The point, Nigel, is that he does not have much voltage to play with. So the voltage dropped across the base resistor further reduces it.Nigel Goodwin said:ljcox said:When the resistor is on the emitter side, you do not need a base resistor.
But there's a difference between 'need' and 'good practice', while the base resistor won't affect the performance of the circuit, it's there to protect the PIC if the transistor goes S/C.
As an added bonus, the resistors also probably simplify PCB layout, avoiding the need for wire links or double sided boards.
ljcox said:The point, Nigel, is that he does not have much voltage to play with. So the voltage dropped across the base resistor further reduces it.Nigel Goodwin said:ljcox said:When the resistor is on the emitter side, you do not need a base resistor.
But there's a difference between 'need' and 'good practice', while the base resistor won't affect the performance of the circuit, it's there to protect the PIC if the transistor goes S/C.
As an added bonus, the resistors also probably simplify PCB layout, avoiding the need for wire links or double sided boards.
I really think that a PNP common emitter driver would be better anyway.
ljcox said:I suggest that you do a static test (ie. steady current, no pulsing) and measure the current through a segment that is necessary to give the required brightness and then measure the voltage across the segment at that current. Report the figures to us so we can calculate the resistor value for you.
Nigel Goodwin said:however your positive driving method is very poor - it's best if you use PNP transistors from the positive rail, but that would require NPN's to drive them (due to your 9V supply rail).
Nigel Goodwin said:the maximum voltage to the displays is only 4.3V, so you're wasting your 9V supply anyway
Nigel Goodwin said:Also the 140 ohm resistor in the collector isn't a very good idea, and will make it difficult to calculate the current to the display - moving it to the emitter would make more sense.
Nigel Goodwin said:I would suggest you use something like 470 ohm resistors between the PIC pins and the bases, to give better base drive.
taengi said:Nigel Goodwin said:however your positive driving method is very poor - it's best if you use PNP transistors from the positive rail, but that would require NPN's to drive them (due to your 9V supply rail).
Hi Nigel,
1. Do you mean that the way I drive the segment is not efficient?
Nigel Goodwin said:the maximum voltage to the displays is only 4.3V, so you're wasting your 9V supply anyway
2. I don't understand how you come up with the maximum voltage = 4.3v? Where did the rest 4.7v go? is it consumed by CE of BJT?
Nigel Goodwin said:Also the 140 ohm resistor in the collector isn't a very good idea, and will make it difficult to calculate the current to the display - moving it to the emitter would make more sense.
3. I thought Ic is obtained by assuming how much current we required, then we calculate the resistance required to give Ic? R= [9Vcc - 1.7v(one segment)]/Ic
Nigel Goodwin said:I would suggest you use something like 470 ohm resistors between the PIC pins and the bases, to give better base drive.
4. May I ask how do you come up with this resistance value. I really wanna learn. Also placing a diode on base doesn't provide protection for PIC?
Thanks William At MyBlueRoom for the suggested diagram. Unfortunately, I have everything soldiered on the common cathode side. So the only way to improve it is on the segment side.
BTW, hereby are my latest schematics based on all your advices. Correct me if I misintepret your meaning
Schematic A [try1.jpg diagram]. -segment connected to collector. which I have to invert the display pattern on the software.
Schematic B [try.jpg diagram]. -segment connect to emitter
One more question: Will the current flow out from common cathode to 7404 and 4017 eventually burned those IC? (based on my design). I have tried once connecting using Schematic A with base resistor 10k and no collector resistor. When connecting all 7 segment using this way, all my ICs burned...at first I thought those BJT are in breakdown region due to high Ib, but they still ok. So I wonder if the Ic may be to high or
something.
taengi said:ljcox said:I suggest that you do a static test (ie. steady current, no pulsing) and measure the current through a segment that is necessary to give the required brightness and then measure the voltage across the segment at that current. Report the figures to us so we can calculate the resistor value for you.
Hi ljcox,
The current through a segment is 6-7mA to achieve considerable brightness, while the forward biase voltage is approx 1.75v.
How do you actually calculate the resistor value? Is it R=[Vcc - 1.75]/6mA ?? I assume that you mean Vcc = 9 V, so the answer is no. Note that it will be 60 mA because of the x 10 muxing.
You need to calculate the voltage that will be across the resistor and then divide by 6 mA. I'll draw a diagram and post it later to show you how to do it.
Oh, so the simulation is using MPLAB....I should've explored earlier. Thanks for sharing.
Nigel Goodwin said:however your positive driving method is very poor - it's best if you use PNP transistors from the positive rail, but that would require NPN's to drive them (due to your 9V supply rail).
Hi Nigel,
1. Do you mean that the way I drive the segment is not efficient?
Nigel Goodwin said:the maximum voltage to the displays is only 4.3V, so you're wasting your 9V supply anyway
2. I don't understand how you come up with the maximum voltage = 4.3v? Where did the rest 4.7v go? is it consumed by CE of BJT? The CE voltage will be 4.7 V.
Nigel Goodwin said:Also the 140 ohm resistor in the collector isn't a very good idea, and will make it difficult to calculate the current to the display - moving it to the emitter would make more sense.
3. I thought Ic is obtained by assuming how much current we required, then we calculate the resistance required to give Ic? R= [9Vcc - 1.7v(one segment)]/Ic See my answer above
Nigel Goodwin said:I would suggest you use something like 470 ohm resistors between the PIC pins and the bases, to give better base drive.
4. May I ask how do you come up with this resistance value. I really wanna learn. Also placing a diode on base doesn't provide protection for PIC? He chose the lowest resistance value consistent with limiting the max current that the PIC can withstand if the transistor developed a short circuit fault
Thanks William At MyBlueRoom for the suggested diagram. Unfortunately, I have everything soldiered on the common cathode side. So the only way to improve it is on the segment side.
BTW, hereby are my latest schematics based on all your advices. Correct me if I misintepret your meaning
Schematic A [try1.jpg diagram]. -segment connected to collector. which I have to invert the display pattern on the software.
Schematic B [try.jpg diagram]. -segment connect to emitter
One more question: Will the current flow out from common cathode to 7404 and 4017 eventually burned those IC? (based on my design). I have tried once connecting using Schematic A with base resistor 10k and no collector resistor. When connecting all 7 segment using this way, all my ICs burned...at first I thought those BJT are in breakdown region due to high Ib, but they still ok. So I wonder if the Ic may be to high or
something. Schematic A is inefficient from a power consumption point of view as the transistors shunt the current. Also, as I wrote in the previous post, the 74LS04 can sink a max of 8 mA. So it would have failed and presumably caused the failure of the 4017.
Thanks for your patience and feedback.
taengi said:Thanks ljcox for your sharing. I need study through it first before I response to you.
taengi said:alright, now I understand how to calculate the value for the resistor. Good
One thing I'm not too sure.
If I don't use 4511 and directly connect the base transistor to PIC output, will it still be ok? As Nigel wrote, if the transistor develops a short circuit, it could destroy your PIC. The PIC data sheet indicates that the max current it can source or sink is 25 mA. So you need to limit the max current to that level in case a transistor fails. I would use PNP transistors with their emitters connected to +5 Volt and resistors from their bases to the PIC. Their collectors would go to the display segments via resistors.
This will prevent any problems with the PIC if a transistor fails and it gives you more voltage for the displays.
base upon the calculation, does it mean that 7404 not able to sink the current of 50mA? Correct. If you look at the 7404 data sheet it cannot sink more than 8 mA.
Nigel Goodwin said:MAJOR ERROR!.
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