Specifying ICL Thermistors
1) Max desired in rush current
2) Energy requirements
3) Steady state current requirements
4) Desired hot working resistance at steady state current
1) Assuming 240V AC rms
10 Ohms thermistor gives 24A rms inrush current
20 Ohms thermistor gives 12A rms inrush current
2) Assuming 100,000uF 0.1R 100,000uF per rail, with 25 volt rails
It is my belief that 90% (guesstimation here) of the work for the thermistor is taken up by the caps before the resistor since those appear as a dead short, the caps after the resistor are charged up in a much softer manner. We will calculate for both scenarios.
Each secondary is loaded by 100,000uF as a dead short so 200,000uF total (that’s 0.2F)
E = (CV^2)/2
= (0.2 x 25^2)/2
= 62.5 Joules
That is being fearless though, if we assume worst case scenario and use total capacitance as dead shot then simply double this figure which is 125 Joules, then allow some overhead if you want the part to last a reasonable amount of time.
3) Steady state current requirements.
This is where I rely on specs of CL60 and Papa’s use in Firstwatt amps. I won’t be using more than 2.6A per mono block. My steady state current requirements will be less.
I am not sure how this is specified. If the steady state current draw is 2A DC by the Amp, the charging currents on the caps provided by the secondary windings on the transformer fed through the rectifier diodes will be much higher but the duty cycle is obviously small in time. If the secondary’s are 19V then the currents spikes will be 19/240, seen on the primaries and seen by the thermistor. However converting a pulsed current with short duty cycle to “steady state current”, is not something I am confident of. I just refer to Papa here given my monoblocks are not drawing more than his stereo amps.
4) Refer to IV curve or RI curve, of data sheet etc, to be sure the resistance figure is ok at steady state.
I am using MegaSurge Series.
MegaSurge - Inrush Current Limiters | Ametherm
1) Max desired in rush current
2) Energy requirements
3) Steady state current requirements
4) Desired hot working resistance at steady state current
1) Assuming 240V AC rms
10 Ohms thermistor gives 24A rms inrush current
20 Ohms thermistor gives 12A rms inrush current
2) Assuming 100,000uF 0.1R 100,000uF per rail, with 25 volt rails
It is my belief that 90% (guesstimation here) of the work for the thermistor is taken up by the caps before the resistor since those appear as a dead short, the caps after the resistor are charged up in a much softer manner. We will calculate for both scenarios.
Each secondary is loaded by 100,000uF as a dead short so 200,000uF total (that’s 0.2F)
E = (CV^2)/2
= (0.2 x 25^2)/2
= 62.5 Joules
That is being fearless though, if we assume worst case scenario and use total capacitance as dead shot then simply double this figure which is 125 Joules, then allow some overhead if you want the part to last a reasonable amount of time.
3) Steady state current requirements.
This is where I rely on specs of CL60 and Papa’s use in Firstwatt amps. I won’t be using more than 2.6A per mono block. My steady state current requirements will be less.
I am not sure how this is specified. If the steady state current draw is 2A DC by the Amp, the charging currents on the caps provided by the secondary windings on the transformer fed through the rectifier diodes will be much higher but the duty cycle is obviously small in time. If the secondary’s are 19V then the currents spikes will be 19/240, seen on the primaries and seen by the thermistor. However converting a pulsed current with short duty cycle to “steady state current”, is not something I am confident of. I just refer to Papa here given my monoblocks are not drawing more than his stereo amps.
4) Refer to IV curve or RI curve, of data sheet etc, to be sure the resistance figure is ok at steady state.
I am using MegaSurge Series.
MegaSurge - Inrush Current Limiters | Ametherm
Last edited:
Probably should have added this for reference.
Thermistor Characteristics of CL60
CL60 (R vs I)
R = 1.08 Ohms, at 1.25A
R = 0.44 Ohms, at 2.5A
R = 0.26 Ohms, 3.75A
R = 0.18 Ohms, at 5.0A
Calculating voltage drops for each of those conditions
1.25 x 1.08 = 1.35V
2.5 x 0.44 = 1.1V
3.75 x 0.26 = 0.975
5 x 0.18 = 0.9V
On 240V mains that represents less than a 0.5% drop under nearly all operating conditions (except the first which is marginally over 0.5%)
It might even be better than this since the ambient temperature of a Class A amp is obviously greater than 25C so that may drop the resistance figure further, given they have a negative temperature coefficient. Either way 0.5% is 3/5 of FA.
I wanted to choose a thermistor with similar resistance characteristics.
The Ametherm specs for the thermistor I am using MS22 20005, is
0.48 Ohms at 2.5 A
0.19 Ohms at 5.0 A
Equivalent to 1.2V and 0.95V drop (respectively) on 240V AC mains, again 0.5% or better, ie much less significant to the loss associated with the rectifier diodes after the secondaries which is a loss of around 1/25 or 4% (just for context), 1/25 since my rails are probably around 25V, and 1 V loss on the diodes (approximately)
So the Ametherm thermistor I am using has very similar resistance characteristics to CL60 under operating conditions, but has a 20 Ohm soft start, and 180J rating, compared with a 10 Ohm soft start, and 36J rating for CL60.
Thermistor Characteristics of CL60
CL60 (R vs I)
R = 1.08 Ohms, at 1.25A
R = 0.44 Ohms, at 2.5A
R = 0.26 Ohms, 3.75A
R = 0.18 Ohms, at 5.0A
Calculating voltage drops for each of those conditions
1.25 x 1.08 = 1.35V
2.5 x 0.44 = 1.1V
3.75 x 0.26 = 0.975
5 x 0.18 = 0.9V
On 240V mains that represents less than a 0.5% drop under nearly all operating conditions (except the first which is marginally over 0.5%)
It might even be better than this since the ambient temperature of a Class A amp is obviously greater than 25C so that may drop the resistance figure further, given they have a negative temperature coefficient. Either way 0.5% is 3/5 of FA.
I wanted to choose a thermistor with similar resistance characteristics.
The Ametherm specs for the thermistor I am using MS22 20005, is
0.48 Ohms at 2.5 A
0.19 Ohms at 5.0 A
Equivalent to 1.2V and 0.95V drop (respectively) on 240V AC mains, again 0.5% or better, ie much less significant to the loss associated with the rectifier diodes after the secondaries which is a loss of around 1/25 or 4% (just for context), 1/25 since my rails are probably around 25V, and 1 V loss on the diodes (approximately)
So the Ametherm thermistor I am using has very similar resistance characteristics to CL60 under operating conditions, but has a 20 Ohm soft start, and 180J rating, compared with a 10 Ohm soft start, and 36J rating for CL60.
Last edited:
Your equations (1) and (2) are what I ran. I've got about 1/2 the capacitance but double the voltage; since V is ^2 I need about 2X your energy requirement. That means the one I picked is barely big enough if being fearless, but easily big enough if only considering the first set of capacitors.
I did find another Ametherm datasheet which stipulated that the rated energy was 1/2 the destruction energy, so that provides some comfort that there's already some level of margin built in....
I did find another Ametherm datasheet which stipulated that the rated energy was 1/2 the destruction energy, so that provides some comfort that there's already some level of margin built in....
Let's see the difference for a simple example of a dual rail supply calculating both ways.
Assuming +/-20V rails and two 10,000uF caps ie just one for each rail.
Calculating by the original method shown above we have a total of 20,000uF, ie 0.02F
E= CV^2
= 0.02 x 20^2
= 8J
Calculating by rail to rail, we now have a total capacitance 1/C = 1/C + 1/C since the caps are in series.
So 1/C = 1/10,000 + 1/10,000
C = 5000uF ie 0.005F
But now we use 40V as the voltage drop
So
E = 0.005 x 40^2
= 8J
So you can do it either way but using rail to rail requires calculating for caps in series.
These things are quite robust, going on my experience with CL60.
I was previously using over 200,000uF per rail with 36J rated CL60, but my cap arrangement was CCRCCCCCC.
You could possibly also use a thermistor as the resistor in the CRC, eg CThermistorC, to reduce the work done by the thermistor on the transformer. Just find one that has the appropriate resistance value at your bias level.
Either way I think you will be fine.
The R in the CRC certainly does soften the load for the thermistor somewhat.
Last edited:
For determing steady state current I considered that it would be reasonable to consider the steady state load of the amp as a function of power then using P = VI to determine steady state current on the primary.
If the steady state power of the amp was say 360W and mains power was 240V, then using P=VI
360=240 x I
I = 1.5A rms
Even though in reality it is going to be much larger but short in duration current spikes, I figured the method above is probably reasonable.
Last edited:
Yeah jfets on front end.
Then laterals without degeneration
Then source follower buffer laterals without degeneration
Then quasi output stage.
If we bias up laterals on 2nd gain stage without degeneration say 100mA we can do away with buffer driver stage.
Edit: You can see I have to try and simplify everything.
Then laterals without degeneration
Then source follower buffer laterals without degeneration
Then quasi output stage.
If we bias up laterals on 2nd gain stage without degeneration say 100mA we can do away with buffer driver stage.
Edit: You can see I have to try and simplify everything.
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.