Build a current source, somewhere between 100ma and 1a. Then just measure the voltage across the resistor (~22mv-1V, depending on resistor value and the current you chose), plop the values into ohm's law and...
You don't even need to calibrate the current source or use ohm's law to find the actual value, if you're just matching emitter resistors. Just match voltages.
You don't even need to calibrate the current source or use ohm's law to find the actual value, if you're just matching emitter resistors. Just match voltages.
Q1 can be any medium/high power transistor. Q2 can be almost anything. R1 should be between 0.1 and 10. R2 should be Vcc/((0.65/R1)/Hfe'q1)*2, or around 10 to 100 times larger.
P.S. Sorry about the image quality, I compressed it rather quickly without thinking and compressed it into nothing (20kb from 420kb).
P.S. Sorry about the image quality, I compressed it rather quickly without thinking and compressed it into nothing (20kb from 420kb).
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ok Tim thanks I think I will not blown more transitors with this circuit.its enought at this month.
am I too naive to thinking this way- I solder input terminals for source resitor can be solder/desolder without dismounting pcb from heatsinks. I just turn on amp after 10 sec measure voltage arround this resistor and write value(lets say 0,4-0,5V) and then try to match pairs . this would be pretty ok?
am I too naive to thinking this way- I solder input terminals for source resitor can be solder/desolder without dismounting pcb from heatsinks. I just turn on amp after 10 sec measure voltage arround this resistor and write value(lets say 0,4-0,5V) and then try to match pairs . this would be pretty ok?
Hi!
Works just fine using any dc source as log you don't exceed power dissipation for resistors. Connect all resistor of same value in series to mentioned source, then measure voltage across every reistor. Those most simular voltage will be the ones with simular resistance. If you need value, use ohm law.
This works perfecly for mathing as current is the same in series connection.
Example: Voltage source of 1V across ten series connected resistors of 0.22Ohm result in about 0.1V across every resistor.
Important is that source is stable during measurment.
Works just fine using any dc source as log you don't exceed power dissipation for resistors. Connect all resistor of same value in series to mentioned source, then measure voltage across every reistor. Those most simular voltage will be the ones with simular resistance. If you need value, use ohm law.
This works perfecly for mathing as current is the same in series connection.
Example: Voltage source of 1V across ten series connected resistors of 0.22Ohm result in about 0.1V across every resistor.
Important is that source is stable during measurment.
I made a precision 9V DC power supply. A 25ohm 25W 1% resistor series 4 pieces of 0.47ohm resistors.It's total 26.8ohm connect to 9V power supply, about 0.335A current through this circuit and the 0.47ohm resistor will have 0.157V voltage drop.I used a digital-multimeter 200mv range to measure this voltage . Compare the voltage to make the match.THis method is simple,but the power supply work stable is very important.I pre-run the power supply one houre when doing the measuring close the room door and used stool to handle the resistors intead touch the hands. I match the resistors for Aleph-X 100W have good result.
This method of matching resistors using a power supply is just reverting back to an ohmmeter with the PS acting like the battery in the meter (albeit more unstable). With an ohmmeter the meter computes Ohm's law using the voltage of the battery and the current the resistor draws. With a PS you do the computing.
The practical way I described in an earlier post with a larger R in series with your <1 ohm resistor should serve the purpose.
The practical way I described in an earlier post with a larger R in series with your <1 ohm resistor should serve the purpose.
I second Tim_X for using a current source and measure voltage across your small-value resistor. If you want absolute value, you can connect an additional multimeter at 10A DC current measurement to measure the current.
You can build a very simple current source for this purpose using a standard LM317 on heatsink. Will give you 1.5A max. assuming you do not allow much more than 5V drop across the 317. Just check the datasheet.
Patrick
You can build a very simple current source for this purpose using a standard LM317 on heatsink. Will give you 1.5A max. assuming you do not allow much more than 5V drop across the 317. Just check the datasheet.
Patrick
Remember resistance of cables and connectors to ohmmeter that makes it almost inpossible to measure very low resistances exactly.
This way of doing it in series connection has several anvantages.
1. Current is the same for whole circuit (all resistors) therefore very exact matching.
2. Current in cables to voltmeter is virtually nonexistent therefore no voltagedrop and corresponding misreading.
Ok, revert back to....hm we are talking same thing but different. The trick here is higher current than ohmmeter will supply, therefore more voltage over resistor(s) and more exact reading.
If you measure low resistance one resistor at a time chances are that you never get the same contact resistance.
Using ohmmeter, how can series resistance inprove reading ?
This way of doing it in series connection has several anvantages.
1. Current is the same for whole circuit (all resistors) therefore very exact matching.
2. Current in cables to voltmeter is virtually nonexistent therefore no voltagedrop and corresponding misreading.
Ok, revert back to....hm we are talking same thing but different. The trick here is higher current than ohmmeter will supply, therefore more voltage over resistor(s) and more exact reading.
If you measure low resistance one resistor at a time chances are that you never get the same contact resistance.
Using ohmmeter, how can series resistance inprove reading ?
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