How do i calculate to get the resistor value?
see circuit attachment.
Input A and Input B is from arduino.
Supply is 24V.
Q1 and Q4 is BDV67B. Q2 and Q3 is BDV66B. Q5 and Q6 is PN2222.
There is this calculator:
http://www.raltron.com/cust/tools/voltage_divider.asp
If your Arduino Output 5V then Transistors gets about 3.7V with this circuit. I may be wrong
You need to apply Thevenin's Theorem.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
So the equivalent drive to the bases of Q5 and Q6 is:
Voltage = 5(10/(10+3.3)) = 3.76V
Resistance = 10k//3.3k = 1(1/10k + 1/3.3k) = 2.48k
The Vbe drop is going to be about 0.7V. Therefore the drive current is:
(3.76 - 0.7)/2.48k = 1.23mA
As far as "calculate to get the resistor value", did you mean "how do I determine all the resistor values?"
R1 and R2 are limited by power dissipation, lower values would allow more efficient
switching but would then need to be big heat-sink mounted jobs.
This kind of design originates from before good power MOSFETs were available.
polymorph:
You need to apply Thevenin's Theorem.So the equivalent drive to the bases of Q5 and Q6 is:
Voltage = 5(10/(10+3.3)) = 3.76V
Resistance = 10k//3.3k = 1(1/10k + 1/3.3k) = 2.48kThe Vbe drop is going to be about 0.7V. Therefore the drive current is:
(3.76 - 0.7)/2.48k = 1.23mA
As far as "calculate to get the resistor value", did you mean "how do I determine all the resistor values?"
Yup. what i meant was how do i determine all the resistor values?
What is wrong with the values already shown?
Grumpy_Mike:
What is wrong with the values already shown?
I don't quite understand how to get the value.
The values are not critical.
Just work out the base current you need in the transistor to ensure it is turned on fully.
You need to know the gain of the transistor and the current load it is going to take.
Base current = load current / gain
Grumpy_Mike:
The values are not critical.
Just work out the base current you need in the transistor to ensure it is turned on fully.
You need to know the gain of the transistor and the current load it is going to take.
Base current = load current / gain
The gain current for the pn2222 is ;
DC Current Gain
(IC = 0.1 mAdc, VCE = 10 Vdc)
(IC = 1.0 mAdc, VCE = 10 Vdc)
How do i get the load current?
Do you have any link that would help me on the calculation?
The load current is determined by the supply voltage and that 1K resistor by using ohms law.
Plus the current into the base of the complementary transistors which is determined by what load the motor draws.
Do you want a link to ohms law?
You work backwards. What is the maximum collector current in the transistor? For a transistor to be saturated, the base current should be no less than 1/20th of the required collector current. Refer to the datasheet, some transistors may get away with less, most require that or perhaps up to 1/10th of the collector current.
The TIP142 and TIP147 are a bit different - they are Darlington Pair. That means two transistors wired so the Beta values multiply. Darlington Pair don't quite go into saturation the same way, so let's refer to the datasheet:
Page 3:
Collector-Emitter Saturation
Ice @ 5A Vce(sat)=2V Ib=10mA
Ice @ 10A Vce(sat)=3V Ib=40mA
After that it is just Ohm's Law to calculate resistors to drive the transistors.
For the 2N3904 transistors, use the 1/20th rule-of-thumb.
polymorph:
You work backwards. What is the maximum collector current in the transistor? For a transistor to be saturated, the base current should be no less than 1/20th of the required collector current. Refer to the datasheet, some transistors may get away with less, most require that or perhaps up to 1/10th of the collector current.The TIP142 and TIP147 are a bit different - they are Darlington Pair. That means two transistors wired so the Beta values multiply. Darlington Pair don't quite go into saturation the same way, so let's refer to the datasheet:
http://www.onsemi.com/pub_link/Collateral/TIP140-D.PDFPage 3:
Collector-Emitter Saturation
Ice @ 5A Vce(sat)=2V Ib=10mA
Ice @ 10A Vce(sat)=3V Ib=40mAAfter that it is just Ohm's Law to calculate resistors to drive the transistors.
For the 2N3904 transistors, use the 1/20th rule-of-thumb.
Which means, to get the transistor to fully saturated, Ib must be 40mA?
Which means, to get the transistor to fully saturated, Ib must be 40mA?
Only if your transistor is drawing 10A, which is unlikely.
If I were to use pn2222 on my Q5, the R3 would be;
5V-0.7V=4.3V
4.3V/30mA (Arduino max is 40mA)=143ohms
is my calculations correct?
is my calculations correct?
Yes the calculation is correct. However your reasoning is not.
You do not need that much current through Q5. This transistor only drives a 1K load plus the base currents of Q1 & Q2. There is no way you want to drive so much current through Q5.
Grumpy_Mike:
is my calculations correct?
Yes the calculation is correct. However your reasoning is not.
You do not need that much current through Q5. This transistor only drives a 1K load plus the base currents of Q1 & Q2. There is no way you want to drive so much current through Q5.
Btw how do you do the calculation for R1 and R4?
what is the function/purpose of R4 and R6?
R4 is a pull down resistor designed to turn off transistor when there is no signal connected to the input. It's effect is to reduce the impedance on the gate so that interference does not turn it on. The voltage developed across it must be less than 0.7V when an interfering voltage is present. So you need to know the maximum interference voltage you will get and also the impedance of that interfering voltage. Then the calculation is similar to R3.
However, the impedance of the interference voltage is a difficult thing to measure and requires a specific environment to measure it in ( an EM shielded chamber ). So in practice 10K will cover virtually all interference situations you will ever encounter.
The maximum value of R1 could be calculated if you know the current gain of Q1 and the collector / emitter current current being passed by the transistor. For this you would need to know the supply voltage and the current draw of the motor. So R1 would be:-
the base current into Q1 = current gain of Q1 / collector to emitter current of Q1.
Grumpy_Mike:
R4 is a pull down resistor designed to turn off transistor when there is no signal connected to the input. It's effect is to reduce the impedance on the gate so that interference does not turn it on. The voltage developed across it must be less than 0.7V when an interfering voltage is present. So you need to know the maximum interference voltage you will get and also the impedance of that interfering voltage. Then the calculation is similar to R3.However, the impedance of the interference voltage is a difficult thing to measure and requires a specific environment to measure it in ( an EM shielded chamber ). So in practice 10K will cover virtually all interference situations you will ever encounter.
The maximum value of R1 could be calculated if you know the current gain of Q1 and the collector / emitter current current being passed by the transistor. For this you would need to know the supply voltage and the current draw of the motor. So R1 would be:-
the base current into Q1 = current gain of Q1 / collector to emitter current of Q1.
Thanks for explanation on the pull down resistor. Regarding the R1, the supply voltage which required for the motor is 24V and the rated current is 3.5A. Im not quite familiar on the datasheet. I cant find the current gain on the datasheet.