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PostPosted: 01 Aug 2018, 00:32 
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Joined: 30 Jul 2018, 00:29
Posts: 4
Location: Mumbai, IN
Hi Gio,
I am new here but not new to DIYing. Stumbled into this awesome thread while searching for a class A HP amp.
Anyway, coming to the point, I want to know whether the 4.7K and 150R resistors in the input signal path are critical for their values or can I exchange them with like 5.6K & 100R respectively.


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PostPosted: 02 Aug 2018, 00:55 
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Joined: 26 Dec 2016, 03:46
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Location: Bayarea
anuragn wrote:
Hi Gio,
I am new here but not new to DIYing. Stumbled into this awesome thread while searching for a class A HP amp.
Anyway, coming to the point, I want to know whether the 4.7K and 150R resistors in the input signal path are critical for their values or can I exchange them with like 5.6K & 100R respectively.


You refer to the schematic of the first post? Both 4.7K and 150 resistors are driving the gate of the MOSFET. It should be OK to change to 5.6K and 100. That's for isolating the gate of the MOSFET from the input to avoid oscillation.


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PostPosted: 02 Aug 2018, 04:20 
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Joined: 30 Jul 2018, 00:29
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Location: Mumbai, IN
Yungman wrote:
You refer to the schematic of the first post? Both 4.7K and 150 resistors are driving the gate of the MOSFET. It should be OK to change to 5.6K and 100. That's for isolating the gate of the MOSFET from the input to avoid oscillation.


Yes the first page. Thanks for clearing my doubt.


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PostPosted: 02 Aug 2018, 12:13 
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Joined: 26 Dec 2016, 03:46
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Location: Bayarea
anuragn wrote:
Yungman wrote:
You refer to the schematic of the first post? Both 4.7K and 150 resistors are driving the gate of the MOSFET. It should be OK to change to 5.6K and 100. That's for isolating the gate of the MOSFET from the input to avoid oscillation.


Yes the first page. Thanks for clearing my doubt.


People put series resistor to protect the input of the transistor of amplifiers to limit the current to the input, also the resistance with the input capacitance of the transistor to form a low pass filter to prevent radio frequency from radios from getting in.

But in this case, it is not protecting the input as the gate of the MOSFET does not draw current. The really protect the input, you need protection diodes. But that's a different story for a different time.


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PostPosted: 16 Sep 2018, 04:45 
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Joined: 30 Jul 2018, 00:29
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Location: Mumbai, IN
I have constructed the amp and it sounds awesome with mobile phone as source. I am running it at 20 volts as it sounds better than at lower voltages like 15v. The only issue is that the mosfets and 317s become very hot. Need to mention that avg day temperature is usually around 25-30℃ here in Mumbai. So thinking of biasing the transistors at lower current like say 160ma instead of 250ma. Will it affect the sound quality in any way? Or shall I go ahead with this mod?


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PostPosted: 16 Sep 2018, 22:47 
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Joined: 23 Feb 2017, 02:02
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0.16Amp biasing should be more than enough for16r headphones or higher
for 8r headphones you want larger output capcitor and lower operating voltages with 0.25ma of biasing


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PostPosted: 17 Sep 2018, 00:18 
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Joined: 30 Jul 2018, 00:29
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Location: Mumbai, IN
Thanks ilovehifi. My headphones are 40r (Audio Technica). I'll try and configure for 170ma by using 3 of 22R in parallel. Maybe I'll increase the supply voltage to 22 or 23. This should also help in reducing the thermal load on Lm317 located in the PSU.


Last edited by anuragn on 17 Sep 2018, 01:09, edited 1 time in total.

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PostPosted: 17 Sep 2018, 00:51 
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Joined: 23 Feb 2017, 02:02
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The peak class a output power to headphones class a would be I^2R = current*current*resistance

Say your biasing at 0.1amps then at max swing current is 0.2amps class a idle at half max current
So then 0.2*0.2*40=1.6W of peak class a power into 40r
and RMS power is half of peak power = 0.8w rms


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PostPosted: 17 Sep 2018, 15:18 
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Joined: 23 Feb 2017, 02:02
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Actually because the amp is set to draw a constant current from lm317.
So at twice idle current, half of it goes into lm317 and another half into speaker

So if your biasing at 0.1amps then output power into 40r class a would be 0.1^2*70=0.7W peak = 0.35W rms

If you do the derviation with RMS current and voltage you will get 0.35w rms which is half of peak power
you get voltage first by V=IR
0.1*40 =x volts

then you can find rms voltage current by simply v or I / √2


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