What up..Ok I made this Circuit(THE LINK) http://home.cogeco.ca/~rpaisley4/ComparatorFLIPFLOPDV.GIF....The output of the comparator "turns on" the transistor, is there another transistor or another way of wireing this so that the same output "turns off" a secend transistor..
audioguru2
- Apr 6, 2004
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Hi Tacamo,
Welcome to our forum. ;D
The PNP transistor turns on when the output of the opamp is low.
Add an NPN transistor and it will be off when the output of the opamp is low.
Welcome to our forum. ;D
The PNP transistor turns on when the output of the opamp is low.
Add an NPN transistor and it will be off when the output of the opamp is low.
Two problems.
1) It the first circuit, the PNP transistor will never be completely off. This may not be a serious problem, but can cause problems depending how this circuit will be used. This can be fixed by adding a 10k resistor between b-e to cancel the leakage current through the diode.
2) By adding the NPN as shown the PNP transistor will be on all the time. The 2.2k resistor should be connected to the PNP collector.
1) It the first circuit, the PNP transistor will never be completely off. This may not be a serious problem, but can cause problems depending how this circuit will be used. This can be fixed by adding a 10k resistor between b-e to cancel the leakage current through the diode.
2) By adding the NPN as shown the PNP transistor will be on all the time. The 2.2k resistor should be connected to the PNP collector.
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audioguru2
- Apr 6, 2004
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If the circuit uses an opamp and not an open-collector comparator, then the output of the opamp will go up to 11V to turn on the NPN transistor and try to turn off the PNP. Then when the opamp output goes low to about 1V then the PNP turns on and the NPN turns off.AN920 said:
The PNP transistor will turn off if your 10k resistor is 1k like mine.
The circuits won't work with a normal opamp and is meant to use a open collector comparator.
The following diagrams show one of the problems I stated earlier
The first two diagrams show the circuit in both conditions. ( I only left the switches in their respective states for clarity). It can be seen from the simulation that LED1 is always on, while LED2 is off in the SET state.
Next is the circuit that uses another PNP transistor driven from the collector of first PNP. Simulation voltages show the results.
You need to zoom in otherwise mA looks like nA. (At least on my monitor)
The following diagrams show one of the problems I stated earlier
The first two diagrams show the circuit in both conditions. ( I only left the switches in their respective states for clarity). It can be seen from the simulation that LED1 is always on, while LED2 is off in the SET state.
Next is the circuit that uses another PNP transistor driven from the collector of first PNP. Simulation voltages show the results.
You need to zoom in otherwise mA looks like nA. (At least on my monitor)
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This shows the potential problem(1) of the circuit. In normal operation with a low load impedance there would be no problems. It is always good to be aware of such problems when used in certain applications.
Example: If the output is sensed by another very high impedance instrument like in automated test setup, problems may show.
First condition show that even though the transistor should be off, there is a leakage current of 170pA flowing in the b-e junction. This causes 74nA to flow in the collector, which will read as 7.47V on our measuring instrument. That error could halt out test process as detecting a fault.
This must be caused by leakage from U1's (off-state open collector impedance) and D1's reverse leakage current. Note that this leakage current will roughly double with each 10deg C increase in temperature. We must also be aware that the leakage of D1 will be aggravated by light shining through the glass body on the internal junction. Shield it from light.
To see the contribution by each, we open U1(pin1) and see that we still have 123pA through D1. Voltage on Q2 collector is still 6.09V
By adding a resistor (10k in this example, but anything up to 100k will do), we cancel or reverse this current flowing out of Q2 base. This is now -58pA, and our measuring instrument measures only 5.96mV on Q2 collector. Note the collector leakage current of Q2 is now about 1000 times smaller.
Example: If the output is sensed by another very high impedance instrument like in automated test setup, problems may show.
First condition show that even though the transistor should be off, there is a leakage current of 170pA flowing in the b-e junction. This causes 74nA to flow in the collector, which will read as 7.47V on our measuring instrument. That error could halt out test process as detecting a fault.
This must be caused by leakage from U1's (off-state open collector impedance) and D1's reverse leakage current. Note that this leakage current will roughly double with each 10deg C increase in temperature. We must also be aware that the leakage of D1 will be aggravated by light shining through the glass body on the internal junction. Shield it from light.
To see the contribution by each, we open U1(pin1) and see that we still have 123pA through D1. Voltage on Q2 collector is still 6.09V
By adding a resistor (10k in this example, but anything up to 100k will do), we cancel or reverse this current flowing out of Q2 base. This is now -58pA, and our measuring instrument measures only 5.96mV on Q2 collector. Note the collector leakage current of Q2 is now about 1000 times smaller.
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Herman the German
- Sep 1, 2006
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Hi AN920,AN920 said:
this has nothing to do with the circuit you are discussing. After having purchased a lousy, but expensive simulation program I'm looking for a better one. Would you kindly give me the name of the one you are using?
Thank you in advance.
Kind regards
HtG
