calculating current draw

[seavote]

i'll be starting an m1 preamp build after i wrap up my MK7 build (few weeks?)
the schematic is posted on the hardy web site: page 7 of the PDF
http://www.johnhardyco.com/pdf/M1_M2_M1p_20031025.pdf

could someone school me as to how to determine the current draw of the circuit? i'll be running 2 channels off of one power transformer and would like to make sure i have enough current + some to spare.
i know ohms law but need to know how to apply it. do i just add the current draw of the i.c.s i'll be using?
ive searched for an answer but i'm thick and often need things spelled out for me. thanks for any help.
                                          chris v

 

[Walrus]

The circuit on page 7 has a PSU using 78L15/79L15 regulators. These are 100mA max current. I'd allow for 200mA max with 2 amps, then you can't go wrong!

 

[Samuel Groner]

The circuit on page 7 has a PSU using 78L15/79L15 regulators. I'd allow for 200 mA.

Na. The regulators just power U2 (and RV2) which will explode if it would draw 200 mA (at +/-15 V that's 6 W, right?). U1 is the only thing which draws significant current here and is powered directly from the +/-24 V rails.

Check the quiescent current for the JH-990 in the datasheet, multiply by 3 for the output current and some headroom and there we are (that is per channel).

Samuel

 

[seavote]

thanks for the replies. ive only sourced transformers for tube projects in the past. i'd check the currennt draw from spec sheets and add this together then throw some extra on top of that. it seems you guys are doing the same thing here?  add the current draw from all active components? just wanted to check to see if this is the correct (standard) method. is there a standard % of current that should be added for headroom? any ideas, discussion, or arguments welcome. thanks

 

[Samuel Groner]

In the end you have to understand the circuit, no way around that. For DIY (where cost/space/weight ain't that important) you can be generous and just add ample headroom. For a low-cost commercial product we need to look closer at what the circuit actually draws.

Samuel

 

[PRR]

> Check the quiescent current for the JH-990 in the datasheet, multiply by 3 for the output current and ...

3?

What if some miracle (or just nasty class B) amp pulled zero or insignificant idle current? What if it were Class A and constant-current?

A totem-pole driving a resistive load with a sine: the power supply "feels" a DC load like 6 times the nominal load.

Driving 8 ohms with 24V rms 35V peak: 8*6= 50 ohms, 70V/50r= 1.4A supply current, 70V*1.4A= 100 watts input for 72 Watts output, 73% efficiency, ballpark for sine through good AB/B outputs.

The MPC-1 plan shows a 990 driving unknown impedances. Not greater than 8K: it has to drive its own NFB and the servo. The 990 is rated to drive 75 ohms.... unlikely in today's desks but possible. When driving modern 10K inputs via the optional transformer T2, we must assume the transformer loading falls below 600 ohms for subsonics.

So at +/-24V or ~~50V total supply, FULL-level Sine, we get these dynamic currents for various loads:

5K = 2mA
1K = 8mA
500 = 16mA
100 = 83mA
75 = 111mA

Pick your load, find your dynamic current.

Hardy's 990 is specced 25mA quiescent, add that.

Add quiescent for 7*L15 and LM11.

It looks like 50mA per channel would be ample for any "normal" load. Being abnormal myself, I have tasked similar line-drivers with 75 ohms, and would want 150mA available.

 

[Samuel Groner]

3? What if some miracle (or just nasty class B) amp pulled zero or insignificant idle current? What if it were Class A and constant-current?

As I said above: In the end you have to understand the circuit. My factor (which is actually just a rounded version of pi) is based on pretty sound understanding of the circuit at hand and not meant to be a general rule-of-thumb.

Samuel

 

[mitsos]

According to the 990 specs, no-load current is 25mA (per rail?).  I've built this thing and maybe I should measure current consumption... but it runs just fine on a PSU running a couple of 312 channels  (I'm running the 990 on 18V temporarily).

Hardy has a 500 series twin servo, which has 2x990 and 2 servos and runs on 130mA per rail max (VPR specs, right?). So, can  we ASSume that an M1 would max out at half of that, even though it has the bias circuit, which I am ASSuming the hardy doesn't (it's not on the basic schem on jensen's site)?

 

[tv]

My personal opinion is that you will need "not-L", i.e. the ordinary TO-220 parts to feed it right.

 

[Samuel Groner]

My personal opinion is that you will need "not-L", i.e. the ordinary TO-220 parts to feed it right.

My apologies for repeating myself but the regulators just power the servo and the bias current cancellation circuit. I really doubt that the few mA for it will require TO-220 parts...

Samuel

 

[Walrus]

The circuit on page 7 has a PSU using 78L15/79L15 regulators. I'd allow for 200 mA.

Na. The regulators just power U2 (and RV2) which will explode if it would draw 200 mA (at +/-15 V that's 6 W, right?). U1 is the only thing which draws significant current here and is powered directly from the +/-24 V rails.

Check the quiescent current for the JH-990 in the datasheet, multiply by 3 for the output current and some headroom and there we are (that is per channel).

Samuel

That'll teach me not to look closely at the circuit!
Re the 200mA, I meant for total draw from the transformer for 2 amps. In my mistaken take on the circuit, I was thinking the 7*L regs would have to be duplicated for the second amp, or replaced with beefier devices.
But as you say they don't power the whole circuit.

Mind you, it looks like I arrived at the right answer, but with the wrong workins out!  ;D