On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote:> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com > wrote: > > >On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: > >> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message > >> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... > >> > On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" > >> > <bperryb@bowdenshobbycircuits.info> wrote: > >> > > >> >>I have a switching application where a single bipolar transistor might be > >> >>used. The AC input is about 300mV and the transistor needs about 700 mV to > >> >>switch on. My idea is to construct a voltage divider using a diode which > >> >>will produce about a 600mV drop across the diode and a voltage divider of > >> >>2 > >> >>equal resistors to set the transistor base voltage at about 300 mV so the > >> >>transistor will switch on with an additional 300mV. I understand the > >> >>temperature problems, but it seems to be minimal. What am I missing? > >> >> > >> > > >> > More detail would help. What is the drive waveform like? What is the > >> > load? Got a schematic? > >> > > >> > A PNP emitter follower can be a neat way to drive an NPN transistor. > >> > It moves the effective threshold near zero volts and does some > >> > temperature compensation. > >> > > >> > Or just add an IC, an opamp or a comparator. > >> > > >> > -- > >> > > >> > John Larkin Highland Technology, Inc > >> > > >> > lunatic fringe electronics > >> > > >> > >> All I want to do is detect a positive half cycle going sine wave at about > >> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 > >> volts. I figure I can do this with one transistor. I'm not interested in > >> adding more parts. I can easily do it with more parts. All I want to know is > >> the circuit stability of biasing a transistor near it's conduction point so > >> it works under all conditions of temperature and supply voltage. It needs to > >> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit > >> using LTSpice. > >> > >[snip LTSpice] > > > > +2V +2V > > -+- -+- > > | | > > .-. R1 .-. RL > > | | 5.6k | | 2k > > '-' '-' > > | | > > | R2 |/ > > +--15k--+--| Q1 > > | | |>. > >D1 V C1 --- | > > --- 2uF--- === > > | | > > === ^ > > Vin > >Six parts. > > > > Remove D1 and R2: four parts.That might be better for his application -- C1 could store a negative voltage to keep Q1 fully 'off' most of the time. Bill showed me a gadget earlier this month. I assume this question is related. I think he wants to produce 5-to-10% duty pulses at about 20Hz, and save battery the rest of the time. A darlington might be worth it to save the R1 idle-current. Cheers, James Arthur
Bipolar transistor bias question
Started by ●December 18, 2016
Reply by ●December 20, 20162016-12-20
Reply by ●December 20, 20162016-12-20
On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote:> On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com >> wrote: >> >>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: >>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message >>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... >>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" >>>>> <bperryb@bowdenshobbycircuits.info> wrote: >>>>> >>>>>> I have a switching application where a single bipolar transistor might be >>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to >>>>>> switch on. My idea is to construct a voltage divider using a diode which >>>>>> will produce about a 600mV drop across the diode and a voltage divider of >>>>>> 2 >>>>>> equal resistors to set the transistor base voltage at about 300 mV so the >>>>>> transistor will switch on with an additional 300mV. I understand the >>>>>> temperature problems, but it seems to be minimal. What am I missing? >>>>>> >>>>> >>>>> More detail would help. What is the drive waveform like? What is the >>>>> load? Got a schematic? >>>>> >>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. >>>>> It moves the effective threshold near zero volts and does some >>>>> temperature compensation. >>>>> >>>>> Or just add an IC, an opamp or a comparator. >>>>> >>>>> -- >>>>> >>>>> John Larkin Highland Technology, Inc >>>>> >>>>> lunatic fringe electronics >>>>> >>>> >>>> All I want to do is detect a positive half cycle going sine wave at about >>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 >>>> volts. I figure I can do this with one transistor. I'm not interested in >>>> adding more parts. I can easily do it with more parts. All I want to know is >>>> the circuit stability of biasing a transistor near it's conduction point so >>>> it works under all conditions of temperature and supply voltage. It needs to >>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit >>>> using LTSpice. >>>> >>> [snip LTSpice] >>> >>> +2V +2V >>> -+- -+- >>> | | >>> .-. R1 .-. RL >>> | | 5.6k | | 2k >>> '-' '-' >>> | | >>> | R2 |/ >>> +--15k--+--| Q1 >>> | | |>. >>> D1 V C1 --- | >>> --- 2uF--- === >>> | | >>> === ^ >>> Vin >>> Six parts. >>> >> >> Remove D1 and R2: four parts. > > > That might be better for his application -- C1 could store a > negative voltage to keep Q1 fully 'off' most of the time.How exactly will that happen? R1 and the BE junction will set the DC voltage at the base. The capacitor can't influence that other than the AC signal that is passed. The DC set point will remain the same, no? I don't think any of these circuits do what the OP wanted, or did he change that? This thread has gotten pretty long. He wanted something that would detect when the AC input got above 300 mV peak, no? The above schematic cuts off when the input goes negative and is on the rest of the time. Removing the diode and R2 just turn it into an AC coupled small signal amplifier. -- Rick C
Reply by ●December 20, 20162016-12-20
"John S" <Sophi.2@invalid.org> wrote in message news:o3aabk$q6c$1@dont-email.me...> On 12/19/2016 8:41 PM, billbowden wrote: >> All I want to do is detect a positive half cycle going sine wave at about >> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 >> volts. I figure I can do this with one transistor. I'm not interested in >> adding more parts. I can easily do it with more parts. All I want to know >> is >> the circuit stability of biasing a transistor near it's conduction point >> so >> it works under all conditions of temperature and supply voltage. It needs >> to >> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit >> using LTSpice. > (snip) > > Okay. Add the following to your schematic using the LTSpice directives: > > .step v1 list 2 3.2 > .step temp list -40 25 70 > > This will give you V1 steps of 2 to 3.2 > And temperature in 3 steps from -40 to 70C > > Check the help section to refine your desired range. >Yes, that worked well and all looked good. The base voltage varies from 360 to 430 to 534mV with temp changes and in all cases the transistor is off. That's what I wanted to know. Thanks.
Reply by ●December 21, 20162016-12-21
On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote:> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: > > On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: > >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com > >> wrote: > >> > >>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: > >>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message > >>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... > >>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" > >>>>> <bperryb@bowdenshobbycircuits.info> wrote: > >>>>> > >>>>>> I have a switching application where a single bipolar transistor might be > >>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to > >>>>>> switch on. My idea is to construct a voltage divider using a diode which > >>>>>> will produce about a 600mV drop across the diode and a voltage divider of > >>>>>> 2 > >>>>>> equal resistors to set the transistor base voltage at about 300 mV so the > >>>>>> transistor will switch on with an additional 300mV. I understand the > >>>>>> temperature problems, but it seems to be minimal. What am I missing? > >>>>>> > >>>>> > >>>>> More detail would help. What is the drive waveform like? What is the > >>>>> load? Got a schematic? > >>>>> > >>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. > >>>>> It moves the effective threshold near zero volts and does some > >>>>> temperature compensation. > >>>>> > >>>>> Or just add an IC, an opamp or a comparator. > >>>>> > >>>>> -- > >>>>> > >>>>> John Larkin Highland Technology, Inc > >>>>> > >>>>> lunatic fringe electronics > >>>>> > >>>> > >>>> All I want to do is detect a positive half cycle going sine wave at about > >>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 > >>>> volts. I figure I can do this with one transistor. I'm not interested in > >>>> adding more parts. I can easily do it with more parts. All I want to know is > >>>> the circuit stability of biasing a transistor near it's conduction point so > >>>> it works under all conditions of temperature and supply voltage. It needs to > >>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit > >>>> using LTSpice. > >>>> > >>> [snip LTSpice] > >>> > >>> +2V +2V > >>> -+- -+- > >>> | | > >>> .-. R1 .-. RL > >>> | | 5.6k | | 2k > >>> '-' '-' > >>> | | > >>> | R2 |/ > >>> +--15k--+--| Q1 > >>> | | |>. > >>> D1 V C1 --- | > >>> --- 2uF--- === > >>> | | > >>> === ^ > >>> Vin > >>> Six parts. > >>> > >> > >> Remove D1 and R2: four parts. > > > > > > That might be better for his application -- C1 could store a > > negative voltage to keep Q1 fully 'off' most of the time. > > How exactly will that happen? R1 and the BE junction will set the DC > voltage at the base. The capacitor can't influence that other than the > AC signal that is passed.C1 charges every time you use it to drive Q1(b). When the drive drops back to zero, the C1-Q1(b) node drops to (Vbe - drive voltage).> The DC set point will remain the same, no? > > I don't think any of these circuits do what the OP wanted, or did he > change that? This thread has gotten pretty long. He wanted something > that would detect when the AC input got above 300 mV peak, no? The > above schematic cuts off when the input goes negative and is on the rest > of the time. Removing the diode and R2 just turn it into an AC coupled > small signal amplifier. > > -- > > Rick CCheers, James Arthur
Reply by ●December 21, 20162016-12-21
On 12/21/2016 10:00 AM, dagmargoodboat@yahoo.com wrote:> On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: >>> On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: >>>> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com >>>> wrote: >>>> >>>>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: >>>>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message >>>>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... >>>>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" >>>>>>> <bperryb@bowdenshobbycircuits.info> wrote: >>>>>>> >>>>>>>> I have a switching application where a single bipolar transistor might be >>>>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to >>>>>>>> switch on. My idea is to construct a voltage divider using a diode which >>>>>>>> will produce about a 600mV drop across the diode and a voltage divider of >>>>>>>> 2 >>>>>>>> equal resistors to set the transistor base voltage at about 300 mV so the >>>>>>>> transistor will switch on with an additional 300mV. I understand the >>>>>>>> temperature problems, but it seems to be minimal. What am I missing? >>>>>>>> >>>>>>> >>>>>>> More detail would help. What is the drive waveform like? What is the >>>>>>> load? Got a schematic? >>>>>>> >>>>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. >>>>>>> It moves the effective threshold near zero volts and does some >>>>>>> temperature compensation. >>>>>>> >>>>>>> Or just add an IC, an opamp or a comparator. >>>>>>> >>>>>>> -- >>>>>>> >>>>>>> John Larkin Highland Technology, Inc >>>>>>> >>>>>>> lunatic fringe electronics >>>>>>> >>>>>> >>>>>> All I want to do is detect a positive half cycle going sine wave at about >>>>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 >>>>>> volts. I figure I can do this with one transistor. I'm not interested in >>>>>> adding more parts. I can easily do it with more parts. All I want to know is >>>>>> the circuit stability of biasing a transistor near it's conduction point so >>>>>> it works under all conditions of temperature and supply voltage. It needs to >>>>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit >>>>>> using LTSpice. >>>>>> >>>>> [snip LTSpice] >>>>> >>>>> +2V +2V >>>>> -+- -+- >>>>> | | >>>>> .-. R1 .-. RL >>>>> | | 5.6k | | 2k >>>>> '-' '-' >>>>> | | >>>>> | R2 |/ >>>>> +--15k--+--| Q1 >>>>> | | |>. >>>>> D1 V C1 --- | >>>>> --- 2uF--- === >>>>> | | >>>>> === ^ >>>>> Vin >>>>> Six parts. >>>>> >>>> >>>> Remove D1 and R2: four parts. >>> >>> >>> That might be better for his application -- C1 could store a >>> negative voltage to keep Q1 fully 'off' most of the time. >> >> How exactly will that happen? R1 and the BE junction will set the DC >> voltage at the base. The capacitor can't influence that other than the >> AC signal that is passed. > > C1 charges every time you use it to drive Q1(b). When the drive drops back > to zero, the C1-Q1(b) node drops to (Vbe - drive voltage).Actually, I don't think that is correct - unless we are saying the same thing. I'm not sure the electrical definition of "use it". The base node will be driven above and below the DC set point no matter the DC content of the input signal. The only effect of the capacitor is blocking the DC content of the input signal. So what ever the set point is (as determined by R1 and the BE characteristics), if the input signal is applied and the voltage on the capacitor is allowed to settle, the bias on the base is going to be determined by that set point. Q1 will *not* be off most of the time. Just the opposite. It is on at the set point and will only be turned off by significantly negative portions of the AC component on the input. -- Rick C
Reply by ●December 21, 20162016-12-21
On Wed, 21 Dec 2016 07:00:51 -0800 (PST), dagmargoodboat@yahoo.com wrote:>On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: >> > On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: >> >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com >> >> wrote: >> >> >> >>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: >> >>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message >> >>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... >> >>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" >> >>>>> <bperryb@bowdenshobbycircuits.info> wrote: >> >>>>> >> >>>>>> I have a switching application where a single bipolar transistor might be >> >>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to >> >>>>>> switch on. My idea is to construct a voltage divider using a diode which >> >>>>>> will produce about a 600mV drop across the diode and a voltage divider of >> >>>>>> 2 >> >>>>>> equal resistors to set the transistor base voltage at about 300 mV so the >> >>>>>> transistor will switch on with an additional 300mV. I understand the >> >>>>>> temperature problems, but it seems to be minimal. What am I missing? >> >>>>>> >> >>>>> >> >>>>> More detail would help. What is the drive waveform like? What is the >> >>>>> load? Got a schematic? >> >>>>> >> >>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. >> >>>>> It moves the effective threshold near zero volts and does some >> >>>>> temperature compensation. >> >>>>> >> >>>>> Or just add an IC, an opamp or a comparator. >> >>>>> >> >>>>> -- >> >>>>> >> >>>>> John Larkin Highland Technology, Inc >> >>>>> >> >>>>> lunatic fringe electronics >> >>>>> >> >>>> >> >>>> All I want to do is detect a positive half cycle going sine wave at about >> >>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 >> >>>> volts. I figure I can do this with one transistor. I'm not interested in >> >>>> adding more parts. I can easily do it with more parts. All I want to know is >> >>>> the circuit stability of biasing a transistor near it's conduction point so >> >>>> it works under all conditions of temperature and supply voltage. It needs to >> >>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit >> >>>> using LTSpice. >> >>>> >> >>> [snip LTSpice] >> >>> >> >>> +2V +2V >> >>> -+- -+- >> >>> | | >> >>> .-. R1 .-. RL >> >>> | | 5.6k | | 2k >> >>> '-' '-' >> >>> | | >> >>> | R2 |/ >> >>> +--15k--+--| Q1 >> >>> | | |>. >> >>> D1 V C1 --- | >> >>> --- 2uF--- === >> >>> | | >> >>> === ^ >> >>> Vin >> >>> Six parts. >> >>> >> >> >> >> Remove D1 and R2: four parts. >> > >> > >> > That might be better for his application -- C1 could store a >> > negative voltage to keep Q1 fully 'off' most of the time. >> >> How exactly will that happen? R1 and the BE junction will set the DC >> voltage at the base. The capacitor can't influence that other than the >> AC signal that is passed. > >C1 charges every time you use it to drive Q1(b). When the drive drops back >to zero, the C1-Q1(b) node drops to (Vbe - drive voltage). > >> The DC set point will remain the same, no? >> >> I don't think any of these circuits do what the OP wanted, or did he >> change that? This thread has gotten pretty long. He wanted something >> that would detect when the AC input got above 300 mV peak, no? The >> above schematic cuts off when the input goes negative and is on the rest >> of the time. Removing the diode and R2 just turn it into an AC coupled >> small signal amplifier. >> >> -- >> >> Rick C > >Cheers, >James ArthurProblem is that a coupling capacitor will get charged by base current on the positive stroke, but then will discharge slowly via the bias, creating a threshold movement. So a simple transistor stage can't maintain a constant threshold. ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 |
Reply by ●December 21, 20162016-12-21
On Wednesday, 21 December 2016 15:59:24 UTC, Jim Thompson wrote:> On Wed, 21 Dec 2016 07:00:51 -0800 (PST), dagmargoodboat@yahoo.com > wrote: > >On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: > >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: > >> > On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: > >> >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com > >> >> wrote: > >> >>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: > >> >>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message > >> >>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... > >> >>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" > >> >>>>> <bperryb@bowdenshobbycircuits.info> wrote: > >> >>>>> > >> >>>>>> I have a switching application where a single bipolar transistor might be > >> >>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to > >> >>>>>> switch on. My idea is to construct a voltage divider using a diode which > >> >>>>>> will produce about a 600mV drop across the diode and a voltage divider of > >> >>>>>> 2 > >> >>>>>> equal resistors to set the transistor base voltage at about 300 mV so the > >> >>>>>> transistor will switch on with an additional 300mV. I understand the > >> >>>>>> temperature problems, but it seems to be minimal. What am I missing? > >> >>>>>> > >> >>>>> > >> >>>>> More detail would help. What is the drive waveform like? What is the > >> >>>>> load? Got a schematic? > >> >>>>> > >> >>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. > >> >>>>> It moves the effective threshold near zero volts and does some > >> >>>>> temperature compensation. > >> >>>>> > >> >>>>> Or just add an IC, an opamp or a comparator. > >> >>>>> > >> >>>>> -- > >> >>>>> > >> >>>>> John Larkin Highland Technology, Inc > >> >>>>> > >> >>>>> lunatic fringe electronics > >> >>>>> > >> >>>> > >> >>>> All I want to do is detect a positive half cycle going sine wave at about > >> >>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 > >> >>>> volts. I figure I can do this with one transistor. I'm not interested in > >> >>>> adding more parts. I can easily do it with more parts. All I want to know is > >> >>>> the circuit stability of biasing a transistor near it's conduction point so > >> >>>> it works under all conditions of temperature and supply voltage. It needs to > >> >>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit > >> >>>> using LTSpice. > >> >>>> > >> >>> [snip LTSpice] > >> >>> > >> >>> +2V +2V > >> >>> -+- -+- > >> >>> | | > >> >>> .-. R1 .-. RL > >> >>> | | 5.6k | | 2k > >> >>> '-' '-' > >> >>> | | > >> >>> | R2 |/ > >> >>> +--15k--+--| Q1 > >> >>> | | |>. > >> >>> D1 V C1 --- | > >> >>> --- 2uF--- === > >> >>> | | > >> >>> === ^ > >> >>> Vin > >> >>> Six parts. > >> >>> > >> >> > >> >> Remove D1 and R2: four parts. > >> > > >> > > >> > That might be better for his application -- C1 could store a > >> > negative voltage to keep Q1 fully 'off' most of the time. > >> > >> How exactly will that happen? R1 and the BE junction will set the DC > >> voltage at the base. The capacitor can't influence that other than the > >> AC signal that is passed. > > > >C1 charges every time you use it to drive Q1(b). When the drive drops back > >to zero, the C1-Q1(b) node drops to (Vbe - drive voltage). > > > >> The DC set point will remain the same, no? > >> > >> I don't think any of these circuits do what the OP wanted, or did he > >> change that? This thread has gotten pretty long. He wanted something > >> that would detect when the AC input got above 300 mV peak, no? The > >> above schematic cuts off when the input goes negative and is on the rest > >> of the time. Removing the diode and R2 just turn it into an AC coupled > >> small signal amplifier. > >> > >> -- > >> > >> Rick C > > > >Cheers, > >James Arthur > > Problem is that a coupling capacitor will get charged by base current > on the positive stroke, but then will discharge slowly via the bias, > creating a threshold movement. > > So a simple transistor stage can't maintain a constant threshold. > > ...Jim ThompsonEither you guys are missing soemthing or I am. This looks like a 0.3v threshold detector: . _________+ . | | . R R . | _| . --R--+---(_) . | . | . ___________| 0v Keep P_diss right down for Vbe stability. NT
Reply by ●December 21, 20162016-12-21
On Wed, 21 Dec 2016 11:07:30 -0800 (PST), tabbypurr@gmail.com wrote:>On Wednesday, 21 December 2016 15:59:24 UTC, Jim Thompson wrote: >> On Wed, 21 Dec 2016 07:00:51 -0800 (PST), dagmargoodboat@yahoo.com >> wrote: >> >On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: >> >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: >> >> > On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: >> >> >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com >> >> >> wrote: >> >> >>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: >> >> >>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message >> >> >>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... >> >> >>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" >> >> >>>>> <bperryb@bowdenshobbycircuits.info> wrote: >> >> >>>>> >> >> >>>>>> I have a switching application where a single bipolar transistor might be >> >> >>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to >> >> >>>>>> switch on. My idea is to construct a voltage divider using a diode which >> >> >>>>>> will produce about a 600mV drop across the diode and a voltage divider of >> >> >>>>>> 2 >> >> >>>>>> equal resistors to set the transistor base voltage at about 300 mV so the >> >> >>>>>> transistor will switch on with an additional 300mV. I understand the >> >> >>>>>> temperature problems, but it seems to be minimal. What am I missing? >> >> >>>>>> >> >> >>>>> >> >> >>>>> More detail would help. What is the drive waveform like? What is the >> >> >>>>> load? Got a schematic? >> >> >>>>> >> >> >>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. >> >> >>>>> It moves the effective threshold near zero volts and does some >> >> >>>>> temperature compensation. >> >> >>>>> >> >> >>>>> Or just add an IC, an opamp or a comparator. >> >> >>>>> >> >> >>>>> -- >> >> >>>>> >> >> >>>>> John Larkin Highland Technology, Inc >> >> >>>>> >> >> >>>>> lunatic fringe electronics >> >> >>>>> >> >> >>>> >> >> >>>> All I want to do is detect a positive half cycle going sine wave at about >> >> >>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 >> >> >>>> volts. I figure I can do this with one transistor. I'm not interested in >> >> >>>> adding more parts. I can easily do it with more parts. All I want to know is >> >> >>>> the circuit stability of biasing a transistor near it's conduction point so >> >> >>>> it works under all conditions of temperature and supply voltage. It needs to >> >> >>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit >> >> >>>> using LTSpice. >> >> >>>> >> >> >>> [snip LTSpice] >> >> >>> >> >> >>> +2V +2V >> >> >>> -+- -+- >> >> >>> | | >> >> >>> .-. R1 .-. RL >> >> >>> | | 5.6k | | 2k >> >> >>> '-' '-' >> >> >>> | | >> >> >>> | R2 |/ >> >> >>> +--15k--+--| Q1 >> >> >>> | | |>. >> >> >>> D1 V C1 --- | >> >> >>> --- 2uF--- === >> >> >>> | | >> >> >>> === ^ >> >> >>> Vin >> >> >>> Six parts. >> >> >>> >> >> >> >> >> >> Remove D1 and R2: four parts. >> >> > >> >> > >> >> > That might be better for his application -- C1 could store a >> >> > negative voltage to keep Q1 fully 'off' most of the time. >> >> >> >> How exactly will that happen? R1 and the BE junction will set the DC >> >> voltage at the base. The capacitor can't influence that other than the >> >> AC signal that is passed. >> > >> >C1 charges every time you use it to drive Q1(b). When the drive drops back >> >to zero, the C1-Q1(b) node drops to (Vbe - drive voltage). >> > >> >> The DC set point will remain the same, no? >> >> >> >> I don't think any of these circuits do what the OP wanted, or did he >> >> change that? This thread has gotten pretty long. He wanted something >> >> that would detect when the AC input got above 300 mV peak, no? The >> >> above schematic cuts off when the input goes negative and is on the rest >> >> of the time. Removing the diode and R2 just turn it into an AC coupled >> >> small signal amplifier. >> >> >> >> -- >> >> >> >> Rick C >> > >> >Cheers, >> >James Arthur >> >> Problem is that a coupling capacitor will get charged by base current >> on the positive stroke, but then will discharge slowly via the bias, >> creating a threshold movement. >> >> So a simple transistor stage can't maintain a constant threshold. >> >> ...Jim Thompson > >Either you guys are missing soemthing or I am. This looks like a 0.3v threshold detector: > >. _________+ >. | | >. R R >. | _| >. --R--+---(_) >. | >. | >. ___________| 0v > > >Keep P_diss right down for Vbe stability. > > >NTMaybe I'm missing something, but suppose there are amplitudes higher than 300mV... is the threshold supposed to stay put? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 |
Reply by ●December 21, 20162016-12-21
On Wednesday, December 21, 2016 at 10:59:24 AM UTC-5, Jim Thompson wrote:> On Wed, 21 Dec 2016 07:00:51 -0800 (PST), dagmargoodboat@yahoo.com > wrote: > > >On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: > >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: > >> > On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: > >> >> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com> >> >>> +2V +2V > >> >>> -+- -+- > >> >>> | | > >> >>> .-. R1 .-. RL > >> >>> | | 5.6k | | 2k > >> >>> '-' '-' > >> >>> | | > >> >>> | R2 |/ > >> >>> +--15k--+--| Q1 > >> >>> | | |>. > >> >>> D1 V C1 --- | > >> >>> --- 2uF--- === > >> >>> | | > >> >>> === ^ > >> >>> Vin > >> >>> Six parts. > >> >>> > >> >> > >> >> Remove D1 and R2: four parts. > >> > > >> > > >> > That might be better for his application -- C1 could store a > >> > negative voltage to keep Q1 fully 'off' most of the time. > >> > >> How exactly will that happen? R1 and the BE junction will set the DC > >> voltage at the base. The capacitor can't influence that other than the > >> AC signal that is passed. > > > >C1 charges every time you use it to drive Q1(b). When the drive drops back > >to zero, the C1-Q1(b) node drops to (Vbe - drive voltage). > > > >> The DC set point will remain the same, no? > >> > >> I don't think any of these circuits do what the OP wanted, or did he > >> change that? This thread has gotten pretty long. He wanted something > >> that would detect when the AC input got above 300 mV peak, no? The > >> above schematic cuts off when the input goes negative and is on the rest > >> of the time. Removing the diode and R2 just turn it into an AC coupled > >> small signal amplifier. > >> > >> -- > >> > >> Rick C > > > > Problem is that a coupling capacitor will get charged by base current > on the positive stroke, but then will discharge slowly via the bias, > creating a threshold movement. > > So a simple transistor stage can't maintain a constant threshold.That's right, but AIUI he doesn't need a stable d.c. threshold. I think it's for a variation on this: http://www.bowdenshobbycircuits.info/motor.htm#new But I'm guessing, & might be mistaken. Cheers, James Arthur
Reply by ●December 21, 20162016-12-21
On Wednesday, December 21, 2016 at 10:36:38 AM UTC-5, rickman wrote:> On 12/21/2016 10:00 AM, dagmargoodboat@yahoo.com wrote: > > On Tuesday, December 20, 2016 at 4:23:52 PM UTC-5, rickman wrote: > >> On 12/20/2016 12:58 AM, dagmargoodboat@yahoo.com wrote: > >>> On Tuesday, December 20, 2016 at 12:22:22 AM UTC-5, John Larkin wrote: > >>>> On Mon, 19 Dec 2016 21:15:45 -0800 (PST), dagmargoodboat@yahoo.com > >>>> wrote: > >>>> > >>>>> On Monday, December 19, 2016 at 9:41:11 PM UTC-5, Bill Bowden wrote: > >>>>>> "John Larkin" <jjlarkin@highlandtechnology.com> wrote in message > >>>>>> news:q0de5ct1a14ppo9ooehjv8g32vk467bk0u@4ax.com... > >>>>>>> On Sun, 18 Dec 2016 16:41:42 -0800, "billbowden" > >>>>>>> <bperryb@bowdenshobbycircuits.info> wrote: > >>>>>>> > >>>>>>>> I have a switching application where a single bipolar transistor might be > >>>>>>>> used. The AC input is about 300mV and the transistor needs about 700 mV to > >>>>>>>> switch on. My idea is to construct a voltage divider using a diode which > >>>>>>>> will produce about a 600mV drop across the diode and a voltage divider of > >>>>>>>> 2 > >>>>>>>> equal resistors to set the transistor base voltage at about 300 mV so the > >>>>>>>> transistor will switch on with an additional 300mV. I understand the > >>>>>>>> temperature problems, but it seems to be minimal. What am I missing? > >>>>>>>> > >>>>>>> > >>>>>>> More detail would help. What is the drive waveform like? What is the > >>>>>>> load? Got a schematic? > >>>>>>> > >>>>>>> A PNP emitter follower can be a neat way to drive an NPN transistor. > >>>>>>> It moves the effective threshold near zero volts and does some > >>>>>>> temperature compensation. > >>>>>>> > >>>>>>> Or just add an IC, an opamp or a comparator. > >>>>>>> > >>>>>>> -- > >>>>>>> > >>>>>>> John Larkin Highland Technology, Inc > >>>>>>> > >>>>>>> lunatic fringe electronics > >>>>>>> > >>>>>> > >>>>>> All I want to do is detect a positive half cycle going sine wave at about > >>>>>> 20Hz and 300mV peak and produce a negative rectified DC change of about 3 > >>>>>> volts. I figure I can do this with one transistor. I'm not interested in > >>>>>> adding more parts. I can easily do it with more parts. All I want to know is > >>>>>> the circuit stability of biasing a transistor near it's conduction point so > >>>>>> it works under all conditions of temperature and supply voltage. It needs to > >>>>>> work between a supply voltage of 2 to 3.2 volts. Here's a typical circuit > >>>>>> using LTSpice. > >>>>>> > >>>>> [snip LTSpice] > >>>>> > >>>>> +2V +2V > >>>>> -+- -+- > >>>>> | | > >>>>> .-. R1 .-. RL > >>>>> | | 5.6k | | 2k > >>>>> '-' '-' > >>>>> | | > >>>>> | R2 |/ > >>>>> +--15k--+--| Q1 > >>>>> | | |>. > >>>>> D1 V C1 --- | > >>>>> --- 2uF--- === > >>>>> | | > >>>>> === ^ > >>>>> Vin > >>>>> Six parts. > >>>>> > >>>> > >>>> Remove D1 and R2: four parts. > >>> > >>> > >>> That might be better for his application -- C1 could store a > >>> negative voltage to keep Q1 fully 'off' most of the time. > >> > >> How exactly will that happen? R1 and the BE junction will set the DC > >> voltage at the base. The capacitor can't influence that other than the > >> AC signal that is passed. > > > > C1 charges every time you use it to drive Q1(b). When the drive drops back > > to zero, the C1-Q1(b) node drops to (Vbe - drive voltage). > > Actually, I don't think that is correct - unless we are saying the same > thing. I'm not sure the electrical definition of "use it". > > The base node will be driven above and below the DC set point no matter > the DC content of the input signal. The only effect of the capacitor is > blocking the DC content of the input signal. So what ever the set point > is (as determined by R1 and the BE characteristics), if the input signal > is applied and the voltage on the capacitor is allowed to settle, the > bias on the base is going to be determined by that set point. Q1 will > *not* be off most of the time. Just the opposite. It is on at the set > point and will only be turned off by significantly negative portions of > the AC component on the input. > > -- > > Rick CThe capacitor is charged rapidly when driving Q1, and discharges slowly, creating a d.c. offset across C1 that depends on the time constants. If you don't understand it, LTSpice might clear things up. Cheers, James Arthur
