
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Solid State Talk all about solid state amplification. 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
21st February 2004, 08:07 PM  #1 
diyAudio Member
Join Date: Feb 2004
Location: Sheffield

Basic question
Hi all,
You may not be the best people to ask about that as it is a little bit basic, and probably sounds like a stupid question but... I understand that with bipolar output stages, given the maximum current flowing through the device, one can work out the maximum base current required of the previous amplifier stage using a minimum value of hfe. This makes sense to me but... if you are using a MOSFET output stage, driven by a common emitter amplifier stage using a current source, MOSFETs do not have a base current or hfe as such, as they conduct according to the voltage at the gate, not current, so how is the required current in this previous section worked out. I gather that it has something to do with charging the input capacitance of the device quickly enough, but this makes little or no sense to me. This is for a project that i have to do, so super high end audio is not exactly on the cards Any help or pointers in the right direction would help thanks Tom 
22nd February 2004, 11:43 AM  #2 
diyAudio Member
Join Date: Nov 2003
Location: Brighton UK

To change the voltage of the mosfet you need to charge the
capacitance and to change it back you need to discharge the capacitance so an AC current flows, but no net DC current. Q=CV, and Q which = charge = current x time. Though normally you look at the impedance of the capacitor at a specific frequency to work out the AC current. Fet capacitance is in the order of 1nF, for 30V AC at 20KHz, (around a 100W/8ohm) this needs ~ 4mA AC current to drive it in a unity gain output stage. For the above the driving stage current needs to be at least 5mA. And a lot more for good performance at 100Khz. sreten. 
22nd February 2004, 12:05 PM  #3 
Account Disabled
Join Date: Apr 2003
Location: US

the fet's equivalent of hfe is gm. but that doesn't apply to the gate current.
I suspect it depends on what devices you use. my simulation with irf540 type devices (2000pf Cgs) seems to suggest that they suck in abou 600ua  1ma of current at 20khz. it is not a good idea to size your driver stage current close to that figure, tho. so I usually run my driver stage at 10x of the gate current. 10ma for the driver is a minimal for me. I suppose that mosfets with higher Cgs will require hotter driver; as does wider frequency; or more output devices; 
22nd February 2004, 12:07 PM  #4 
diyAudio Member
Join Date: Jan 2002
Location: Earth

It is a little hard to picture what's going on in these devices but it can be modelled fairly simply.
First, let's assume everything is LINEAR. A bipolar has a baseemitter impedance that looks much like a resistor in parallel with a capacitor. The collector current (Ic) will be related to the portion of the baseemitter current (Ib) that flows through the resistive element. At low frequencies Ib mostly flows through the resistive element. At very high frequencies the capacitive element steals some of Ib and causes a phase shift between Ic and Ib. The frequency at which so much current flows through the capacitive element that Ic has the same amplitude as Ib is called the cutoff frequency or fT. A MOSFET is similar except that the resistive element can be considered infinite and the capacitive element is, all other things being equal, much larger. So right from the start Id is related to the integral of Ig because Id is related to Vgs. The current flowing into a capacitor is I=CdV/dt. Capacitance in farads multiplied by the rate of change of voltage across the capacitor in volts/second. The upshot is that a mosfet device demands more current from the driver stage as the frequency of the signal increases from dc. A bipolar requires a steady driver current up to a certain frequency when it begins to need more current with frequency like the mosfet. For low frequency systems mosfets are very good because they require less drive current capability. Ex: you require a 10A peak Id at 20kHz sinewave. The Gm is a constant 5 siemens and Cgs is a constant 3nF, Cdg=0. The gate voltage will be a 2Vpk sinewave, given by Vgs = 2sin(2pi.20000.t). This needs to be differentiated to give the rate of change of Vgs, dVgs/dt = 4pi.20000.cos(2pi.20000.t). The maximum value of dVgs/dt is when cos(...)=1 or when dVgs/dt=4pi.20000 or 251kV/s. Now the peak Ig will be this number multiplied by the Cgs, giving the max Ig = 754uA. The other problem driving either device is that the Ib or Ig will include Icb and Idg respectively. These two junctions are also capacitive, moreso for the mosfet, and a current related to the rate of change of voltage across these junctions will also need to be supplied. The size of this current depends entirely on the topology. But you just need to apply the same calculation to these junctions. When you take NONLINEARITIES into account everything changes. Both bipolar and mosfets are very nonlinear, but in different ways. The scientific artistry of good audio design is to understand what linearity characteristic is best in a particular circuit. There is no single answer for this. The hfe, the resistive element, the capacitances all change their value with current and voltage. This is what makes design difficult. 
22nd February 2004, 12:13 PM  #5 
diyAudio Member
Join Date: Nov 2003
Location: Brighton UK

my quick calculation is wrong, sorry .......
(I thought it looked too high) sreten. 
22nd February 2004, 12:29 PM  #6 
diyAudio Member
Join Date: Feb 2004
Location: Sheffield

Thanks
Thanks for all the help on this,
I suppose in many ways this will be a case of trying values to see what gives the best results  like pretty much everything else on the amp. Having said this though, i think it is quite important for me to understand the scientific principals behind the operation of these amps... more useful stuff to write in my report too cheers Tommy 
22nd February 2004, 03:08 PM  #7  
diyAudio Member
Join Date: Jun 2003
Location: Connecticut

Quote:
Input capacitance = Cgs + (1+Av) * Cdg Not so significant in a unity gain output stage, but very significant in high speed switchers.
__________________
dave 

Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Really basic TL question  weinstro  Full Range  3  11th October 2007 05:16 AM 
amp question basic rec  rebith75  Solid State  3  16th March 2007 05:43 PM 
Basic RC Value Question  SRMcGee  Tubes / Valves  6  13th August 2006 05:39 PM 
really basic question  cuallito  Tubes / Valves  6  15th April 2004 08:25 PM 
really basic question  MbayAQ  Digital Source  3  7th January 2003 12:57 PM 
New To Site?  Need Help? 