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DIY mesa boogie mark iic+

Pel22

Nov 11, 2016
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Nov 11, 2016
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Hello, I'm making a mark iic+ guitar amp, I need some help with the current draw, and tube bias.

In the attached schematic (don't mind the reverb circuit or the graphic equalizer circuit), there are 2 input voltage connections, they are named C and D in the schematic. C's voltage is 397v and D's voltage is 392v. I'm planning to get these voltages from 230v AC using a transformer and a rectifier. The problem is, I don't know how much current I should input (I'll burn the tubes if I get the current draw wrong).

Another problem I've got is the tube bias. The schematic has got some voltages written down on some tubes (beside some of its 3 connections). Do I need to setup certain things to get these voltages or are they generated automatically when I input 397v into C and 392v into D?

I really don't want to burn the tubes.

Thanks
 

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duke37

Jan 9, 2011
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I cannot read the diagram properly but it looks as if the tubes are low power amplifiers such as ECC83. They all have cathode resistors which will set the current. You can look up the tube characteristics to calculate the current but if you assume say 100V on the anodes, it is easy to estimate the current. (I=V/R)

390V seems far too high, I would think 250V would be more reasonable. The current will be under 10mA in each tube so the total current drain will be low.
390V is probably used because the power amplifier needs this.

What are the tube types?
The current 'draw' depends on the circuit shown and not on the power supply, this needs to provide the correct voltage at a larger current than the tubes take.

The power supply can be a transformer,and bridge rectifier as you say but will need smoothing with a reservoir capacitor (33μF) a dropper resistor (1kΩ?) and another 33μF capacitor. The capacitors should be capable of higher voltage than is being used, 390V on a 400V capacitor I would say is too much.

Note that most resistors these days are limited to less than 100V. You could use big1W or 2W resistors to get the voltage capability.
 

Pel22

Nov 11, 2016
3
Joined
Nov 11, 2016
Messages
3
I cannot read the diagram properly but it looks as if the tubes are low power amplifiers such as ECC83. They all have cathode resistors which will set the current. You can look up the tube characteristics to calculate the current but if you assume say 100V on the anodes, it is easy to estimate the current. (I=V/R)

390V seems far too high, I would think 250V would be more reasonable. The current will be under 10mA in each tube so the total current drain will be low.
390V is probably used because the power amplifier needs this.

What are the tube types?
The current 'draw' depends on the circuit shown and not on the power supply, this needs to provide the correct voltage at a larger current than the tubes take.

The power supply can be a transformer,and bridge rectifier as you say but will need smoothing with a reservoir capacitor (33μF) a dropper resistor (1kΩ?) and another 33μF capacitor. The capacitors should be capable of higher voltage than is being used, 390V on a 400V capacitor I would say is too much.

Note that most resistors these days are limited to less than 100V. You could use big1W or 2W resistors to get the voltage capability.

Thanks for your reply!

The tubes are indeed ECC83 (12AX7), they are dual triode tubes.
The ECC83 tubes have a max. anode voltage of around 310v.

I am going to input around 250v into connections C and D from the power supply. So, how much current do you think should be coming from the power supply?

I will make the power supply using 2W resistors.
 

duke37

Jan 9, 2011
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As I said, I cannot read the digram properly but assuming an anode resistor of 100k and 100V across it, that is 1mA. Four valves equals 4mA. You can calculate better if you know the valve voltages.
Make a supply capable of 20mA or so to give plenty of leeway.
 

Pel22

Nov 11, 2016
3
Joined
Nov 11, 2016
Messages
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As I said, I cannot read the digram properly but assuming an anode resistor of 100k and 100V across it, that is 1mA. Four valves equals 4mA. You can calculate better if you know the valve voltages.
Make a supply capable of 20mA or so to give plenty of leeway.

Alright, thanks a lot! I'll start soldering soon.
 
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