Re: Amplify 1.5v DC to 5v DC?

From: KM (kianmeng.tey_at_gmail.com)
Date: 10/20/04 

Date: 20 Oct 2004 06:23:23 -0700

HI,
I am not sure if it works for you, you can try this

> 5V
> |
> |
> \
> / R2
> \
> |
> +-----> PIC port |
> R1 |/c
> 1.5V>--/\/\---| Q1
> |>e
> |
> |
> Digital line of IC (0 or 1.5V)

When the digital line is logic low at 0 volt the transistor turn on
and current flow via the resistor R2. (choose R1 & R2 to be loage
value eg. 47k not to stress the IC sink cabability ) and when the the
line turn to high the transistor turn off.

Good luck!
KM

Jonathan Kirwan <jkirwan@easystreet.com> wrote in message news:<a1u7n01rigiau830jucfecs2u2p8knsp7l@4ax.com>...
> On 18 Oct 2004 03:03:25 -0700, "mark.mcgee@csfb.com" <mark.mcgee@csfb.com>
> wrote:
>
> >I need to amplify a signal (actually, 4 of them, one being a clock
> >pulse) from a digital line from an IC - it's either off - 0v or on at
> >1.5v, nothing complicated. I need to feed this in to a PIC
> >microcontroller, so it needs to be at 5v when high.
> >
> >I assume I'd want to use a transistor, feeding the 1.5v signal in to
> >the base.
> >
> >Can I take the amplified 5v signal from a resistor connecting the
> >emitter to gnd?
>
> I don't think so, as the NPN emitter will have a voltage that is lower than the
> base, if it's operating normally. It doesn't invert, but the lower voltage will
> be the problem.
>
> >I've had a bit of a google, and I never see this - only a resistor from
> >+v to collector, but this will invert my signal - something I don't
> >want to do.
>
> Ah. Okay. Then you probably need a second transistor, perhaps a PNP, to
> reinvert it, again. A little more complexity.
>
> >I have had a line-level converter chip recommended to me - MAX3001E,
> >but this is only comes in very, very tiny TSSOP format, which would be
> >difficult to solder up, but I'm interested in the transistor solution
> >as this will help me to 'swat up' a bit on my A-level electronics.
>
> Understood. It's about how I'd feel, too, I think. (TI and others do make ICs
> for level shifting, which include two power pins for the purpose, in fact. But
> doing it with BJTs means you can understand the basic ideas.)
>
> As you already know, the inverter is:
>
> 5V
> |
> |
> \
> / R2
> \
> |
> +-----> out
> |
> R1 |/c
> in >--/\/\----| Q1
> |>e
> |
> |
> gnd
>
> The above circuit assumes that 'in' is an active, low impedance drive for both
> 0V and 1.5V and just uses Q1 as a saturated switch, more or less, allowing the
> collector to get very close to the base voltage of 0V, or probably about .2V or
> so. But that only happens when 'in' is at 1.5V and turns on Q1's switching
> function. And when that happens, as you already know, 'out' gets very close to
> 0V, which is the opposite of what you want happening.
>
> To invert the inverter:
>
> 5V
> |
> |
> \ 5V
> / R2 |
> \ |
> | |<e
> +----| Q2
> | |\c
> |/c |
> in >----| Q1 +-----> out
> |>e |
> | \
> \ / R3
> / R1 \
> \ |
> | |
> gnd gnd
>
> There are actually several different arrangements. One would look more like the
> first case, with a base resistor like before for Q1. But this works, too, and
> it does things a little differently. In this case, Q1's base is "tacked" hard
> to the low-impedance drive and thus the emitter is forced to follow closely.
> This impresses a voltage on R1 which allows you to calculate a fairly precise
> current that you want to flow -- roughly a 'programmable constant current' that
> is presented via the collector of Q1 and is driven through R2 and Q2's base.
> Now, you could add another resistor to Q2's base, too, to allow the base to
> 'find it's place'. But if you are careful in designing R2 and R3, it's not
> really necessary -- you just make sure that there is slightly more than enough
> current to force Q2's base down adequately, when calculated through R2, and
> allow enough margin to operate Q2's collector current via some modestly
> predicted beta.
>
> For example,
>
> Q1 = 2N3904
> Q2 = 2N3906
> R3 = 47k (perhaps stiff enough for PIC inputs)
>
> Now, you might decide to lower R3 to make it somewhat stiffer or raise it. What
> you choose here will depend some on the loading that needs to be driven. But
> PIC inputs (like most micro inputs) don't require more than 10uA and 5V across a
> 47k yields about 100uA from which 10uA won't hurt much. But read the data ***
> and make sure about this and set the expected current through R3 to be some 10X
> over the worst case you have to deal with.
>
> Okay, so to continue. When Q2 is 'on', we'll project that the beta will be
> about 50 (normally, when V(CE) is a few volts the 2N3906 is often showing a beta
> 4-6 times this much beta, so we can expect a V(CE) of say 0.3V at a beta of 50,
> I think -- but I'm guessing here.) This suggests that the voltage across the
> 47k is 4.7V for a collector current of 100uA.
>
> With a collector current of 100uA and a guessed beta of about 50, the base
> current needed will be 100uA/50 or about 2uA. So, let's plan to use 4uA drive
> from Q1, doubling this estimate. We know that when Q1 is 'on', the emitter
> voltage will be something circa 0.7V. But with such low currents (4uA) it will
> probably be closer to more like 0.5V. So, let's guess at that number for now.
> The emitter of Q1 will then be 0.5V less than the base, which we know to be
> 1.5V, so the emitter voltage will be 1.5V - 0.5V or 1.0V. We know that we want
> about 4uA emitter current (this transistor's beta will be pretty high, so the
> base current will be negligible) and thus, R1 = 1.0V/4uA or 250k.
>
> Now, R2 will need to provide about 0.6V (another guess, I'm making) at 2uA (it's
> 'half' of the allotted 4uA.) So, this implies R2 = 0.6V / 2uA or 300k.
>
> So, let's try:
>
> R1 = 220k (a little more drive current, just in case)
> R2 = 270k (to suck up just a little more of that drive current)
>
> We have a design?!
>
> 5V
> |
> |
> R2 \ 5V
> 270k / |
> \ |
> | |<e
> +----| Q2
> 2N3904 | |\c 2N3906
> |/c |
> in >----| Q1 +-----> out
> |>e |
> | \
> \ / R3
> R1 / \ 47k
> 220k \ |
> | |
> gnd gnd
>
> How does this simulate? Well, I get about 0V to 4.9V swings on the output,
> given 0V to 1.5V swings on the input. And it looks pretty clean. Simulated
> estimates are:
>
> Q1 base current average of some 20pA with spikes on the transitions reaching as
> much as 1.5uA (chances are, your drive to this circuit can handle that.) I(R1)
> is more like 4.5uA instead of the 4uA we planned. But R1 is 220k instead of
> 250k, too. So we expected something extra here. I(R2) is about 2.2uA, instead
> of the planned 2uA, but again we had adjusted it down a bit to suck up some
> extra -- and it does. Base current on Q2 is about the same, 2.2uA (the
> unaccounted for 0.1uA is actually just rounding errors I'm making.) Collector
> current on Q2 is very close, too, at a bit less than 105uA.
>
> In other words, it's seems to hold up.
>
> But none of this takes care of speed issues. You didn't mention how fast your
> logic and clock pulses may be.
>
> If you want something with active drive HI and LO and that is faster, try
> something along these lines:
>
> R3 || 2200pF
> ,------------/\/\----||--------------,
> | 470 || C2 |
> | |
> | 5V 5V |
> | | | |
> | | | |
> | | 1N4148 | |
> | \ D1 --- |
> | / R1 / \ | 5V
> | \ 22k --- | |
> | / | | |
> | | | | |
> | | | | |<e Q2
> | +-------------+------+-----| 2N3906
> | | |\c
> | 1.5V | |
> | | |/c Q3 |
> | '------| 2N3904 |
> | |>e |
> | | |
> | | |
> | ,-----+ +---> OUT
> | | | |
> | | \ |
> | | / R2 |
> | C1 --- \ 22k |
> | 470pF --- / |
> | | | |
> | | | |
> | | | R4 |/c Q1
> IN >---++--------+-----+----------/\/\------+-----| 2N3904
> | 10k | |>e
> | | |
> | C3 | |
> | R5 || 1000pF | |
> '-----------/\/\----||--------------' gnd
> 750 ||
>
> In this case, R3+C2 as well as R5+C3 are "speed up" sections to help propagate
> an edge forward. The rest is interesting to read through and get an eye to how
> it works.
>
> I haven't been careful about the designing the values here, because I've no idea
> if you really care about something like this. Also, it again is an inverter and
> you'd need to take some care about making this into a non-inverting circuit.
> The point is mainly that for fast circuits you may need to do a little more than
> something dead simple.
>
> Jon