should allow around 11Vpk output into your 8r0 load. I suspect you'll get a bit more.
11Vpk = 7.5W and = 1.4Apk.
The bias of a ClassA push pull amplifier must be set to around 700mA to achieve that 7.5W of classA.
Each output device will pass 700/4 = 175mA of bias current in the quiescent state.
Look at the voltage drop on that 4r7 collector resistor on the first output device pair.
822mV across the the Base Emitter junction OOPS
your protection scheme prevents you setting up the bias current. even if you reduce the bias current to get ClassAB in the output stage the protection will start to trigger @ ~ 85mA. That's just 340mA to the 8ohm load. That's the protection triggering at less than 1W output when in ClassAB and preventing operation in ClassA.
Learn to design on the back of a postage stamp before building a prototype.
Put in those emitter resistors.
Nelson told you, do it.
BTW,
an 8ohm severe reactance speaker can draw peak currents approaching 3.9Apk on fast transients from 11Vpk.
Your output stage must be designed to meet this demand or it's not going to sound nice. Yes, it goes into ClassAB on transient peaks, but it'll sound better if that capability is built in.
4pair of MJE can just about manage this if you keep them cold.
Thanks Andrew.
Q11 and Q12 aren't protection. They set the bias. Note that the emitters are not connected to ground or output.
With the 4.7 ohm resistors this gives about 150mA on the first output devices but slightly more on the others. This gives about 600mA of bias.
My concern here is not squeezing the most power I can out of what I have. I am more interested in making a low-THD circuit that sounds good and is loud enough. 10V is more than 6W.
The diodes on the collectors of Q11/Q12 can go. I figured out that the problem wasn't in the output arrangement, but with the LTP. Something's going wonky and I'm not sure what. If I shift parameters sometimes the output will stick to one rail and stay there. One end of the LTP is completely off, the other full on, drawing 3mA from the base of the VAS.
- keantoken
Q11 and Q12 aren't protection. They set the bias. Note that the emitters are not connected to ground or output.
With the 4.7 ohm resistors this gives about 150mA on the first output devices but slightly more on the others. This gives about 600mA of bias.
My concern here is not squeezing the most power I can out of what I have. I am more interested in making a low-THD circuit that sounds good and is loud enough. 10V is more than 6W.
The diodes on the collectors of Q11/Q12 can go. I figured out that the problem wasn't in the output arrangement, but with the LTP. Something's going wonky and I'm not sure what. If I shift parameters sometimes the output will stick to one rail and stay there. One end of the LTP is completely off, the other full on, drawing 3mA from the base of the VAS.
- keantoken
150mA of bias current sets the pair of transistors to ~705mVbe.
Now swing some output signal/current.
Let's allow +-75mA through each output device for 600mA of total output ~=1.4W.
The currents in the collector resistors are now 225mA and 75mA.
the volts drop across the pair are 1.058V and 0.353V giving a total of 1.41V for the pair of biasing transistors. How do these transistors apportion their base voltage requirement to allow Ic of 225mA and 75mA to pass to the emitters?
If the transistors were truly linear and gave a current that is proportional to the Vbe then all is OK. But transistors do not give proportionality of Ic/Ie with Vbe. They are generally off at ~400mVbe and full on (saturated) @ ~1000mVbe.
Now swing some output signal/current.
Let's allow +-75mA through each output device for 600mA of total output ~=1.4W.
The currents in the collector resistors are now 225mA and 75mA.
the volts drop across the pair are 1.058V and 0.353V giving a total of 1.41V for the pair of biasing transistors. How do these transistors apportion their base voltage requirement to allow Ic of 225mA and 75mA to pass to the emitters?
If the transistors were truly linear and gave a current that is proportional to the Vbe then all is OK. But transistors do not give proportionality of Ic/Ie with Vbe. They are generally off at ~400mVbe and full on (saturated) @ ~1000mVbe.
It's hard to understand your explanation because you don't refer to transistor numbers, but I'm assuming you're talking Q11 and Q12 when you say "biasing transistors".
The emitters of Q11 and Q12 are not attached to ground or anything, so there will be equal Vbe for both transistors. This gives about 1.3V across R5 and R6, regardless of whether the resistor currents are equal.
Vcb for both Q11 and Q12 will always be the same value depending on the darlington Vbe's.
- keantoken
The emitters of Q11 and Q12 are not attached to ground or anything, so there will be equal Vbe for both transistors. This gives about 1.3V across R5 and R6, regardless of whether the resistor currents are equal.
Vcb for both Q11 and Q12 will always be the same value depending on the darlington Vbe's.
- keantoken
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